A man is known to speak truth 3 out of 4 times. He throws a die and reports that it is four. The probability that it is actually four is

A man throws a die and reports that it is four. The man speaks truth 3 out of 4 times.

Probability he tells the truth = $\frac{3}{4}$

Probability he lies = $\frac{1}{4}$

Probability of getting four on a die = $\frac{1}{6}$

Probability of not getting four = $\frac{5}{6}$

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The man can report "four" in two scenarios:

Scenario A: Die actually shows 4 and he tells the truth

Probability = $\frac{1}{6} \times \frac{3}{4}$

$= \frac{3}{24}$

Scenario B: Die does not show 4 and he lies (claims it is 4)

Probability = $\frac{5}{6} \times \frac{1}{4}$

$= \frac{5}{24}$

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Total probability of hearing "four":

$P(\text{Reports 4}) = P(\text{Scenario A}) + P(\text{Scenario B})$

$= \frac{3}{24} + \frac{5}{24}$

$= \frac{8}{24}$

$= \frac{1}{3}$

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Using Bayes' Theorem:

$P(\text{Actually 4} | \text{Reports 4}) = \frac{P(\text{Actually 4 and Reports 4})}{P(\text{Reports 4})}$

$= \frac{3/24}{8/24}$

$= \frac{3}{24} \times \frac{24}{8}$

$= \frac{3}{8}$

Therefore, the probability that it is actually four is $\frac{3}{8}$.