Question

A, B, and C are independent witness of an event which is known to have occurred. Aspeaks the truth three times out of four, B four times out of five and C five times out of six. What is the probability that the occurrence will be reported truthfully by majority of three witnesses?

Answer 1

\[P\left(\text{  A speaks truth }  \right) = \frac{3}{4}\]

\[P\left( \text{ B speaks truth } \right) = \frac{4}{5}\]

\[P\left( \text{ C speaks truth } \right) = \frac{5}{6}\]

\[P\left( \text{ majority speaks truth } \right) = P\left( \text{ two speak truth } \right) + P\left( \text{ all speak truth } \right)\]

\[ = P\left( A \right) \times P\left( B \right)\left[ 1 - P\left( C \right) \right] + P\left( A \right) \times P\left( C \right)\left[ 1 - P\left( B \right) \right] + P\left( C \right) \times P\left( B \right)\left[ 1 - P\left( A \right) \right] + P\left( A \right) \times P\left( B \right) \times P\left( C \right)\]

\[ = \frac{3}{4} \times \frac{4}{5}\left( 1 - \frac{5}{6} \right) + \frac{3}{4} \times \frac{5}{6}\left( 1 - \frac{4}{5} \right) + \frac{4}{5} \times \frac{5}{6}\left( 1 - \frac{3}{4} \right) + \frac{3}{4} \times \frac{4}{5} \times \frac{5}{6}\]

\[ = \frac{12}{120} + \frac{15}{120} + \frac{20}{120} + \frac{60}{120}\]

\[ = \frac{107}{120}\]