Probability 2 Question 1

1. For three events A,B and C, if P (exactly one of A or B occurs )=P (exactly one of B or C occurs )=P (exactly one of C or A occurs )=14 and P (all the three events occur simultaneously) =116, then the probability that atleast one of the events occurs, is

(a) 732

(b) 716

(c) 764

(d) 316

(2017 Main)

Show Answer

Answer:

Correct Answer: 1. (b)

Solution:

  1. We have, P (exactly one of A or B occurs)

=P(AB)P(AB)=P(A)+P(B)2P(AB)

According to the question,

P(A)+P(B)2P(AB)=14P(B)+P(C)2P(BC)=14P(C)+P(A)2P(CA)=14

and

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

2[P(A)+P(B)+P(C)P(AB)P(BC)P(CA)]=34P(A)+P(B)+P(C)P(AB)P(BC)P(CA)=38=P(A)+P(B)+P(C)P(AB)P(BC)P(CA)+P(ABC)=38+116=716P(ABC)=116



जेईई के लिए मॉक टेस्ट

एनसीईआरटी अध्याय वीडियो समाधान

दोहरा फलक