Exercise – 13.1
Q.1: Q and R are events such as P(Q) = 0.6 , P(R) = 0.3, P(Q∩R) = 0.2. Determine the value of P(QR) and P(RQ)
Sol:
According to the Q., it is given that P(Q) = 0.6 , P(R) = 0.3, P (Q∩R) = 0.2
P(QR) = P(Q∩R)P(R) = 0.20.3 = 23
P(RQ) = P(Q∩R)P(Q) = 0.20.6 = 13
Q.2: Determine the value of P (QR) if Q & R are the events such that P(Q) = 0.5 and P(Q∩R) = 0.32
Sol:
According to the Q., it is given that P(Q) = 0.5 , P(Q∩R) = 0.32.
P(QR) = P(Q∩R)P(R)
= 0.320.5 = 1625
Q.3: If Q and R are events such that P(Q) = 0.8 , P(R) = 0.5 , P(RQ) = 0.4. Determine the value of
(a) P(Q∩R)
(b) P(QR)
(c) P(Q∪R)
Sol:
According to the Q, it is given that:
P(Q) = 0.8 , P(R) = 0.5 , P(RQ) = 0.4
(a) P(RQ) = P(Q∩R)P(Q)
i.e. 0.4 = P(Q∩R)0.8
Therefore, P(Q∩R) = 0.32
(b) P(QR) = P(Q∩R)P(R) = 0.320.5 = 0.64
(c) P(Q∪R)=P(Q)+P(R)–P(Q∩R)
= P(Q∪R)=0.8+0.5–32 = 0.98
Q.4: Determine the value of P(Q∩R) if 2P(Q) = P(R) = 513 & P(QR)=25
Sol:
According to the Q, it is given that:
2P(Q) = P(R) = 513 & P(QR)=25
P(Q) = 526
P(R) = 513
P(QR)=25
= Q∩RR = 25
= P(Q∩R) = 25×513
= 213
We know that:
P(Q∪R)=P(Q)+P(R)–P(Q∩R)
= Q∩R = 526+513–213
= Q∩R = 5+10–426
= Q∩R = 1126
Q.5: If Q and R are events such that P(Q) = 611 , P(R) = 511, P(Q∪R = 711. Determine the value of
(a) P(Q∩R)
(b) P(QR)
(c) P(RQ)
Sol:
According to the Q., it is given that:
P(Q) = 611 , P(R) = 511 & P(Q∩R) = 711
(a) P(Q∪R)=P(Q)+P(R)–P(Q∩R)
= 711 = 611 + 511 – P(Q∩R)
= P(Q∩R) = 1111 –711
= P(Q∩R) = 411
(b) P(QR) = P(Q∩R)P(R)
= 411511 = 45
(c) P(RQ) = P(Q∩R)P(Q)
= 411611 = 23
Q.6: An experiment consists of tossing up of a coin three times. Determine the following:
(a) Q: obtaining heads on third toss & R: obtaining heads from the first consecutive two tosses.
(b) Q: obtaining at least two heads & R: obtaining at most two heads.
(c) Q: obtaining at most two tails & R: obtaining at most one tail.
Sol:
According to the Q., a coin is tossed thrice in an order to conduct an experiment.
Sample space (S) = {TTT , TTH ,THT , THH , HTT , HTH , HHT , HHH }
(a) Number of favourable outcomes of event Q = { TTH , THH ,HTH , HHH }
Number of favourable outcomes of event F = {HHT , HHH}
Q∩R = { HHH }
P(R) = 28=14
P(QR) = P(Q∩R)P(R) = 1814 = 12
(b) Number of favourable outcomes of event Q = { THH ,THH ,HHT , HHH }
Number of favourable outcomes of event F = {TTT , TTH , THT , THT , THH , HTT , HTH , HHT}
Q∩R = { HHT , HTH , HHT }
P(R) = 78
P(QR) = P(Q∩R)P(R) = 3878 = 37
(c) Number of favourable outcomes of event Q = {TTH , THT , THH , HTH , HTT ,HHT , HHH }
Number of favourable outcomes of event F = {TTT , TTH , THT , THT , THH , HTT , HHT}
Q∩R = {TTH , THT , THH , HTH , HTT , HTT , HHT }
P(R) = 78
P(QR) = P(Q∩R)P(R) = 6878 = 67
Q.7: An experiment consists of tossing up of a coin where:
(a) Q: obtaining tail on one coin & R: head appears in one coin
(b) Q: tail does not appear & R: head does not appears
Sol:
An experiment consists of tossing up of a coin only once.
Sample space (S) = {TT, TH, HT, HH}
(a) Number of favourable outcomes of event Q = {TH, HT}
Number of favourable outcomes of event F = {TH, HT}
Q∩R = { TH , HT }
P(R) = 28=28=14
P(QR) = P(Q∩R)P(R)
= 22 = 1
(b) Number of favourable outcomes of event Q = { HH }
Number of favourable outcomes of event F = {TT }
Q∩R = { ϕ }
P(R) = 1 =
P(QR) = P(Q∩R)P(R) = 01 = 0
Q.8: An experiment consists of throwing up of a dice three times .Find:
Q: the number 4 appears during the third toss
R: 6 & 5 appears consecutively during the first two tosses
Sol:
According to the Q., it is given that a die is being tossed thrice.
Total number of elements in the sample space = 216
Favorable outcomes of event Q =
{ (6,1,4) , (6,2,4) , (6,3,4) , (6,4,4) , (6,5,4) , (6,6,4)
(5,1,4) , (5,2,4) , (5,3,4) , (5,4,4) , (5,5,4) , (5,6,4)
(4,1,4) , (4,2,4) , (4,3,4) , (4,4,4) , (4,5,4) , (4,6,4)
(3,1,4) , (3,2,4) , (3,3,4) , (3,4,4) , (3,5,4) , (3,6,4)
(2,1,4) , (2,2,4) , (2,3,4) , (2,4,4) , (2,5,4) ,(2,6,4)
(1,1,4) , (1,2,4) , (1,3,4) , (1,4,4) , (1,5,4) , (1,6,4) }
Favorable outcomes of event R = { (6,5,6) , (6,5,5) , (6,5,4) , (6,5,3) , (6,5,2) , (6,5,1) }
Q∩R = { (6,5,4) }
P(R) = 6216
Q∩R = 1216
P(QR) = P(Q∩R)P(R) = 12166216 = 16
Q.9: An experiment consists of taking into account the line up all the members in a family namely the father, mother and the child.
Q: the son is placed in one of the end
R: the position of the father is in the middle
Sol:
An experiment consists of taking into account all the members of a family namely the father , mother and the child.
Let us consider the father , mother and the child are denoted by F , M and S respectively.
Sample space = { SFM , SMF , FSM , FMS , MSF , MFS }
Favourable outcomes of event Q = { SFM , SMF , FMS , MFS }
Favourable outcomes of event R = { SFM , MFS }
P(R) = 26=13
Q∩R = 26=13
P(QR) = P(Q∩R)P(R) = 1313 = 1
Q.10: An experiment consists of a black and a red die which are rolled.
(a) Determine the conditional probability of getting a sum of numbers greater than 9 such that the black die results in a 5
(b) Determine the conditional probability of getting a sum of numbers 8 such that red die results in numbers less than 4
Sol:
According to the Q., it is given that the a black and a red die are rolled
Total number of elements in the sample space = 36
(a) Favourable outcomes of event Q = { (4,6) , (5,5) , (5,6) , (6,4) ,(6,5) ,(6,6) }
Favourable outcomes of event R = { (5,1) , (5,2) , (5,3) , (5,4) , (5,5) , (5,6) }
Q∩R = { (5,6) , (5,5) }
Conditional probability of getting a sum greater than 9 , so that the black die results in 5 .
P (QR) = P(Q∩R)P(R)
= 2361836 = 19
Q.11: An experiment consists of rolling up of a dice including events such as Q = { 5,3,1 } , R = { 3,2} , S = { 5,4,3,2}. Determine:
(a) P(QR) and P(RQ)
(b) P(QS) and P(SQ)
(c) P(Q∪R)S and P(Q∪S)S
Sol:
According to the Q., it is given that a die is rolled in which events namely Q, R and S are recorded.
Sample space = { 6 , 5 , 4 , 3 , 2 , 1}
Favourable outcomes of event Q = { 5 , 3 , 1 }
Favourable outcomes of event R = { 3 , 2 }
Favourable outcomes of event S = { 5 , 4 , 3 , 2 }
Therefore , P(Q) = 36=12
P(R) = 26=13
P(S) = 46=23
(a) Q∩R = {3}
P (Q∩R) = 16
P(QR)=Q∩RR=1613=12 P(RQ)=Q∩RQ=1612=13
(b) Q∩S = {5 , 3}
P (Q∩R) = 16
P(QS)=Q∩SS=1323=12 P(SQ)=Q∩SQ=1312=23
(c) Q∪R = {5 , 3 , 2 , 1}
(Q∪R)∩S=5,3,2,1∩5,4,3,2 = {5,3,2}
Q∩R = {3}
(Q∩R)∩S=3∩5,4,3,2 = { 3 }
P(Q∪S)=46=23 P[(Q∪R)∩S]=36=12 P(Q∩R)=16 P[(Q∪R)S]=P[(Q∪R)∩S]P(S)=1223=34 P[(Q∩R)S]=P[(Q∩S)∩S]P(S)=1623=14
Q.12: Assume that each of the children born in a family can either be a boy or a girl. If a family having 2 children, determine the conditional probability that both of them are girls given:
(a) The youngest child is a girl
(b) At least one is a girl.
Sol:
According to the Q., it is given that a family is having 2 children where both of them are girls.
Let us represent boy and the girl child with the letter (b) and (g) respectively.
Sample space = {(g, g), (g, b), (b, g), (b, b)}
Let us consider Q be the event which indicates that both of the child born to a family are girls.
Q = {(g, g)}
(a) Let us consider R be the event that the youngest child born in the family is a girl.
R = {(g, g), (b, g)}
Q∩R=(g,g)
P (Q) = 24=12
P (Q∩R=14
According to the Q., both of the children are girls and the youngest being a girl child.
P (QR)=P(Q∩R)P(R)=1412=12
The required probability is 12
(b) Let us consider S be the event that at least one child born in the family is a girl.
S = {(g, g), (b, g), (g, b)}
Q∩S=(g,g)
P(S) = 34
P (Q∩S=14
According to the Q., both of the children are girls and the youngest being a girl child.
P (QS)=P(Q∩S)P(S)=1434=13
The required probability is 13
Q.13: A Q. bank consisting of 500 easy multiple choice Q.s , 400 difficult multiple choice Q.s , 300 easy False / True Q.s & 200 difficult False / True Q.s . Determine the probability that a Q. chosen from this Q. bank will be an easy multiple choice Q…
Sol:
| False / True | Multiple choice | Sum total | |
| Difficult | 200 | 400 | 600 |
| Easy | 300 | 500 | 800 |
| Sum total | 500 | 900 | 1400 |
Let us consider:
Easy Q. = E
Difficult Q. = D
Multiple choice Q. = M
False / True Q. = T
Sum total of all Q.s in the Q. ban k = 1400
Number of multiple choice Q.s = 900
Number of False / True Q.s = 500
Probability of getting an easy multiple choice Q. in the Q. bank =
P (E∩M)=5001400=514
Probability of selecting a multiple choice Q. be it easy or difficult
P (M) = 9001400=914
The conditional probability of getting an easy multiple choice Q. from the Q. bank = P (EM)=P(E∩M)P(M)=5494=59
The required probability is 59
Q.14: On rolling two dice simultaneously two different numbers appears on both the faces of the dice. Find the probability that the sum of two different numbers is 4.
Sol:
Total number of elements in the sample space = 36
Let Q be the event that the sum of two different numbers is 4 & R be the event that two numbers appearing on both the faces of the dice are different.
Q = {(3, 1), (2, 2), (1, 3)}
R = { (6 , 1) , (6 , 2) , (6 , 3) , (6 , 4) , (6 , 5) , (6 , 6)
(5 , 1) , (5 , 2) , (5 , 3) , (5 , 4) , (5 , 5) , (5 , 6)
(4 , 1) , (4 , 2) , (4 , 3) , (4 , 4) , (4 , 5) , (4 , 6)
(3 , 1) , (3 , 2) , (3 , 3) , (3 , 4) , (3 , 5) , (3 , 6)
(2 , 1) , (2 , 2) , (2 , 3) , (2 , 4) , (2 , 5) , (2 , 6)
(1 , 2) , (1 , 3) , (1 , 4) , (1 , 5) , (1 , 6) }
Q∩R=(1,3),(3,1))
P(R) = 3036=56
Q∩R=236=118
Let P(QR) = P(Q∩R)P(R) = 11856 = 115
The required probability is 115
Q.15: A die is rolled and if any number multiples of 3 come up then it is rolled again. This time if any other number appears then a coin is tossed. Determine the conditional probability of the event ‘tail appears ‘given that ‘at least one die shows a 3 ‘.
Sol:
Sample space of the above conducted experiment = { (6 , 6) , (6 , 5) , (6 , 4) , (6 , 3) , (6 , 2) , (6 , 1) , (5 , T) , (5 , H) , (4 , T) , (4 , H) , (3, 6) , (3 , 5) ,(3 , 4) , (3 , 3) , (3 , 2) , (3 , 1) , (2 , T) , (2 , H) , (1, T) , (1, H) }
Let Q be the event that a tail appears
R be the event at least one die shows 3
Q = {(1, T), (2, T), (4, T), (5, T)}
R = { (3, 6) , (3 , 5) ,(3 , 4) , (3 , 3) , (3 , 2) , (3 , 1) , (6 , 3) }
Q∩R=ϕ P(Q∩R)=0
Then,
P (R) ={ { P (3, 6) , P (3 , 5) ,P (3 , 4) , P (3 , 3) , P (3 , 2) , P (3 , 1) , P (6 , 3) }}
= 136 + 136 + 136 + 136 + 136 + 136 + 136 = 736
Probability that the die shows a tail given that at least one die shows 3
= P (QR) = 0736 = 0
Q.16: If P (Q) = 12
P (R) = 0,
Then P (QR) =?
(i) 0 (ii) 12 (iii) not defined (iv) 1
Sol:
It is given that:
P (Q) = 12
P (R) = 0
P (QR)
= Q∩R0
= not defined
Hence, the correct answer is C
Q.17: If Q and Rare events such that P( QR) = P( RQ) , then :
(a) Q⊂R , Q ≠ R
(b) Q = R
(c) Q∩R=ϕ
(d) P (Q) = P(R)
Sol:
According to the Q., it is given that:
P( QR) = P( RQ)
= P(Q∩R)P(R) = P(Q∩R)P(Q)
= P (Q) = P(R)
The correct answer is D
Exercise – 13.2
Q.1: If P (Q) = 35 & P (R) = 15 . Considering events Q and R are independent.
Sol:
Given, P (Q) = 35 & P (R) = 15
Q and R are the events which are independent in nature.
P(Q capR)=P(Q)×P(R)
= 35×15 = 325
Q.2: 2 cards are drawn at random from a well shuffled deck of 52 cards. Determine the probability that the drawn cards are black.
Sol:
In a well shuffled deck of 52 cards there are only 26 black cards (13 spade cards and 13 clubs cards)
Let Q be the event black card is drawn from a deck of cards.
P (Q) = 2652=12
Let R be the probability a black card on the second draw.
P (R) = 2552
Thus, the probability of getting both black cards = (2652)(2552) = 25102
Q.3: A bag full of oranges is inspected by selecting 3 random oranges without replacement. If it is seen that all the 3 drawn oranges are good then the box is sent for sale, otherwise it is sent for rejection. Determine the probability if a box containing 15 oranges out of which 12 of which are good and rest are bad will be approved for sale or not?
Sol:
Let us consider events Q, R and S be the events of drawing first, second and third drawing of orange.
P (Q) be the event of first drawing of orange = 1215
P(R) be the event of second drawing of orange = 1114
P(S) be the event of third drawing of orange = 1013
Thus , getting the probability of all good apples = (1215)(1114)(1013)
= 4491
Thus, the probability of getting approved for sale is 4491
Q.4: An unbiased coin is tossed. An EVENT Q and R be the head appears on the coin & 3 appears on the face of the die. Find out whether Q and R are independent events or not?
Sol:
Sample space = { (T, 6), (T, 5), (T, 4), (T, 3), (T, 2), (T, 1), (H, 6), (H, 5), (H, 4), (H, 3), (H, 2), (H, 1) }
Let Q be the event of head appearing on the face of the coin
Q = { (H , 6) , (H, 5) , (H , 4) , (H , 3) , (H , 2) , (H , 1) }
P (Q) = 612=12
Let R be the event that the number 3 appears on the face of the die
R = {(T, 3), (H, 3)}
P (R) = 212=16
Therefore,
Q∩R = {(H, 3)}
P (Q∩R) = 112
P(Q)×P(R) = 12×16 = P(Q)∩P(R)
Therefore, Q & R are independent events.
Q.5: An experiment consists of marking a die 1, 2, 3 in red and 4, 5, 6 in green and then rolled. Let Q be the event that the number is even and R be the event that color of the face of the die is red. Determine the probability that both of the events are independent.
Sol:
Sample space (S) when a die is rolled = {6, 5, 4, 3, 2, 1}
According to the Q., it is given that:
Q = Event that the number appears is even
R = the colour of the face of the die is red.
Q = {6, 4, 2}
P (Q) = 36=12
R = {3, 2, 1}
P (R) = 36=12
Therefore, Q∩R=2
P(QR)=P(Q∩R)=16 P(Q)⋅P(R)=12×12=14≠16
Hence, events Q and R are not independent events.
Q.6: Q and R be the events such that P (Q) = 35 , P (R) = 310 & Q∩R=15 . Determine whether events Q and R are independent or not /
Sol:
According to the Q., it is given that:
P (Q) = 35 , P (R) = 310 & Q∩R=15 .
P(Q)⋅P(R)=35×310=950≠15
Hence, events Q and R are not independent events.
Q.7: Q and R be the events such that P (Q) = 12 , Q∩R=35 , P (R) = p . Determine the value of p if Q and R are:
(a) Mutually exclusive
(b) Independent.
Sol:
(a) Let us consider Q and R are mutually exclusive.
Q∩R=ϕ, P (R) = p
P(Q∩R)=0
We know, that:
P(Q∩R)=P(Q)+P(R)–P(Q∪R)
= 35=12+p–0
= p = 110
(b) Considering events Q and R are independent events.
P(Q∩R)=P(Q)×P(R)=12p P(Q∩R)=P(Q)+P(R)–P(Q∪R)
= 35=12+p–12p
= p = 210=15
Q.8: Let Q and R be the independent events such that P (Q) = 0.3, P (R) = 0.4
Determine:
(a) P (P(Q∩R))
(b) P (P(Q∪R))
(c) P(QR)
(d) P(RQ)
Sol:
(a) Considering the events Q and R independent events,
P(Q∩R)=P(Q)⋅P(R)=(0.3)(0.4)=0.12
(b) P(Q∩R)=P(Q)+P(R)–P(Q∪R)
P (P(Q∪R)) = 0.3 + 0.4 – 0.12 = 0.58
(c) P(QR)=P(Q∩R)P(R)
= P(QR)=0.120.4 = 0.3
(d) P(RQ)=P(Q∩R)P(Q)
= P(RQ)=0.120.3 = 0.4
Q.9: Let Q and R be the two events such that P(Q)=12, P(R)=12, P(Q∩R)=18. Determine the value of P (neither Q nor R)
Sol:
According to the Q., it is given that:
P(Q)=12, P(Q∩R)=18, P(R)=12
P (neither Q or R) = P (Q′∩R′)
P (neither Q or R) = P (Q∩R)′
= 1 – P(Q∩R)
= 1 – P(Q)+P(R)–P(Q∩R)
= 1 – 14+12–18
= 1 – 58 = 38
Q.10: Q and R are events such that P(Q) = 12 , P(R) = 12 & P(neither Q nor R) = 14
Sol:
P(Q′∪R′)=14
= P((Q∩R)′)=14
= 1 – P(Q∩R)=14
= P(Q∩R)=34
= P(Q∩R)=P(Q)⋅P(R)=(12)(712)=724
34≠724
Hence, Q & R are not independent events.
Q.11: Q and R are two independent events such that P (Q) 0.3 & P(R) = 0.6. Determine:
(a) P (Q and R)
(b) P (Q and not R)
(c) P (Q or R)
(d) P (neither Q nor R)
Sol:
(a) P (Q and R) = P(Q∩R)=P(Q)⋅P(R)=(0.3)(0.6)=0.18
(b) P (Q and not R)
= P(Q∩R′)
= P(Q)–P(Q∩R)
= 0.3 – 0.18 = 0.12
(c) P (Q or R)
= (Q∩R)
= P(Q)+P(R)–P(Q∩R)
= 0.3 + 0.6 – 0.18 = 0.72
(d) P (neither Q nor R)
= P(Q′∩R′)
= P((Q∩R)′)
= 1 – P(Q∩R)
= 1 – 0.72 = 0.28
Q.12: An experiment consists of tossing up of die thrice. Determine the probability of getting an odd number at least once.
Sol:
Probability of getting an odd number in a single throw = 36=12
Probability of getting an even number = 36=12
Probability of getting an even number = 12×12×12 = 18
According to the Q.,
Determining the probability of getting an odd number at least once
= 1 – Probability of getting an odd number
= 1 – probability of getting an even number thrice
= 1 – 18 = 78
Q.13: From a bag of 10 black balls and 8 red balls 2 balls are drawn at a random without random. Determine the probability:
(a) Both the drawn balls are red
(b) First and the second drawn ball are black & red respectively.
(c) One of them is black and the other one is red.
Sol:
According to the Q., it is given that:
The bag contains a total number of number balls of = 18
No. of red balls = 8
No. of black balls = 10
(a) Probability that a red ball appears in the first draw = 818=419
The ball so obtained is replaced.
Probability that a red ball appears in the second draw = 818=419
Probability that both the red balls appear = 49×49=1681
(b) Probability that a black ball appears in the first draw = 1018=519
The ball so obtained is replaced.
Probability that a black ball appears in the second draw = 818=419
Probability that both the black balls appear = 59×49=2081
(c) Probability that a red ball appears in the first draw = 818=419
The ball so obtained is replaced.
Probability that a black ball appears in the second draw = 818=59
Probability that both the red balls appear = 49×59=2081
Hence, the probability of getting one black and one red ball =
Probability of getting first red ball and then black ball + Probability of getting first black ball and then red ball
= 2081+2081 = 4081
Q.14: The probability of solving particular problems by methods Q and R are 12 and 13 respectively. Considering that the problem is solved independently, determine the probability:
(a) The problem is solved
(b) At least one of them solves the problem correctly.
Sol:
Probability that the problem is solved by Q, P (Q) = 12
Probability that the problem is solved by R, P (R) = 13
According to the Q., it is given that both problems are solved independently by Q and R:
= P (QR) = P(Q)×P(R)=12×13=16
P (Q’) = 1 – P (Q) = 1 – 12=12
P (R’) = 1 – P(R) = 1 – 13=23
(a) Probability that the problem:
= P(Q∩R)
= P (Q) + P(R) – P (QR)
= 12+13–16 = 23
(b) Probability that exactly one solves the above mentioned problem:
= P(A)×P(B′)+P(B)×P(A′)
= (12×23)+(12×13)
= 13+16
= 12
Q.15: An experiment consists of drawing up of a card at random from a well shuffled deck of 52 cards. Events Q and R associated with the following experiment are independent in nature
(a) Q: the drawn card is a spade
R: the drawn card is an ace
(b) Q: it is a black card
R: the drawn card is a king
(c) Q: the drawn card can either be a king or queen
R: the drawn card is either a queen or jack.
Sol:
(a) Total number of spades = 13
Total number of ace cards in a pack of cards = 4
Probability of drawing a spade card P (Q) = 1352=14
Probability of drawing an ace card P(F) = 14
In a pack of cards, there is only card which is an ace of spades. Hence probability of getting an ace of spades = 152
= P(Q)×P(R)
= 14×113 = P (QR)
P (QR) = P(Q)×P(R)
Hence, events Q and R are independent in nature.
(b) In a pack of 52 cards, there are 26 black cards and 4 cards which are kings.
P (Q) = P (the drawn card is black) = 2652=12
P (R) = P (the drawn card is a king) = 452=113
In a pack of 52 cards, there are 2 black cards which are kings.
P (QR) = P (the drawn card is black in colour as well as a king) = 252=126
= P(Q)×P(R)
= 12×113=126 = P (QR)
Hence, events Q and R are independent in nature.
(c) In a pack of 52 cards, there are 4 kings, 4 jacks and 4 queens.
P (Q) = P (the drawn card can either be a king or queen) = 852=213
P (R) = P (the drawn card is a queen or a jack) = 852=213
There are 4 cards which are king or queen and queen or jack = 452=113
= P(Q)×P(R)
= 213×213≠4169 = P (QR)
Hence, events Q and R are not independent in nature.
Q.16: Among the students residing in a hostel, 60% of them read Hindi newspaper, 20% of them read both Hindi and English newspapers. A student is selected as a random:
(a) Determine the probability the students reads neither Hindi nor English newspapers.
(b) Determine the probability that the student reads Hindi newspapers, but also reads English newspaper too.
(c) If the students read English newspapers, find the probability that they reads Hindi newspaper too.
Sol:
According to the Q. it is given that Q be the event who reads Hindi newspapers and R be the event who reads English newspapers.
P (Q) = 60% = 60100=35
P (R) = 40% = 40100=25
P (Q∩R) = 20% = 20100=15
(a) Probability that the student reads Hindi or English newspapers:
= (Q∪R)′
= 1 – (Q∪R)
= 1 – { P(Q) + P(R) – (Q∪R) }
= 1 – { 35 + 25 – 15 }
= 1 – 45 = 15
(b) Probability that a guy chose to read English newspaper , if they already reads Hindi newspaper = P(QR)
= Q∩RR = 1535 = 13
(c) Probability that a guy chose to read Hindi newspaper, if they already reads English newspaper = P(RQ)
= Q∩RQ = 1525 = 12
Q.17: Find the probability of obtaining an even number, when two dice are rolled at once:
(i) 0
(ii) 13
(iii) 112
(iv) 136
Sol:
Total number of outcomes when two dice are rolled at once = 36
2 is the only even prime number.
Let Q be the event of getting an even prime number.
Therefore, Q = {2, 2}
P (Q) = 136
Hence, the correct answer to the above mentioned Q. is (iv)
Q.18: 2 events Q and R will be independent, such as:
(i) Q and R are mutually exclusive.
(ii) P (Q’R’) = [1 – P (Q) ][1 – P(R) ]
(iii) P(Q) = P (R)
(iv) P (Q) + P (R) = 1
Sol:
2 events Q and R are said to be independent, if
P (QR) = P(Q)×P(R)
= P (Q’R’) = [ 1 – P(Q)][1 – P(R)]
= P(Q′∩R′) = 1–P(Q)–P(R)+P(Q)⋅P(R)
= 1- P(Q′∩R′) = P(Q)–P(R)+P(Q)⋅P(R)
= P(Q∪R) = P(Q)–P(R)+P(Q)⋅P(R)
= P (Q) + P (R) – P (QR) = P(Q)–P(R)+P(Q)⋅P(R)
= P (QR) = P(R)+P(Q)⋅P(R)
This implies that event Q and R are independent event,
P (Q’R’) = [1 – P(Q)][1 – P(R)]
Let P (Q) = a
P (R) = b
0 ˂ a, b ˂ 1
Q and R are both mutually exclusive events.
Therefore,
= P(Q∪R)=ϕ
= P (QR) = 0
Therefore, P(Q) × P (R) = ab ≠ 0
Therefore, P(Q)×P(R)≠P(QR)
(ii) Event of getting an even number = {6, 4, 2}
P (R) = 36 = 12
= P(Q∪R)=ϕ
= P(Q)×P(R)=14≠0
(iii) Let Q be the event of getting a odd number = {5 , 3, 1 }
P (Q) = 36 = 12
(iv) P(Q) + P(R) = 12 + 12 = 1
From the above solution it cannot be concluded that Q and R are independent events.
Therefore, the correct answer to the above mentioned Q. is (ii).
Exercise – 13.3
Q.1: An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted and is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. What is the probability that the second ball is red?
Answer I:
The urn contains 5 red and 5 black balls.
Let a red ball be drawn in the first attempt.
So, P (drawing a red ball) = 5/10 = 1/2
If two red balls are added to the urn, then the urn contains 7 red and 5 black balls.
P (drawing a red ball) = 7/12
Let a black ball be drawn in the first attempt.
P (drawing a black ball in the first attempt) = 5/10 = 1/ 2
If two black balls are added to the urn, then the urn contains 5 red and 7 black balls.
P (drawing a red ball) = 5/12
Therefore, probability of drawing second ball as red is 12×712+12×512=12[712+512]=12×1=12
- 2: A bag contains 4 red and 4 black balls; another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag.
Answer 2:
Let E1 and E2 be the events of selecting first bag and second bag respectively.
P(E1) = P(E2) = 1/2
Let A be the met of getting a red ball
P(A
E1) = P (drawing a red ball from first bag) = 4/8 = 1/2
P(AlE2) = P(drawing a red ball from second bag) = 2/8 = 1/4
The probability of drawing a ball from the first bag, given that it is red, is given by P(E2
A)
By using Bayes’ theorem, we obtain
P(E1
A) = P(E1).P(A
E2)P(E1).P(A
E1)+P(E2).O(A
E2)
= 12×1212×12+12×14
= 1414+18 = 1438 = 23
Q.3: Of the students in a college, it is known that 6 0% reside in hostel and 40% are day scholars (not residing in hostel) Previous year results report that 30% of all students who reside in hostel attain A grade and 20% of day scholars attain A grade in their annual examination At the end of the year, one student is chosen at random from the college and he has an A grade, what is the probability that the student is hostler?
Answer 3:
Let E1 and E2 be the events that the student is a hostler and a day scholar respectively and A be the event that the chosen student gets grade A.
P(E1) = 60% = 0.6
P(E2) = 40% = 0.4
P(A
E1))= P(student getting an A grade is a hostler)= 30% = 0.3
P(A
E2)= P(student getting an A grade is a day scholar)= 20%= 0.2
The probability that a randomly chosen student is a hostler, given that he has an A grade, is given by P(E1
A)
By using Bayes’ theorem, we obtain
P(E1
A)=P(E1.P(A
E1)P(E!.P(A
E1)+P(E2).P(A
E2)
= 0.6×0.30.6×0.3+0.4×0.2
= 0.180.26 = 1826 = 913
Q.4: In answering a Q. on a multiple choice test, a student either knows the answer or guesses. Let 3/4 be the probability that he knows the answer and ¼ be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability 1/4. What is the probability that the student knows the answer given that he answered it correctly?
Answer 4:
Let E1 and E2 be the respective events that the student knows the answer and he guesses the answer.
Let A be the event that the answer is correct.
P(E1) = 3/4
P(E2) = 1/4
The probability that the student answered correctly, given that he knows the answer, is 1
P(A
E1) = 1
Probability that the student answered correctly, given that he guessed, is 1/4.
P(A
E2) = 1/4
The probability that the student knows the answer, given that he answered it correctly, is given by P(E1
A)
By using Bayes’ theorem, we obtain
P(E1
A)=P(E1.P(A
E1)P(E!.P(A
E1)+P(E2).P(A
E2)
= 34×134×1+1414
= 3434+116 = 341316 = 1213
Q.5: A laboratory blood test is 99% effective in detecting a certain disease when it is in fact, present. However, the test also yields a false positive result for 0 5% of the healthy person tested (that is, if a healthy person is tested, then, with probability 0.005, the test will imply he has the disease) If 0.1 percent of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive?
Answer 5:
Let E1 and E2 be the respective events that a person has a disease and a person has no disease.
Since E1 and E2 are events complimentary to each other,
P (E1) + P (E2) = 1
P(E2) = 1 -P(E1)= 1 – 0.001= 0.999
Let A be the event that the blood test result is positive
P(E1) = 0.1% = 1/100= 0.001
P(AlE1) = P(result is positive given the person has disease) = 99% = 0.99
P(AlE2) = P(result is positive given that the person has no disease) = 0.5% = 0.005
Probability that a person has a disease, given that his test result is positive, is given by
P(E1
A)
By using Bayes’ theorem, we obtain
P(E1
A)=P(E1.P(A
E1)P(E!.P(A
E1)+P(E2).P(A
E2)
= 0.001×0.990.001×0.99+0.000×0.005
= 0.000990.005985 = 110665 = 22133
Q.6: There are three coins. One is two headed coin (having head on both faces), another is a biased coin that comes up heads 73% of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin?
Answer 6:
Let E1, E2, and E3 be the respective events of choosing a two headed coin, a biased coin, and an unbiased coin
P (E1) + P (E2) + P (E3) = 1/3.
Let A be the event that the coin shows heads.
A two-headed coin will always show heads
P (AlE1) = P(coin showing heads, given that it is a two-headed coin) = 1
Probability of heads coming up, given that it is a biased coin= 75%
P(AlE2)= P(coin showing heads. given that it is a biased coin) = 3/4
Since the third coin is unbiased, the probability that it shows heads is always 1/2
P(AlE3) = P(coin showing heads, given that it is an unbiased coin) = 1/2.
The probability that the coin is two-headed, given that it shows heads, is given by
P(E1
A)
By using Bayes’ theorem, we obtain
P(E1
A)=P(E1.P(A
E1)P(E!.P(A
E1)+P(E2).P(A
E2)
= 13×113×1+1334+13×12
= 1313(1+34+12
= 194 = 49
Q.7: An insurance company insured 2000 scooter drivers, 4000 Car drivers and 6000 truck drivers. The probability of accidents are 0 01, 0.03 and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?
Answer 7:
Let E1, E2, and E3 be the respective events that the driver is a scooter driver, a car driver, and a truck driver.
Let A be the event that the person meets with an accident.
There are 2000 scooter drivers, 4000 car drivers, and 6000 truck drivers.
Total number of drivers = 2000 + 4000 + 6000 = 12000
P (E1) = P (driver is a scooter driver) = 2000/ I2000 = 1/6
P (E2) = P (driver is a car driver) = 4000/12000 = 1/3
P (E3) = P (driver is a truck driver) = 6000/1 2000 = 1/2
P(AlE1) = P(scooter driver met with an accident) = 0.01= 1/100
P(A
E2)= P (car driver met with an accident) = 0.03 = 3/100
P(A
E3) = P(truck driver met with an accident) = 0.15 = 15/100
The probability that the coin is two-headed, given that it shows heads, is given by P(E1
A)
By using Bayes’ theorem, we obtain
P(E1
A)=P(E1.P(A
E1)P(E1.P(A
E1)+P(E2).P(A
E2)+P(E3).P(A
E3)
= 16×110016×1100+133100+12×15100
= 161100110016+1+152
= 1610412 = 152
Q.8: A factory has two machines A and B. Past record shows that machine A produced 60% of the Items of output and machine B produced 40% of the items. Further, 2% of the items produced by machine A and 1% produced by machine B were defective. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that was produced by machine B?
Answer 8:
Let E1 and E2 be the respective events of items produced by machines A and B. Let X be the event that the produced item was found to be defective.
Probability of items produced by machine A, P (E1) = 60% = 3/5
Probability of items produced by machine B, P (E2) = 40% = 2/5
Probability that machine A produced defective items, P (X
E1) = 2% = 2/100
Probability that machine B produced defective items. P (X
E2) = 1% = 1/100
The probability that the randomly selected item was from machine B, given that it is defective, is given by P (E2
X)
By using Bayes’ theorem, we obtain
P(E1
A)=P(E2.P(X
E2)P(E1.P(X
E1)+P(E2).P(X
E2)
= 25×110035×2100+251100
= 25006500+2500
= 28 = 14
Q.9: Two groups are competing for the position on the board of directors of a corporation. The probabilities that the first and the second groups will win are 0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducing a new product is 0.7 and the corresponding probability is 0 3 if the second group wins. Find the probability that the new product introduced was by the second group.
Answer 9:
L.et E1 and E2 be the respective events that the first group and the second group win the competition. Let A be the event of introducing a new product.
P (E1) = Probability that the first group wins the competition = 0.6
P (E2) = Probability that the second group wins the competition = 0.4
P (A
E1) = Probability of introducing a new product if the first group wins =0.7
P(A
E2) = Probability of introducing a new product if the second group wins =0.3
The probability that the new product is introduced by the second group is given by P(E2
A)
By using Hayes’ theorem, we obtain
P(E2
A)=P(E2.P(A
E2)P(E1.P(A
E1)+P(E2).P(A
E2)
= 0.4×0.330.6×0.7+0.4×0.3
= 0.120.42+0.12 = 0.1254 = 29
Q.10: Suppose a girl throws a die if she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets 1, 2, 3 or 4, she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw 1, 2, 3 or 4 with the die?
Answer 10:
Let E1 be the event that the outcome on the die is 5 or 6 and E2 be the event that the outcome on the die is 1, 2, 3, or 4.
P(E1) = 2/6 = 1/3 and P(E2)= 4/6 = 2/3.
Let A be the event of getting exactly one head.
P (A
E1) =Probability of getting exactly one head by tossing the coin three times if she gets 5 or 6 = 3/8
P (A
E2) = Probability of getting exactly one head in a single throw of coin if she gets 1, 2, 3, or 4 =1/2.
The probability that the girl threw 1, 2, 3, or 4 with the die, if she obtained exactly one head, is given by P (E2
A)
By using Bayes’ theorem, we obtain
P(E2
A)=P(E1.P(A
E1)P(E!.P(A
E1)+P(E2).P(A
E2)
= 23×1213×38+2312
= 1313(1+38
= 1118 = 811
Q.11: A manufacturer has thee machine operators A, B and C. The first operator A produces 1% defective items, whereas the other two operators B and C produce 5% and 7% defective items respectively. A is on the Job for 50% of the time, B is on the job for 30% of the time and C is on the Job for 20% of the time A defective item is produced. What is the probability that was produced by A?
Answer 11:
Let E1, E2, and E3 be the respective events of the time consumed by machines A, B, and C for the job
P(E1)= 50% = 50/100 = 1/2
P(E2) = 30% = 30/100 = 3/10
P(E3) = 20% = 20/100 = 1/5
Let A be the event of getting exactly one head.
P (A
E1) =1% = 1/100
P (A
E2) = 5% = 5/100
P (A
E3) = 7% = 7/100
The probability that the defective item was produced by A is given by P (E1
A)
By using Bayes’ theorem, we obtain
P(E1
A)=P(E1.P(A
E1)P(E1.P(A
E1)+P(E2).P(A
E2)+P(E3).P(A
E3)
= 12×110012×1100+3105100+15×7100
= 1100121100(12+32+75
= 12175 = 534
Q.12: A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond.
Answer 12:
Let E1 and E2 be the respective events of choosing a diamond card and a card which is not diamond.
Let A denote the lost card
Out of 52 cards, 13 cards are diamond and 39 cards are not diamond
P(E1) = 13/52 = 1/4
P(E2) = 39/52 = 3/4
When one diamond card is lost, there we 12 diamond cards out of 51 cards
Two cards can be drawn out of 12 diamond cards in 12C2 ways.
Similarly, 2 diamond cards can be drawn out of 51 cards in 51C2 ways. The probability of getting two cards, when one diamond card is lost, is given by P(A
E1).
P(A
E1)=12C251C2=12!2!×10!×2!×49!50×51!=11×1250×51=22425
When the lost card is not a diamond, there are 13 diamond cards out of 51 cards.
Two cards can be drawn out of 13 diamond cards in 13C2 ways whereas 2 cards can be drawn out of 51 cards in 51C2 ways.
The probability of getting two cards, when one card is lost which is not diamond, is given by P (E1
A)
By using Bayes’ theorem, we obtain
P(E1
A)=P(E1.P(A
E1)P(E1.P(A
E1)+P(E2).P(A
E2)
= 14×2242514×22425+3426425
= 14252242514(22425)+3426425
= 11225 = 1150
Q.13: Probability that A speaks troth is 4/5. A coin is tossed a reports that a head appears. The probability that actually there was head 5 is:
(A) 4/5
(B) ½
(C) 1/5
(D) 2/5
Answer 13:
Let E1 and E2 be the events such that
E1: A speaks truth
E2: A speaks false
Let X be the event that a head appears
P (E1) = 4/5
P(E2) =
1 – 4
E1) = P(X
E2)= 1/2
The probability that there is actually a head is given by P (E1
X)
P(E1
X)=P(E1.P(X
E1)P(E1.P(X
E1)+P(E2).P(X
E2)
= 45×1245×12+1512
= 124512(45+15) = 451 = 45
Therefore, the correct answer is A.
Q.14; If A and B are two events such that A⊂BandP(B)≠0, then which of the following is correct?
(A) P(A
B)=P(B)P(A)
(B) P(A
B) < P(A)
(C) P(A
B) >= P(A)
(D) None of these
If A⊂B,then,A∩B=A.
⇒P(A∩B)=P(A)
Also, P(A) < P(B)
Consider, P(A
B)=P(A∩B)P(B)=P(a)P(B)≠P(B)P(A) . . . . (1)
Consider, P(A
B)=P(A∩B)P(B)=P(A)P(B) . . . . . . . . . (2)
It is known that P(B) <= 1
⇒1P(B)≥1 ⇒P(A)P(B)≥P(A)
From (2), we obtain
P(A
B)≥P(A) . . . . . . . . . (3)
Hence, P(A
B) is not less than P(A)
Thus, from (3), it can be concluded that the relation given in alternative C is correct.
Exercise – 13.4
Q-1: Check whether the following are the probability distributions of a random variable or not. Also, justify your answer.
(i)
| X | 0 | 1 | 2 |
| P(x) | 0.3 | 0.5 | 0.2 |
(ii)
| X | 0 | 1 | 2 | 3 | 4 |
| P(X) | 0.2 | 0.6 | 0.3 | -0.2 | 0.3 |
(iii)
| Y | 0 | 1 | 2 |
| P(Y) | 0.7 | 0.2 | 0.3 |
(iv)
| Z | 3 | 2 | 1 | 0 | -1 |
| P(Z) | 0.4 | 0.3 | 0.5 | 0.2 | 0.06 |
Solution:
We know that the sum of all of the probabilities in the probability distribution should be one.
(i)
| X | 0 | 1 | 2 |
| P(x) | 0.3 | 0.5 | 0.2 |
Sum of the probabilities = 0.3 + 0.5 + 0.2 = 1.0
Hence, this given table is a probability distribution of the random variables.
(ii)
| X | 0 | 1 | 2 | 3 | 4 |
| P(X) | 0.2 | 0.6 | 0.3 | -0.2 | 0.3 |
Here, from the table we can see that for X = 3, P(X) = -0.2
We know that, the probability for any of the observation cannot be negative.
Hence, the table given is not a probability distribution for the random variables.
(iii)
| Y | 0 | 1 | 2 |
| P(Y) | 0.7 | 0.2 | 0.3 |
Sum of all the probabilities given in the Q. = 0.7 + 0.2 + 0.3 = 1.2 ≠ 1
Hence, the table given is not a probability distribution for the random variables.
(iv)
| Z | 3 | 2 | 1 | 0 | -1 |
| P(Z) | 0.4 | 0.3 | 0.5 | 0.2 | 0.06 |
Here, sum of all of the probabilities given in the Q. = 0.4 + 0.3 + 0.5 + 0.2 + 0.06 = 1.46 ≠ 1
Hence, the table given is not a probability distribution for the random variables.
Q.2: There are 5 red balls and 2 black balls. From that, two balls are drawn at random. Let X represents the number of the black balls. Find the values for X. Check whether X is a random variable or not?
Solution:
Let, the two balls drawn randomly are represented as RR, RB, BR, and BB, where R is for a black ball and B is for a red ball.
According to Q,
X represents the number of the black balls.
Thus,
X (RR) = 0
X (RB) = 1
X (BR) = 1
X (BB) = 2
Hence, the possible values of X are 0, 1 and 2.
Therefore, C is a random variable.
Q.3: Let us consider X which represents the difference between the total number of tails and the number of heads which can be obtained when a single coin is tossed for 5 times. Find the possible values for X.
Solution:
As per the data given in the Q,
A coin will be tossed for 5 times and every time the result is observed. Also, X represents the difference between the total number of heads and tails observed after each toss.
Thus,
X (5H, 0T) =
5 – 0
= 5
X (4H, 1T) =
4 – 1
= 3
X (3H, 2 T) =
3 – 2
= 1
X(2H, 3T) =
2 – 3
= 1
X(1H, 4T) =
= 3
X(0H, 5T) =
0 – 5
= 5
Hence, the possible values for X are 5, 3 and 1.
Q.4: What will be the probability distribution for?
(i) Number of tails after tossing a coin for twice.
(ii) Number of heads after simultaneously tossing a coin for three times.
(iii) Number of tails after tossing a coin for four times.
Solution:
(i) Once a coin is tossed for two times, the chances are:
{TT, TH, HT, HH}, where H is for heads and T is for tails.
Let, the total number of tails be represented by X.
Then,
X(TT) = 2, X(TH) = 1, X(HT) = 1 and X(HH) = 0.
Hence, X can take the values of 0, 1 and 2.
We know that,
P(TT) = P(TH) = P(HT) = P(HH) = 14
P(X = 0) = P(HH) = 14
P(X = 1) = P (HT) + P(TH) = 14+14 = 1+14=24=12
P(X = 2) = P(TT) = 14
Hence, the probability distribution which is required as per the Q.’s demand is given below:
| X | 0 | 1 | 2 |
| P(X) | 14 | 12 | 14 |
(ii) Once a coin is tossed for three times, the chances are:
{TTT, TTH, THH, HHH, HHT, HTT, HTH, THT}, where H is for heads and T is for tails.
Let, the total number of tails be represented by X.
Then,
X(HHH) = 3, X(HTH) = 2, X(HHT) = 2, X(THH) = 2, X(TTH) = 1, X(HTT) = 1, X(THT) = 1 and X(TTT) = 0.
Hence, X can take the values of 0, 1, 2 and 3.
Now,
P(X = 0) = P(TTT) = 18
P(X = 1) = P (TTH) + P(HTT) + P(THT) = 18+18+18 = 1+1+18=38
P(X = 2) = P (HTH) + P(THH) + P(HHT) = 18+18+18 = 1+1+18=38
P(X = 3) = P(HHH) = 18
Hence, the probability distribution which is required as per the Q.’s demand is given below:
| X | 0 | 1 | 2 | 3 |
| P(X) | 18 | 38 | 38 | 18 |
(iii) Once a coin is tossed for four times, the chances are:
{TTTT, TTTH, TTHT, TTHH, THTT, THTH, THHT, THHH, HTTT, HTTH, HTHH, HTHT, HHTT, HHTH, HHHT, HHHH}, where H is for heads and T is for tails.
Let, the total number of tails be represented by X.
Then,
X(TTTT) = 4, X(TTTH) = 3, X(TTHT) = 3, X(THTT) = 3, X(HTTT) = 3, X(TTHH) = 2, X(THTH) = 2, X(THHT) = 2, X(HTTH) = 2, X(HTHT) = 2, X(HHTT) = 2, X(THHH) = 1, X(HTHH) = 1, X(HHTH) = 1, X(HHHT) = 1 and X(HHHH) = 0.
Hence, X can take the values of 0, 1, 2, 3 and 4.
Now,
P(X = 0) = P(HHHH) = 116
P(X = 1) = P (THHH) + P(HTHH) + P(HHTH) + P(HHHT) = 116+116+116+116 = 1+1+1+116=416=14
P(X = 2) = P (TTHH) + P(THTH) + P(THHT) + P(HTTH) + P(HTHT) + P(HHTT) = 116+116+116+116+116+116 = 1+1+1+1+1+116=616=38
P(X = 3) = P (TTTH) + P(TTHT) + P(THTT) + P(HTTT) = 116+116+116+116 = 1+1+1+116=416=14
P(X = 0) = P(TTTT ) = 116
Hence, the probability distribution which is required as per the Q.’s demand is given below:
| X | 0 | 1 | 2 | 3 | 4 |
| P(X) | 116 | 14 | 38 | 14 | 116 |
Q.5: What will be the probability distribution of the numbers for the two success tosses of a die, where a success will be defined as?
(i) Obtained number which is greater than 3.
(ii) On the die, six appears for at least once.
Solution:
When we will toss a die for two times, we will obtain at least (6 × 6) = 36 number of the observations.
Let us consider X which is a random variable, which represents total number of successes.
(i) In this case, numbers which is greater than 3 is referred to as success.
P(X = 0) = P (Number which are less than or equal to 3 after having both the tosses) = 36×36=14
P(X = 1) = P(Numbers which are less than or equal to 3 in first toss and greater than 3 in the second toss) + P(Numbers which are greater than 3 in first toss and less than or equal to 3 in the second toss)
= 36×36+36×36
= 12×12+12×12
= 14+14
= 12
P(X = 2) = P(Numbers which are greater than 3 after both toss) = 36×36
= 12×12
= 14
Hence, the probability distribution which is required as per the Q.’s demand is given below:
| X | 0 | 1 | 2 |
| P(X) | 14 | 12 | 14 |
(ii) In this case, success will be when six appeared at least for once after tossing the dice simultaneously.
P(Y = 0) = P(six doesn’t appeared on either of the dice) = 56×56=2536
P(Y = 1) = P(six appeared at least on one of the dice) = 16×56+56×16=536+536=1036=518
Hence, the probability distribution which is required as per the Q.’s demand is given below:
| Y | 0 | 1 |
| P(Y) | 2536 | 518 |
Q.6: Consider a lot of 34 bulbs among which 10 bulbs are defective. From that lot, a sample of 3 bulbs is drawn at random with having replacement. What is the probability distribution for the number of the defective bulbs?
Solution:
As per the data given in the Q., we have
A lot of 34 bulbs among which 10 bulbs are defective.
So, the number of non- defective bulbs = 34 – 10 = 24 bulbs
Now,
A sample of 3 bulbs is drawn at random from the lot with having replacement.
Let us consider X be a random variable which denotes the total number of defective bulbs among the sample drawn.
P(X = 0) = P (3 non- defective and 0 defective) = 3C0. 34×34×34 = 3!0!3!×2764 = 1×2764=2764
P(X = 1) = P (2 non- defective and 1 defective) = 3C1. 14×34×34 = 3!1!2!×964 = 3×2!2!×964=3×964=2764
P(X = 2) = P (1 non- defective and 2 defective) = 3C2. 14×14×34 = 3!2!1!×364 = 3×2!2!×364=3×364=964
P(X = 3) = P (0 non- defective and 3 defective) = 3C3. 14×14×14 = 3!3!0!×164 = 1×164=164
Hence, the probability distribution which is required as per the Q.’s demand is given below:
| X | 0 | 1 | 2 | 3 |
| P(X) | 2764 | 2764 | 964 | 164 |
Q.7: Consider a situation where a coin will be biased in such a manner that the tail will occur likely 4 times the head. What will be the probability distribution of the number of heads, by considering the situation that the coin is tossed for two times?
Solution:
Let us consider that the probability of getting the head in the biased coin is x.
Then,
P(H) = x
⟹ P(T) = 4x
For the biased coin, P(H) + P(T) = 1
⟹ x + 4x = 1
⟹ 5x = 1
⟹ x = 15
Hence, P(H) = 15 and P(T) = 45
When the coin will be tossed for two times, then the sample space will be { TT, TH, HT, HH }
Let, X be the random variable which represents the number of heads.
Then,
P(X = 0) = P(no heads) = P(T) × P(T) = 45×45=1625
P(X = 1) = P(one heads) = P(HT) + P(TH) = 15×45+45×15=425+425=825
P(X = 2) = P(two heads) = P(H) × P(H) = 15×15=125
Hence, the probability distribution which is required as per the Q.’s demand is given below:
| X | 0 | 1 | 2 |
| P(X) | 1625 | 825 | 125 |
Q.8: The following probability distribution is for the random variable X.
| X | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| P(X) | 0 | k | 2k | 2k | 3k | k2 | 2k2 | 7k2 + k |
Find:
(i) The value of k
(ii) P(X < 4)
(iii) P(X > 5)
(iv) P (0 < X < 2)
Solution:
We know that, the sum of all of the probabilities of the probability distribution of a random variable is 1.
Thus,
k + 2k + 2k + 3k + k2 + 2k2 + 7k2 + k = 1
⟹ 10k2 + 9k = 1
⟹ 10k2 + 9k – 1 = 0
⟹ 10k2 + 10k – k – 1 = 0
⟹ 10k(k + 1) – 1(k + 1) = 0
⟹ (k + 1)(10k – 1) = 0
⟹ k = -1, 110
Now,
k = -1 is not possible because the probability for any of the event cannot ever be negative.
So, k = 110
(ii) P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
= o + k + 2k + 2k = 5k
= 5×110 = 12
(iii) P(X > 5) = P(X = 6) + P(X = 7)
= 2k2 + 7k2 + k
= 9k2 + k
= k (9k + 1)
= 110(9×110 + 1)
= 110(910+1)
= 110 (9+1010)
= 110(1910)
= 19100
(iv) P(0 < X < 2) = P(X = 1)
= k = 110
Q.9: Consider P(Y) be the probability distribution of the random variable Y for the following forms, where some number is.
P(Y) =
2a, if a = 0
3a, if a = 1
4a, if a = 2
0, otherwise
(i) Find the value of a.
(ii) Determine P(Y < 2), P(Y ≥ 2), and P (Y ≤ 2).
Solution:
(i) As we know that the sum for all of the probabilities of the probability distribution of random variables is given by 1.
Thus,
2a + 3a + 4a + 0 = 1
⟹ 9a = 1
⟹ a = 19
(ii) P(Y < 2) = P(Y = 0) + P(Y = 1)
= 2a + 3a
= 5a = 5×19 = 59
P(Y ≥ 2) = P(Y = 2) + P(Y > 2)
= 4a + 0
= 4a = 4×19 = 49
P(Y ≤ 2) = P(Y = 2) + P(Y = 1) + P(Y = 0)
= 4a + 3a + 2a
= 9a = 9×19 = 1
Q.10: After tossing a fair coin, what will be the mean number for tails?
Solution:
Let us consider X which denotes the chances of success to get tails.
Thus,
For this case, the sample space will be
S = {TTT, TTH, THT, THH, HTT, HTH, HHT, HHH}
Let, the total number of tails be represented by X.
Then,
X(HHH) = 0, X(HTH) = 1, X(HHT) = 1, X(THH) = 1, X(TTH) = 2, X(HTT) = 2, X(THT) = 2 and X(TTT) = 3.
Here, we can see that X will take values 0, 1, 2 or 3.
Now,
P(X = 0) = P(HHH)
= 12×12×12
= 18
P(X = 1) = P(HTH) + P(HHT) + P(THH)
= 12×12×12 + 12×12×12 + 12×12×12
= 18+18+18
= 38
P(X = 2) = P(TTH) + P(HTT) + P(THT)
= 12×12×12 + 12×12×12 + 12×12×12
= 18+18+18
= 38
P(X = 0) = P(TTT)
= 12×12×12 = 18
Hence, the probability distribution which is required as per the Q.’s demand is given below:
| X | 0 | 1 | 2 | 3 |
| P(X) | 18 | 38 | 38 | 38 |
Mean of X E(X), µ = ∑XiP(Xi)
= 0×18+1×38+2×38+3×18
= 38+68+38
= 128
= 32 = 1.5
Q.11: Take two dices which are thrown simultaneously. If Y denotes the number of times we will get sixes, then what will be the expectation of Y?
Solution:
As per the data given in the Q., Y represents the total number of sixes which will be obtained after throwing two dice simultaneously.
Thus, Y will take the values of 0, 1 or 2.
Hence,
P(Y = 0) = P (no sixes will be obtained in either of the throw) = 56×56=2536
P(Y = 1) = P (six at the first dice and other number rather than 6 on the second dice) + P(other number rather than six on the first dice and six on the second dice)
= 2 (56+16)
= 2×536
= 518
P(Y = 2) = P (sixes will be obtained in both of the throw) = 136
Hence, the probability distribution which is required as per the Q.’s demand is given below:
| Y | 0 | 1 | 2 |
| P(Y) | 2536 | 1036 | 136 |
Therefore,
Expectation of X = E(X) = ∑XiP(Xi)
= 0×2536+1×1036+2×136
= 0+518+118
= 618 = 13
Q.12: From the starting six positive integers, two numbers will be selected at random, without any replacement. Let, Y denotes the larger number among those two numbers obtained. Find E(Y).
Solution:
As per the Q.’s demand, from the starting six positive integers, two numbers will be selected, without having replacement which will be done in 6 × 5 = 30 ways.
Let us consider Y which represents the two numbers obtained which are larger. Thus,
Y will take values 2, 3, 4, 5 or 6.
For Y = 2, the possible observations will be (1, 2) and (2, 1).
Then,
P(Y = 2) = 230=115
For Y = 3, the possible observations will be (1, 3), (2, 3), (3, 1) and (3, 2)
Then,
P(Y = 3) = 430=215
For Y = 4, the possible observations will be (1, 4), (2, 4), (3, 4), (4, 1), (4, 2) and (4, 3)
Then,
P(Y = 4) = 630=15
For Y = 5, the possible observations will be (1, 5), (2, 5), (3, 5), (4, 5), (5, 4), (5, 3), (5, 2), (5, 1)
Then,
P(Y = 5) = 830=415
For Y = 6, the possible observations will be (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 5), (6, 4), (6, 3), (6, 2), (6, 1)
Then,
P(Y = 6) = 1030=13
Hence, the probability distribution which is required as per the Q.’s demand is given below:
| X | 2 | 3 | 4 | 5 | 6 |
| P(X) | 115 | 215 | 15 | 415 | 13 |
Therefore,
Expectation of X = E(X) = ∑XiP(Xi)
= 2×115+3×215+4×15+5×415+6×13
= 215+615+4×315+2015+6×515
= 215+615+1215+2015+3015
= 7015 = 143
Q.13: Y denotes the sum of all the numbers which are obtained when two fair dice are rolled. What will be the variance and the standard deviation of Y.?
Solution:
Number of observations which are obtained when two dice is rolled is 6 × 6 = 36
P(Y = 2) = P(1, 1) = 136
P(Y = 3) = P(1, 2) + P(2, 1) = 236
P(Y = 4) = P(1, 3) + P(2, 2) + P(3, 1) = 336
P(Y = 5) = P(1, 4) + P(2, 3) + P(3, 2) + P(1, 4) = 436
P(Y = 6) = P(1, 5) + P(2, 4) + P(3, 3) + P(4, 2) + P(5, 1) = 536
P(Y = 7) = P(1, 6) + P(2, 5) + P(3, 4) + P(4, 3) + P(5, 2) + P(6, 1) = 636
P(Y = 8) = P(2, 6) + P(3, 5) + P(4, 4) + P(5, 3) + P(6, 2) = 536
P(Y = 9) = P(3, 6) + P(4, 5) + P(5, 4) + P(3, 6) = 436
P(Y = 10) = P(4, 6) + P(5, 5) + P(6, 4) = 336
P(Y = 11) = P(5, 6) + P(6, 5) = 236
P(Y = 12) = P(6, 6) = 136
Hence, the probability distribution which is required as per the Q.’s demand is given below:
| Y | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
| P(Y) | 136 | 236 | 336 | 436 | 536 | 636 | 536 | 436 | 336 | 236 | 136 |
Therefore,
Expectation of X = E(X) = ∑XiP(Xi)
= 2×136+3×236+4×336+5×436+6×536+7×636+8×536+9×436+10×336+11×236+12×136
= 236+636+1236+2036+3036+4236+4036+3636+3036+2236+1236
= 25236 = 7
Expectation of X2 = E(X2) = ∑X2iP(Xi)
= 4×136+9×236+16×336+25×436+36×536+49×636+64×536+81×436+100×336+121×236+144×136
= 436+1836+4836+10036+18036+29436+32036+32436+30036+24236+14436
= 197436 = 54.8333
Then,
Var(X) = E(X2) – [ E(X)2 ]
= 54.8333 – 49 = 5.8333
Hence,
Standard derivation = Var(X)−−−−−−−√
= 5.8333−−−−−√ = 2.415
Q.14: There is a class having 15 students with ages 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16, 19 and 20 years. One student will be selected in such a way that each of them have the same chance of being chosen and Y is recorded as the age of the selected student. Find the probability distribution for the random variable Y. What will be the mean, variance and standard derivation of Y?
Solution:
A class has 15 students with ages 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16, 19 and 20 years.
Every student has the same chance of being selected.
Hence, the probability for every student being selected is 115
The above information will be compiled in the frequency table which is given below:
| Y | 14 | 15 | 16 | 17 | 18 | 19 | 20 | 21 |
| f | 2 | 1 | 2 | 3 | 1 | 2 | 3 | 1 |
P(X = 14) = 215
P(X = 15) = 115
P(X = 16) = 215
P(X = 17) = 315
P(X = 18) = 115
P(X = 19) = 215
P(X = 20) = 315
P(X = 21) = 115
Hence, the probability distribution which is required as per the Q.’s demand is given below:
| Y | 14 | 15 | 16 | 17 | 18 | 19 | 20 | 21 |
| P(Y) | 215 | 115 | 215 | 315 | 115 | 215 | 315 | 115 |
Thus,
Expectation of X = E(X) = ∑XiP(Xi)
= 14×215+15×115+16×215+17×315+18×115+19×215+20×315+21×115
= 2815+1515+3215+5115+1815+3815+6015+2115
= 26315 = 17.53
Expectation of X2 = E(X2) = ∑X2iP(Xi)
= (14)2×215+(15)2×115+(16)2×215+(17)2×315+(18)2×115+(19)2×215+(20)2×315+(21)2×115
= 39215+22515+51215+86715+32415+72215+120015+44115
= 468315 = 312.2
Then,
Var(X) = E(X2) – [ E(X)2 ]
= 312.2 – (17.53)2 = 4.78
Hence,
Standard derivation = Var(X)−−−−−−−√
= 4.78−−−−√ = 2.186
Q.15: 70% members are in the favour and 40% are opposing a certain proposal in a meeting. A member will be randomly selected and we will take Y = 0, if the person opposed, and X = 1 if the person favours. What will be E(Y) and var(Y) in such a situation?
Solution:
As per the data given in the Q., we have
P(Y = 0) = 40% = 40100 = 0.4
P(Y = 1) = (100 – 40)% = 60% = 60100 = 0.6
Hence, the probability distribution which is required as per the Q.’s demand is given below:
| X | 0 | 1 |
| P(X) | 0.4 | 0.6 |
Thus,
Expectation of X = E(X) = ∑XiP(Xi)
= 0 × 0.4 + 1 × 0.6 = 0.6
Expectation of X2 = E(X2) = ∑X2iP(Xi)
= 02 × 0.4 + 12 × 0.6 = 0.6
We know that,
Var(X) = E(X2) – [ E(X)2 ]
= 0.6 – (0.6)2 = 0.6 – 0.36 = 0.24
Q.16: The mean of the numbers which will be obtained after throwing a die on which 1 is written on two of the faces, 2 on three faces and 5 on one face is
(a) 1
(b) 2
(c) 5
(d) 136
Solution:
Let us consider Y which is a random variable representing any number on the die.
Since, it’s a die, then the total number of observations will be six.
Thus, P(Y = 1) = 26=13
P(Y = 2) = 36=12
P(Y = 5) = 16
Hence, the probability distribution which is required as per the Q.’s demand is given below:
| Y | 1 | 2 | 5 |
| P(Y) | 13 | 12 | 16 |
Thus,
Expectation of X = E(X) = ∑XiP(Xi)
= 1×13+2×12+5×16
= 13+1+56 = 2+6+56 = 136
Hence, the correct answer is (d).
Q.17: Let Y be the number of aces obtained. Let any of the two cards are drawn at random from the deck of the cards. Then, the value of E(X) will be
(a) 37221
(b) 513
(c) 213
(d) 113
Solution:
Let, Y be the number of aces obtained.
Therefore, Y will take any of the values of 0, 1, or 2
As we know, there are 52 cards in a deck. Among them 4 cards are aces.
Hence, there are 48 non- ace cards.
P(Y = 0) = P(2 non- ace cards and 0 ace card) = 48C2×4C052C2=11281326
P(Y = 1) = P(1 non- ace cards and 1 ace card) = 48C1×4C152C2=1921326
P(Y = 2) = P(0 non- ace cards and 2 ace card) = 48C0×4C252C2=61326
Hence, the probability distribution which is required as per the Q.’s demand is given below:
| Y | 0 | 1 | 2 |
| P(Y) | 11281326 | 1921326 | 61326 |
Thus,
Expectation of X = E(X) = ∑XiP(Xi)
= 0×11281326+1×1921326+2×61326
= 1921326+121326 = 192+121326 = 2041326 = 213
Hence, the correct answer is (c).
EXERCISE – 13.5
Q.1: A die will be thrown for 6 times. Let us consider that, “getting any odd number “will be the success, then what the probability is of
(a) Getting 5 successes? (b) Getting at least 5 successes? (c) Getting at most 5 successes?
Solution:
Odd numbers on the die are 1, 3 and 5.
When a die is tossed repeatedly, it is Bernoulli trials. Let us consider Y which represents the total number of successes of getting any odd numbers in the experiment of 6 toss trials.
In a single throw of the die, the probability of getting any odd number is a = 36=12
Thus,
b = 1 – A = 12
Y has the binomial distribution.
Hence, P(Y = r) = nCrbn–rar, where n = 0, 1, 2, 3, …………, n
= 6Cr126–r12r
= 6Cr126
(a) Probability of getting 5 successes
P(5 successes) = P(Y = 5)
= 6C5126
= 6!5!×1!126
= 6×164
= 332
(b) Probability of getting at least 5 successes
P(at least 5 successes) = P(Y ≥ 5)
= P(Y = 5) + P(Y = 6)
= 6C5126+6C6126
= 6!5!×1!126+6!6!×0!126
= 6×164+1×164
= 764
(c) Probability of getting at most 5 successes
P(at most 5 successes) = P(Y ≤ 5)
= 1 – P(Y > 5)
= 1 – P(Y = 6)
= 1 – 6C6126
= 1 – 6!6!×0!126
= 1 – 1×164
= 1 – 164 = 6364
Q.2: A pair of dice will be thrown for 4 times. Let us consider that, “getting a doublet” will be the success, then what is the probability for getting two successes?
Solution:
When a die is tossed repeatedly, it is Bernoulli trials. Let us consider Y which represents the total number of successes of getting any doublets in the experiment of 4 toss trials.
In a single throw of the die, the probability of getting any odd number is a = 636=16
Thus,
b = 1 – A = 1–16=6–16=56
Y has the binomial distribution with n = 4.
Hence, P(Y = r) = nCrbn–rar, where n = 0, 1, 2, 3, …………, n
= 4Cr564–r16r
4Cr(54–r64)
Probability of getting 2 successes
P(2 successes) = P(Y = 2)
= 4C2(54–264)
= 4!2!×2!521296
= 4×3×2!2!×2!×251296
= 122×251296
= 6×251296 = 25216
Q.3: In a large bulk of different items, there are 5% of the defective items. Find the probability for the sample having 10 items, which will include at most one defective item only.
Solution:
As the drawing of items will be done with replacement, so the trial will be Bernoulli trials.
Let us consider X which denotes the total number of defective items of the sample from which 10 items are drawn successively.
Thus,
p = 5100=120
⟹ q = 1 – p
⟹ q = 1 – 120=1920
Y has the binomial distribution with n = 4 and p = 120.
Hence, P(Y = r) = nCrbn–rar, where n = 0, 1, 2, 3, …………, n
= 4Cr19204–r120r
P(at most 1 defective item) = P(X ≤ 1)
= P(X = 0) + P(X = 1)
= 10C0(1920)10–0.(120)0+10C1(1920)10–1.(120)1
= 10!10!0!(1920)10.×1+10!9!1!(1920)9.(120)
= 1×(1920)10.×1+10×9!9!1!(1920)9.(120)
= (1920)10×1+10×(1920)9.(120)
= (1920)9[1920+1020]
= (1920)9(2920) = (2920).(1920)9
Q.4: From a deck of 52 cards, well- shuffled, five cards will be drawn successively with having replacement. Find the probability that-
(a) All the five cards chosen will be shades?
(b) Among the five selected cards, only three of them will be spade?
(c) None of them is a spade?
Solution:
As the drawing of cards will be done with replacement, so the trial will be Bernoulli trials.
Let us consider X which denotes the number of spade cards in the deck from which 5 cards are drawn successively.
We know that in a deck of cards, there are 52 cards and among them 13 cards are spade.
Thus,
a = 1352=14
⟹ b = 1 – a
⟹ b = 1 – 14=34
Y has the binomial distribution.
Hence, P(Y = r) = nCrbn–rar, where n = 0, 1, 2, 3, …………, n
= 4Cr19204–r120r
(a) P(all the five cards chosen will be spades) = P(Y = 5) having n = 5 and p = 14
= 5C5.(34)5–5.(14)5
= 5!5!0!.(34)0.(14)5
= 1×1×11024
= 11024
(b) P(among them 3 cards chosen will be spades) = P(Y = 3) having n = 5 and p = 14
= 5C3.(34)5–3.(14)3
= 5!3!2!.(34)2.(14)3
= 5×2×916×164
= 45512
(iii) P(none of them will be spades) = P(Y = 0) having n = 5 and p = 14
= 5C0.(34)5–0.(14)0
= 5!0!5!.(34)5×1
= 1×2431024
= 2431024
Q.5: If the probability that a bulb which is produced by the factory will fuse at least after 150 days of its use, is 0.10. What will be the probability that out of 5 such bulbs,
(a) None
(b) Not more than one
(c) More than 1
(d) At least 1
Bulb will fuse after 150 days of its use.
Solution:
As the drawing of bulbs will be done with replacement, so the trial will be Bernoulli trials.
Let us consider X which denotes the number of bulbs which will fuse at least after 150 days of its use in the experiment on 5 trial bulbs.
As per the data in the Q,
a = 0.1
⟹ b = 1 – 0.1 = 0.9
Y has the binomial distribution.
Hence, P(Y = r) = nCrbn–rar, where n = 0, 1, 2, 3, …………, n
Here, n = 5 and p = 0.1.
Thus,
5Cr(0.9)5–r(0.1)r
(a) P(none) = P(Y = 0)
= 5C0(0.9)5–0(0.1)0
= 5!0!5!(0.9)5×1
= (0.9)5 = 0.59049 = 0.59
(b) P(not more than 1) = P(Y ≤ 1)
= P(Y = 0) + P(Y = 1)
= 5C0(0.9)5–0(0.1)0+5C1(0.9)5–1(0.1)1
= 5!0!5!(0.9)5×1+5!1!4!(0.9)4×0.1
= (0.9)5 + 5×4!1!4!(0.9)4×0.1
= (0.9)5 + 5 × (0.9)4 × 0.1
= 0.59049 + 0.5 × (0.9)4
= 0.59049 + 0.32805 = 0.91854
(c) P(more than 1) = P(Y ≥ 1)
= 1 – P(Y ≤ 1) = 1 – 0.91854 = 0.08146
(d) P(at least one) = P(Y > 1)
= 1 – P(Y < 1) = 1 – P(Y = 0) = 1 – 0.59 = 0.41
Q.6: There are 10 balls in the bag, each marked with any one of the digit from 0 to 9. Find the probability that none of them will be marked with the digit 0, if the four balls will be drawn successively from the bag, with replacement?
Solution:
As the drawing of balls will be done with replacement, so the trial will be Bernoulli trials.
Let us consider X which denotes the number of balls which will be marked with any one of the digits from 0 to 9 from which 4 balls are drawn successively from the bag.
As per the data in the Q.,
a = 110
⟹ b = 1 – 110=910
Y has the binomial distribution.
Hence, P(Y = r) = nCrbn–rar, where n = 0, 1, 2, 3, …………, n
Here, n = 4 and p = 110.
Thus,
4Cr9104–r110r
Now,
P(none of them is marked with 0) = P(Y = 0)
= 4C0(910)4–0.(110)0
= 4!0!4!.(910)4×1
= 1×(910)4×1 = (910)4
Q.7: There are 20 true and false type Q.s which are asked in the Q. paper in an examination. Let us consider a situation that a student picks up his option by tossing a fair coin every time. If its heads on the coin, the student answered true for the Q., and if its tails then the student answered false. What will be the probability that he can answer at least 12 Q.s.
Solution:
Let, Y represents the total number of correctly answered Q.s from those 20 Q.s.
As the picking of answer from the repeated tossing will be done with replacement, so the trial will be Bernoulli trials.
As per the student’s observation, whenever there is head on the coin it represents that the answer is true and whenever there is tails on the coin it represents that the answer is false.
Thus,
a = 12
⟹ b = 1 – p = 1 – 12 = 12
Y will have the binomial distribution with n = 20, a = 12 and b = 12
Thus,
P(Y = r) = nCrbn–rar, where n = 0, 1, 2, 3, …………, n
= 20Cr(12)20–r(12)r
= 20Cr(12)20
P(at least 12 Q.s from those 20 Q.s) = P(Y ≥ 12)
= P(Y = 12) + P(Y = 13) + P(Y = 14) + …….. + P(Y = 20)
= 20C12(12)20 + 20C13(12)20 + 20C14(12)20 + …………………. + 20C20(12)20
= (12)2.[20C12+20C13+20C14+………+20C20]
Q.8: Let us consider Y being a binomial distribution R(6, 12). Prove that, Y = 3 will be the most likely outcome.
Solution:
Here, as per the data given in the Q.,
Y is the binomial distribution with having binomial distribution as R(6, 12).
Hence,
n = 6 and a = 12
⟹ b = 1 – p = 1 – 12 = 12
Y will have the binomial distribution with n = 20, a = 12 and b = 12
Thus,
P(Y = r) = nCrbn–rar, where n = 0, 1, 2, 3, …………, n
= 6Cr(12)6–r(12)r
= 6Cr(12)6
Here, we can see that P(Y = r) can be the maximum if, 6Cr can be maximum.
Thus,
6C0=6C6=6!0!6!=1 6C1=6C5=6!1!5!=6×5!1!5!=6 6C2=6C4=6!2!4!=6×5×4!2!4!=3×5=156C3=6C3=6!3!3!=6×5×4×3!3!3!=2×5×2=15
Here, we can see that the value of 6C3 is maximum.
Hence, for y = 3, P(Y = r) is maximum.
Therefore, Y = 3 is the most likely outcome.
Q.9: In a multiple choice Q.s examination which having three possible answers for every five Q.s, find the probability for the candidate that he will get 4 or more correct answers by having a guess every time.
Solution:
Let, Y represents the total number of correctly answered Q.s among those 5 multiple choice Q.s by having a guess.
As the picking of answer from the repeated guessing for the correct answer from the multiple choices will be done with replacement, so the trial will be Bernoulli trials.
Now,
Probability of getting the correct answer is given by,
a = 13
⟹ b = 1 – a = 1 – 13 = 23
Hence,
Y will have the binomial distribution with n = 5, a = 13 and b = 23
Thus,
P(Y = r) = nCrbn–rar, where n = 0, 1, 2, 3, …………, n
= 5Cr(23)5–r(13)r
P(guessing at least 4 correct answer) = P(Y ≥ 4)
= P(Y = 4) + P(Y = 5)
= 5C4(23)5–4(13)4 + 5C5(23)5–5(13)5
= 5C4(23)1(13)4 + 5C5(23)0(13)5
= 5C4.23.(13)4 + 5C5.1.(13)5
= 5!4!1!.23.(13)4 + 5!5!0!.1.(13)5
= 5×4!4!.23.(13)4 + 1.1.(13)5
= 5.23.(13)4 + 1.1.(13)5
= 10243+1243 = 11243
Q.10: A person purchased a lottery ticket from 50 different lotteries, among each of which his chance to win a prize is 1100. Find the probability that the person will win a prize
(i) Exactly once
(ii) at least once
(iii) at least twice?
Solution:
Let us consider Y which represents the number of winning prizes among 50 lotteries.
As the picking of winner will be done with replacement, so the trial will be Bernoulli trials.
Now,
Y is a binomial distribution with n = 50 and a = 1100
⟹ b = 1 – a = 1 – 1100 = 99100
Thus,
P(Y = r) = nCrbn–rar, where n = 0, 1, 2, 3, …………, n
= 50Cr(99100)50–r(1100)r
(a) P (winning exactly once) = P(X = 1)
= 50C1(99100)50–1(1100)1
= 50!1!49!(99100)49(1100)1
= 50(1100)(99100)49
= 12(99100)49
(ii) P(winning at least once) = P(Y ≥ 1)
= 1 – P(Y < 1)
= 1 – P(X = 0)
= 1 – 50Cr(99100)50
= 1 – 50!0!50!.(99100)50
= 1 – 1.(99100)50
= 1 – (99100)50
(iii) P (at least twice) = P(Y ≥ 2)
= 1 – P(Y < 2)
= 1 – P(Y ≤ 1)
= 1 – [P(Y = 0) + P(Y = 1)]
= [1 – P(Y = 0)] – P(Y = 1)
= 1 – (99100)49. [99100+12]
= 1 – (99100)49. (99+50100)
= 1 – (99+50100) . (99100)49
Q.11: What will be the probability of getting 5 exactly twice in 7 throws of a die?
Solution:
As the repeatedly tossing of a die will be done with replacement, so the trial will be Bernoulli trials.
Let us consider Y which represents the number of times to get 5 in 7 throws of the die.
As per the data given in the Q., a = 16
⟹ b = 1 – a = 1 – 16 = 56
Y is the probability distribution with n = 7, a = 16 and b = 56
Then,
P(Y = r) = nCrbn–rar, where n = 0, 1, 2, 3, …………, n
= 7Cr(56)7–r(16)r
P (getting 5 exactly twice) = P (Y = 2)
= 7C2(56)7–2(16)2
= 7!2!5!(56)2(16)2
= 7×6×5!2×5!(56)2(16)2
= 7×62(56)2(136)
= 21(56)2(136)
= (56)2(712)
Q.12: 10% of the certain articles which are manufactured will be defective. Find the probability for the samples having 12 such articles among which 9 of them will be defective.
Solution:
As the repeated selection of the articles in a random sample space will be done with replacement, so the trial will be Bernoulli trials.
Let us consider Y which represents the number of times to select the defective articles in the random sample space of 12 articles.
As per the data given in the Q., a = 10% = 10100=110
⟹ b = 1 – a = 1 – 110 = 910
Y is the probability distribution with n = 12, a = 110 and b = 910
Then,
P(Y = r) = nCrbn–rar, where n = 0, 1, 2, 3, …………, n
= 12Cr(910)12–r(110)r
P(selecting 9 defective articles) = 12C9(910)12–9(110)9
= 12!9!3!(910)3(110)9
= 12×11×10×9!9!×3×2×1(910)3(110)9
= 12×11×103×2×1(910)3(110)9
= 220.(910)3(110)9
= 22×931011
Q.13: Among the 100 bulbs in the box, 10 bulbs are defective. The probability that out of the sample of 5 bulbs, none of the bulbs are defective is,
(a) 10-1
(b) (12)5
(c) (910)
(d) (910)5
Solution:
As the repeated selection of the defective bulbs in the box will be done with replacement, so the trial will be Bernoulli trials.
Let us consider Y which represents number of the defective bulbs in a random sample of 5 bulbs.
As per the data given in the Q., a = 10100=110
⟹ b = 1 – a = 1 – 110 = 910
Y is the probability distribution with n = 5, a = 110 and b = 910
Then,
P(Y = r) = nCrbn–rar, where n = 0, 1, 2, 3, …………, n
= 5Cr(910)5–r(110)r
P(none of the bulbs are defective) = P(Y = 0)
5C0(910)5–0(110)0
= 5!5!0!(910)5×1
= 1×(910)5×1
= (910)5
Hence, the correct option is (d).
Q.14: The probability that the student will not be a swimmer is 15. The probability that from a group of 5 students, 4 of them will be a swimmer is
(a) (45)415
(b) 5C4(45)415
(c) 5C1(45)415
(d) None of these
Solution:
As the repeated selection of the students in the group who all are swimmers will be done with replacement, so the trial will be Bernoulli trials.
Let us consider Y which represents number of the students from the group of 5 students, who all are swimmers.
As per the data given in the Q., b = 15
⟹ a = 1 – b = 1 – 15 = 45
Y is the probability distribution with n = 5, a = 45 and b = 15
Then,
P(Y = r) = nCrbn–rar, where n = 0, 1, 2, 3, …………, n
= 5Cr(15)5–r(45)r
P(none of the bulbs are defective) = P(Y = 4)
= 5C4(15)5–4(45)4
= 5C4(15)(45)4
Hence, the correct answer is (b).
