Probability Question 232
Question: The probabilities of three mutually exclusive events are 2/3, 1/4 and 1/6. The statement is
[MNR 1987; UPSEAT 2000]
Options:
A) True
B) Wrong
C) Could be either
D) Do not know
Show Answer
Answer:
Correct Answer: B
Solution:
Since here $ P(A+B+C)=P(A)+P(B)+P(C) $
$ =\frac{2}{3}+\frac{1}{4}+\frac{1}{6}=\frac{13}{12} $ , which is greater than 1. Hence the statement is wrong.
BETA
