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 there was actually 3 is

• A. 2/15
• B. 3/15
• C. 1/15
• D. 4/15

Answer 1

Answer: (C)

1/15

Answer 2

P(E₁) = $\frac { 1 } { 6 }$ ; P(E₂) = P(E₃) ; P(E₄) = P(E₅) ; P(E₆) = $\frac { 1 } { 6 }$

Let E be event of reporting 6

P(E/E₁) = $\frac { 1 } { 3 } \times \frac { 1 } { 5 }$P(E/E₂) = $\frac { 1 } { 3 } \times \frac { 1 } { 5 }$; P(E/E₃) = $\frac { 1 } { 15 }$

P(E/E₄) = $\frac { 1 } { 15 }$P(E/E₅) = $\frac { 1 } { 15 }$; P(E/E₆) = $\frac { 2 } { 3 }$

P(E₃/E) =$\frac { \frac { 1 } { 6 } \times \frac { 1 } { 15 } } { \frac { 1 } { 6 } \times \frac { 1 } { 15 } + \frac { 1 } { 6 } \times \frac { 1 } { 15 } + \frac { 1 } { 6 \times 15 } + \frac { 1 } { 6 \times 15 } + \frac { 1 } { 6 } \times \frac { 2 } { 3 } }$

= $\frac { \frac { 1 } { 90 } } { \frac { 5 } { 90 } + \frac { 10 } { 90 } } = \frac { 1 } { 15 }$ .