A man is known to speak the truth 2 out of 3 times. If he throws a die and reports that it is six, then the probability that it is actually five, is

Step-by-step solution

Let the events E₁= six occurs E₂= six does not occurs A= the mean reports that it is six. We have, aligned & P (E₁ )= 1 6 & P (E₂ )= 5 6 aligned Now, P ( A E₁ )= 2 3 , P ( A E₂ )= 1 3 aligned P ( E₁ A ) & = P (E₁ ) P ( A E₁ ) P (E₁ ) P ( A E₁ )+P (E₂ ) P ( A E₂ ) & = 1 6 2 3 1 6 2 3 + 5 6 1 3 = 2 7 aligned The probability that it is actually six = 2 7 The probability that it is not actually six =1- 2 7 = 5 7 The probability that is actually five = 1 5 5 7 = 1 7 .