Question

A man speaks truth 3 out of 4 times. He throws a die and reports that it is a six. Find the probability that it is actually a six.

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

option (c)

For it to be an actual six two things are required to happen together.

1. He is speaking the truth. Let this be an event A.

2. The die shows 6. Let this be the event B.

We have: P(A) =$\frac{3}{4}$ P(B) = $\frac{1}{6}$

The probability that both events happen together = P(A) $\times$ P(B) =$\frac{3}{4}$$\times$$\frac{1}{6}$ =$\frac{1}{8}$.