Question

State true or false:

The probabilities of three mutually exclusive events A, B, C are $P\left(A\right)=\frac{2}{3},P\left(B\right)=\frac{1}{4},P\left(C\right)=\frac{1}{6}.$

• A. True
• B. False

Answer 1

The correct option is B False

Since the events A, B, C are mutually exclusive, we have
$P(A\cup B\cup C)=P\left(A\right)+P\left(B\right)+P\left(C\right)$
$=\frac{2}{3}+\frac{1}{4}+\frac{1}{6}=\frac{13}{12}>1$
which is impossible since the probability of any event cannot be greater than 1.