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14.Probability
easy
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