A man is known to speak truth 3 out of 4 times. He throws a die and report that it is a 6. Find the probability that it is actually 6.

Answer 1

Toolbox:

• Given $E_1, E_2, E_3.....E_n$ are mutually exclusive and exhaustive events, we can find the conditional probability $P(E_i/A)$ for any event A associated w/ $E_i$ using the Bayes theorem as follows: \(\;P(E_i/A)\)=\(\large \frac{P(E_i)P(A/E_i)}{\sum_{i=1}^{n}\;P(E_i)P(A/E_i)}\)

Step 1:

Let $E$ be the event that the man reports that six occurs in the throwing of the die and let $S_1$ be the event that six occurs and $S_2$ be the event that six does not occur.

$P(S_1)=\large\frac{1}{6}$

$P(S_2)=1-\large\frac{1}{6}=\frac{5}{6}$

$P(\large\frac{E}{S_1})$=probability that the man reports that six occurs when 6 has actually occurred on the die.

$P(\large\frac{E}{S_1})$=probability that the man speaks the truth=$\large\frac{3}{4}$

$P(\large\frac{E}{S_2})$=probability that the man reports that six occurs when 6 has not actually occurred on the die.

$P(\large\frac{E}{S_2})$=probability that the man does not speak the truth

$\Rightarrow 1-\large\frac{3}{4}=\frac{1}{4}$

Step 2:

Hence by Baye's theorem we get,

$P(\large\frac{S_1}{E})$=Probability that the report of the man that six has occurred is actually a six.

$P(\large\frac{S_1}{E})=\frac{P(S_1).P(E/S_1)}{P(S_1)P(E/S_1)+P(S_2).P(E/S_2)}$

$\qquad\quad=\large\frac{1/6\times 3/4}{1/6\times 3/4+5/6\times 1/4}$

$\qquad\quad=\large\frac{1}{8}\times \frac{24}{8}$

$\qquad\quad=\large\frac{3}{8}$

Hence the required probability is $\large\frac{3}{8}$