Hello students welcome to this new chapter probability. Where we are going to speak about condition of probability in this module. Now we should have strong knowledge about probability what you studied in your 11th std. You were studied about definition of probability also you have studied about addition theorem rich to tell them. We have proceed further same concept of probability which is chapter for you in the 12th. Lets consider the definition of conditional probability , if E and F are two events associated with a same sample space of a random experiment the condition of probability of the event E given that F has occurred is given by P of E given by F is equal to P of E intersection F divided by P of F where P of F is not equal to zero remember this is a formula of conditional probability now I will note taken on that E given F already seen what about P of F given E what happen what will be happen there P of F given E is equal to P of F intersection E divided by what dived by P of F making sure the denominator is not zero so that is a note you have let me proceed further there are some properties of conditional probability let see one by one and we see proof of it as well the first property says probability of S given F is equal to P of F given F is equal to one . The second one says probability A union B given F is equal to probability P of A given F + probability B given F – probability A intersection B given F And third one says probability E compliment given F is equal to 1 – probability E given F this are the three properties let see solution one by one. Let me start with the first probability S given F I understand the condition of probability e given F is equal to E intersection by F what will S given F it will be P of S intersection F divided by P of F remember this what is after that should be in denominator S given F what is second should be in denominator. now what about the proceeding part of this S intersection F what you understand for this S stands for sample space and I know that F is a subset of sample space F is an event is a subset of sample space and what about the intersection obviously it is F P of S intersection F, P of F divided by P of F it is 1 so I have got first part of first property , what about second part of first property P of F given F it is according the formula of condition probability I can write it as P of F Intersection F divided by P of F and F intersection F because F is Same as F it is contained so I can write F itself P of F divided by P of F is equal to 1 so we have got proof of 1st property . let me go with the proof of second property . let me start P of A union B given F according to condition probability formula we can write it as P of A union B intersection F divided by P of F now what about the numerator so I can use distributive law A union B intersection C is equal to A union C intersection B union C, same thing I will write here P of A union F intersection B union c divided by as it is P of A . now what about numerator use the addition theorem of probability and split the numerator as it is what I get that I will get it as P of A intersection F + P of B intersection F – P of A intersection B intersection F divided by P of F . now I will split it into three parts I will get it as P of A intersection F divide by P of A + P of B intersection F divided by P of A – P of A intersection B intersection F divided by P of F now I can recall the condition of probability . earlier we have seen left hand side right in the right hand side now see the right hand side right the left hand side according to that the first term P of A given F what is the second term P of B given F third term will be p of A intersection B giving F and this is the proof of second property I can take a note regarding the particular property when A and B are mutually exclusive intersection is zero their for I can get P of A union B giving P of A giving F + P of B giving F only A and B are mutually exclusive events . let me go the property no. 3 proof , we know that the first property P of S given F is equal to 1 also I know E union E compliment any event union is compliment same as the sample space let me replace s by E union E’ do you think E union E’ are mutually exclusive yes they are mutually exclusive according to note mentioned earlier A union B giving F is equal to P of A giving F + P of B giving F , and I use that particular note P of E Giving F + P of E’ giving F is equal to 1 there for the taking 1st term to right hand side I get P of E’ giving F is equal to 1- P of E giving F that’s a required proof of the property so these are the 3 properties what we have conditional probability , Thank you .