Question

A man is known to speak truth 5 out of 7 times. He, the man , throws a fair dice and a number appears, What is the probability that the number appears , is an even number and the man lies it to be an odd number, at the same time?

Answer 1

Hint: They have given that the man speaks 5 times truth out of 7 times so find the probability that he speaks false through the true probability. As it is a fair dice find the probability that the output number is an even number i.e. 2 or 4 or 6. As they asked that the both conditions should happen at the same time multiply both the probabilities in order to get the required answer.

Complete step by step answer:
Given that the man speaks 5 times truth out of 7 times
Which implies the probability that he lies will be (1-probability that he speaks truth)=$1 - \dfrac{5}{7}$ = $\dfrac{2}{7}$
Given that a fair dice is thrown one time ,
Which implies probability that the outcome is even number
i.e. Probability that the outcome is 2 or 4 or 6 out of 1, 2, 3, 4, 5, 6= $\dfrac{3}{6} = \dfrac{1}{2}$;
The question is asked for the probability that the outcome is even and he lies that it is an odd number.
Which implies the multiplication of probability that the outcome is even and probability that he is lying as an odd number.
i.e the answer will be $\dfrac{1}{2} \times \dfrac{2}{7} = \dfrac{1}{7}$
Therefore, the probability that the number appears , is an even number and the man lies it to be an odd number, at the same time

Note:
Read the question properly and confirm whether they said the man is lying it as an odd number of any number other than the resulted outcome. And check the condition whether they asked that at the same time or either one of them.