A man speaks truth 3 out of 4 times. He throws a die and reports that

1.	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.
Answer» Let E₁, E₂ and A be the events defined as follows: E₁ = Six occurs, E₂ = Six does not occur A = man reports it is a six Then, P(E₁) = \(\frac{1}{6}\) , P(E₂) = 1- \(\frac{1}{6}\) = \(\frac{5}{6}\) P(A/E₁) = Probability of man reporting it a six when six occurs = Probability of speaking truth = \(\frac{3}{4}\) P(A/E₂) = Probability of man reporting a six when six does not occur = Probability of not speaking truth = 1-\(\frac{3}{4}\) = \(\frac{1}{4}\) ∴ P(Throw is actually a six) =\(\frac{P(E_1)\times P(A/E_1)}{P(E_1)\times P(A/E_1)+P(E_2)\times P(A/E_2)}\) ... (Baye’s Theorem) = \(\frac{\frac{1}{6}\times \frac{3}{4}}{\frac{1}{6}\times \frac{3}{4}+\frac{5}{6}\times \frac{1}{4}}\) = \(\frac{\frac{3}{24}}{\frac{8}{24}}\) = \(\frac{3}{8}\)

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.

Answer»

Let E₁, E₂ and A be the events defined as follows:

E₁ = Six occurs,

E₂ = Six does not occur

A = man reports it is a six

Then, P(E₁) = \(\frac{1}{6}\) , P(E₂) = 1- \(\frac{1}{6}\) = \(\frac{5}{6}\)

P(A/E₁) = Probability of man reporting it a six when six occurs = Probability of speaking truth = \(\frac{3}{4}\)

P(A/E₂) = Probability of man reporting a six when six does not occur

= Probability of not speaking truth = 1-\(\frac{3}{4}\) = \(\frac{1}{4}\)

∴ P(Throw is actually a six) =\(\frac{P(E_1)\times P(A/E_1)}{P(E_1)\times P(A/E_1)+P(E_2)\times P(A/E_2)}\) ... (Baye’s Theorem)

= \(\frac{\frac{1}{6}\times \frac{3}{4}}{\frac{1}{6}\times \frac{3}{4}+\frac{5}{6}\times \frac{1}{4}}\) = \(\frac{\frac{3}{24}}{\frac{8}{24}}\) = \(\frac{3}{8}\)

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