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

   

Answer 1

 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.