Question

A man is known to speak the truth on average $3$ out of $4$ times. He throws a fair dice and reports that its a $6.$ what is the probability it is actually a $6$ .
A. ${\text{ }}\dfrac{3}{5}$
B. ${\text{ }}\dfrac{3}{8}$
C. ${\text{ }}\dfrac{1}{5}$
D. ${\text{ }}\dfrac{3}{4}$

Answer 1

Hint: First of all we will find the probability of the individual man which speaks truth and the favourable probability of getting six in a dice. Then will use Bayes theorem to find the resultant probability.

Complete step by step answer:
Probability is the state of being probable and the extent to which something is likely to happen in the particular situations or the favourable outcomes. Probability of any given event is equal to the ratio of the favourable outcomes with the total number of the outcomes.

$P(A) = $ Total number of the favourable outcomes / Total number of the outcomes
The probability that the man speaks the truth is $P(A) = \dfrac{3}{4}$
The probability that man do not speak the truth is $P(A') = 1 - \dfrac{3}{4} = \dfrac{1}{4}$
Also, the probability of getting $6$ in a fair dice is $P(B) = \dfrac{1}{6}$
Similarly the probability of not getting $6$ in a fair dice is $P(B') = 1 - \dfrac{1}{6} = \dfrac{5}{6}$

Applying the Bayes theorem – $P(A/B) = \dfrac{{P(B/A)P(A)}}{{P(B)}}$
Where, $P(B/A)\;{\text{and P(A / B)}}$are the conditional probability
The required probability can be expressed as
$\Rightarrow \dfrac{{\dfrac{1}{6}.\dfrac{3}{4}}}{{\dfrac{1}{6}.\dfrac{3}{4} + \dfrac{5}{6}.\dfrac{1}{4}}}$
Simplify the above expression –
$\Rightarrow\dfrac{{\dfrac{3}{{24}}}}{{\dfrac{3}{{24}} + \dfrac{5}{{24}}}}$
When there are the same denominators we can add a numerator directly.
$\Rightarrow\dfrac{{\dfrac{3}{{24}}}}{{\dfrac{8}{{24}}}}$
Numerator’s denominator and denominator’s denominator are the same and hence they cancel each other.
$\dfrac{3}{8}$

Hence, option B is correct.

Note: The probability of any event always ranges between zero and one. It can never be the negative number or the number greater than one. The probability of impossible events is always equal to zero whereas, the probability of the sure event is always equal to one.