Question

A person speaks truth 2 out of 3 times. He throws a die and reports that there was a 6. Then the probability that it is actually, a six.

• A. 1/7
• B. 2/7
• C. 3/7
• D. 4/7

Answer 1

Answer: (B)

2/7

Answer 2

Let A and B be the events of throwing & not throwing 6.

P(1) = $\frac { 1 } { 6 }$. P(2) = $\frac { 5 } { 6 }$

E be the event of reporting there was a 6

P(E/A) = $\frac { 1 } { 3 }$

P(A/E) = $\frac { \frac { 1 } { 6 } \times \frac { 2 } { 3 } } { \frac { 1 } { 6 } \times \frac { 2 } { 3 } + \frac { 5 } { 6 } \times \frac { 1 } { 3 } } = \frac { 2 } { 7 }$ .