Question

A speaks the truth 8 times out of 10 times. A die is tossed. He reports that it was 5. What is the probability that it was actually 5?

Answer 1

Let us assume U₁, U₂ and A be the events as follows:

U₁ = Getting 5 on throwing a die

U₂ = Getting other than 5 on throwing a die

A = Reporting 5 after throwing the die

From the problem,

⇒ P(A|U₁) = P(Reporting 5 on actually getting 5 on throwing a die)

⇒ P(A|U₁) = P(Telling the truth)

⇒ P(A|U₁) = \(\cfrac8{10}\)

⇒ P(A|U₂) = P(Reporting 5 but not getting 5 on throwing a die)

⇒ P(A|U₂) = P(Not telling the truth)

⇒ P(A|U₂) = \(\cfrac2{10}\)

Now we find

P(U₁|A) = P(the die actually shows 5 given that man reports 5)

Using Baye’s theorem:

∴ The required probability is \(\cfrac49\).