Question

A man speaks the truth 3 out of 4 times. He rolls a dice and reports that the outcome is 6. What is the probability that the number is actually 6 ?

Answer 1

The probability is `3/4`.

Explanation:

Let `S` be the event that the roll is a 6.
Let `R` be the event that the man reports a 6.

Then we want to find `"P"(S|R).`

`"P"(S|R)=("P"(S nn R))/("P"(R))`

`color(white)("P"(S|R))=["P"(R|S)"P"(S)]/("P"(R))`

`color(white)("P"(S|R))=["P"(R|S)"P"(S)]/("P"(R|S)"P"(S)+"P"(R|S^C)"P"(S^C))`

Here's where it gets a bit tricky. We might think that `"P"(R|S^C)=1/4,` because he has a 1 in 4 chance of lying. But let's say the roll is a '1' and he lies. He has 5 choices of number to falsely report, and only one of them is '6'. So, we really should say that `"P"(R|S^C)=1/4xx1/5.`

So:

`"P"(S|R)=[(3/4)(1/6)]/((3/4)(1/6)+(1/20)(5/6))`

`color(white)("P"(S|R))=[1/8]/(1/8+1/24) color(blue)(xx24/24)`

`color(white)("P"(S|R))=[3]/(3+1)`

`color(white)("P"(S|R))=3/4`