Class 12 Maths Probability Properties of conditional probability

Properties of conditional probability:

• P(S|F) = 1
• P(F|F) = 1
• P((A∪B)|F) = P(A|F) + P(B|F) – P((A∩B)|F)
• P(E'|F) = 1- P(E|F)

Example: Evaluate P(A ∪ B), if 2P(A) = P(B) = 5/13 and P(A|B) =2/5
Solution: ∵ 2P(A) = P(B) = 5/13
=> P(B) = 5/13 and P(A) = 1/2 x P(B) = 1/2  × 5/13 = 5/26

Now the formula of conditional probability is P(A|B) = P( A∩B)/ P(B)
∴ 2/ 5 = P( A∩B) /P(B)  [Since it was given in question that P(A|B)=2/5 ]

=> P(A∩B) = 2/5 X P(B
= 2/5 × 5/13
= 2/13

So, P(A∩B)= 2/13 , but we need to find P(A ∪ B)
Now is there any formula relating P(A ∪ B) and P(A∩B) ?
Yes , the formula is P(A ∪ B)=P(A)+P(B)−P(A∩B)
Now you know values of P(A)= 5/ 26 , P(B)= 5/13 and P(A∩B)= 2/13.

So substitute these values in the formula, you will get value of P(A∪B).
That is P(A∪B)=P(A)+P(B)−P( A∩B)
=> P(A∪B)= 5/26 + 5/13 - 2/13
=> P(A∪B)= (5+10−4 )/26
=> P(A∪B)=11/26

∴ The value of P(A∪B) is 11/26

Example: Mother, father and son line up at random for a family picture
E : son on one end, F : father in middle , what is the probability that the son is on one end given that the father is in middle? Solution:

let Mother be denoted with M, Father be denoted with F and Son be denoted with S
If the sample space for all the possible ways of arranging them is denoted by S
Then the possible elements in S are, S = {MFS, MSF, FMS, FSM, SMF, SFM}
Now E denotes son on one end as given in question so A={MFS, FMS, SMF, SFM}
and F denoted Father in middle as given in question , so B={MFS, SFM}
Since the common elements between A and B are MFS and SFM ∴
A∩B ={MFS, SFM}
Now, P(E∩F) = probability of getting two elements of A∩B I.e; MFS and SFM out of the total 6 elements MFS, MSF, FMS, FSM, SMF, SFM.
= 2/6 = 1/3
and P(F)= probability of getting two elements of B i.e; MFS and SFM out of the total 6 elements MFS, MSF, FMS, FSM, SMF, SFM.
= 2/6 = 1/3
∴ P(E|F) = P(E∩F) /P(F)
= (1/3)  / (1/3)
= 1
∴The probability that the son is on one end given that the father is in middle is 1.

Example: If A and B are events such that P(A|B) = P(B|A), then
(A)A ⊂ B but A ≠ B

(B)A = B

(C)A ∩ B = Φ

(D)P(A) = P(B)

Solution:
Given in question that P(A|B) = P(B|A)
But we know P(A|B) = P( A∩B) /P(B)
and P(B|A) = P( A∩B)/ P(A)
since P(A|B) = P(B|A)
=> P( A∩B)/ P(B) = P( A∩B)/ P(A)
=> P(A) = P(B)
Thus the correct option is (D).

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