Probabilities of three mutually exclusive events are frac{2}{3} , frac{1}{4} and frac{1}{6} .

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14.Probability

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The probabilities of three mutually exclusive events are $\frac{2}{3} , \frac{1}{4}$ and $\frac{1}{6}$. The statement is

A

$1$

B

Wrong

C

Could be either

D

Do not know

Solution

(b) 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.

Standard 11

Mathematics

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