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 .
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