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Question
The probabilities of three mutually exclusive events A, B and C are given by 2/3, 1/4 and 1/6 respectively. The statement
Options
is true
is false
nothing can be said
could be either
Solution
is false
Since the events A, B and C are mutually exclusive, we have:
\[P\left( A \cup B \cup C \right) = \frac{2}{3} + \frac{1}{4} + \frac{1}{6} = \frac{13}{12} > 1\], which is not possible.
Hence, the given statement is false.
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