Question

A man is known to speak the truth 3 out of 4 times. He throws a die and reports that it is a six. The probability that it is actually a six, is

• A. $\frac { 3 } { 8 }$
• B. $\frac { 1 } { 5 }$
• C. $\frac { 3 } { 4 }$
• D. None of these

Answer 1

Answer: (A)

$\frac { 3 } { 8 }$

Answer 2

Let E denote the event that a six occurs and A the event that the man reports that it is a ‘6’, we have

$P \left( A / E ^ { \prime } \right) = \frac { 1 } { 4 }$.

By Baye’s theorem,

$P ( E / A ) = \frac { P ( E ) \cdot P ( A / E ) } { P ( E ) \cdot P ( A / E ) + P \left( E ^ { \prime } \right) P \left( A / E ^ { \prime } \right) } = \frac { \frac { 1 } { 6 } \times \frac { 3 } { 4 } } { \frac { 1 } { 6 } \times \frac { 3 } { 4 } + \frac { 5 } { 6 } \times \frac { 1 } { 4 } } = \frac { 3 } { 8 }$ .