JEE Main Solved Paper 2017 Question 29
Question: For three events A, B and C, P(Exactly one of A or B occurs)= P(Exactly one of B or C occurs) = P(Exactly one of C or A occurs) $ =\frac{1}{4} $ and P(All the three events occur simultaneously) $ =\frac{1}{16}. $ Then the probability that at least one of the events occurs, is :- JEE Main Solved Paper-2017
Options:
A) $ \frac{3}{16} $
B) $ \frac{7}{32} $
C) $ \frac{7}{16} $
D) $ \frac{7}{64} $
Show Answer
Answer:
Correct Answer: C
Solution:
- P (exactly one of A or B occurs) $ =P(A)+P(B)-2P(A\cap B)=\frac{1}{4} $ P(Exactly one of B or C occurs) $ =P(B)+P(C)-2P(B\cap C)=\frac{1}{4} $ P(Exactly one of C or A occurs) $ =P(C)+P(A)-2P(C\cap A)=\frac{1}{4} $ Adding all, we get $ 2\sum P(A)-2\sum P(A\cap B)=\frac{3}{4} $
$ \therefore $ $ \sum P(A)-\sum P(A\cap B)=\frac{3}{8} $ Now, $ P(A\cap B\cap C)=\frac{1}{16} $ (given)
$ \therefore $ $ P(A\cup B\cup C) $ $ =\sum P(A)-\sum P(A\cap B)+P(A\cap B\cap C) $ $ =\frac{3}{8}+\frac{1}{16}=\frac{7}{16} $
