Probability 2 Question 4

4. For the three events A,B and C,P( exactly one of the events A or B occurs )=P (exactly one of the events B or C occurs )=P( exactly one of the events C or A occurs ) =p and P (all the three events occurs simultaneously) =p2, where 0<p<12. Then, the probability of atleast one of the three events A,B and C occurring is

(1996, 2M)

(a) 3p+2p22

(b) p+3p24

(c) p+3p22

(d) 3p+2p24

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Answer:

Correct Answer: 4. (c)

Solution:

  1. We know that,

P (exactly one of A or B occurs)

=P(A)+P(B)2P(AB)P(A)+P(B)2P(AB)=p Similarly, P(B)+P(C)2P(BC)=p and P(C)+P(A)2P(CA)=p

On adding Eqs. (i), (ii) and (iii), we get

2[P(A)+P(B)+P(C)P(AB)P(BC)P(CA)]=3pP(A)+P(B)+P(C)P(AB)P(BC)P(CA)=3p2(v)

It also given that, P(ABC)=p2

P (at least one of the events A,B, and C occurs)

=P(A)+P(B)+P(C)P(AB)P(BC)P(CA)+P(ABC)=3p2+p2 [from Eqs. (iv) and (v)] =3p+2p22



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