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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