Question

$A$ speaks the truth \[2\] out of $3$ times, and $B$ $4$ times out of $5$ ; they agree in the assertion that from a bag containing $6$ balls of different colours a red ball has been drawn: find the probability that the statement is true.

Answer 1

Hint: For solving this particular question , we have to consider two events that are $E$ be an event that the ball drawn is red and both agreeing that it is red , ${E_1}$ be the event that the ball drawn is non-red (any of the remaining five different colours) and both agreeing that it is red but telling a lie.

Complete step by step solution:
Now, the ball drawn may be red or non-red (may be any of the remaining five different colours) and both may say that it is red.
Let $E$ be an event that the ball drawn is red and both agreeing that it is red ,
$P(E) = \dfrac{1}{6} \times \dfrac{2}{3} \times \dfrac{4}{5} = \dfrac{4}{{45}}$
Let ${E_1}$ be the event that the ball drawn is non-red (any of the remaining five different colours) and both agreeing that it is red but telling a lie.
Both can tell a lie in five different ways.
Suppose the non- red ball drawn is blue, they may say it is black, white, red, yellow, green (all lies) of which the probability of telling red is $\dfrac{1}{5}$ on which they agree.
$A$ may tell a lie that it is red in $\dfrac{1}{5} \times \dfrac{1}{3} = \dfrac{1}{{15}}$​ ways.
$B$ can tell it in $\dfrac{1}{5} \times \dfrac{1}{5} = \dfrac{1}{{25}}$ ways.
Also, there is $\dfrac{5}{6}$​ probability of picking up a non-red ball.
Thus, probability that a non-red ball is drawn, and both agree that it is red thus asserting that it is red but telling a lie is ,
$P({E_1}) = \dfrac{5}{6} \times \dfrac{1}{{15}} \times \dfrac{1}{{25}} = \dfrac{1}{{450}}$
Now, we have found a probability of two events in which both are asserting that the ball is red, but when event $E$ occurs the assertion is with speaking the truth and when event ${E_1}$ occurs, assertion is by telling a lie.
The probability of their asserting that the ball is red and speaking truth,
$
 \Rightarrow \dfrac{{P(E)}}{{P(E) + P({E_1})}} \\
 \Rightarrow \dfrac{{\dfrac{4}{{45}}}}{{\dfrac{4}{{45}} + \dfrac{1}{{450}}}} \\
 \Rightarrow \dfrac{{40}}{{41}} \\
 $

Therefore, the answer is $\dfrac{{40}}{{41}}$

Note: Questions similar in nature as that of above can be approached in a similar manner and we can solve it easily. The probability of their asserting that the ball is red and speaking truth is
$\dfrac{{P(E)}}{{P(E) + P({E_1})}}$ where ${E_1}$ is the event that the ball drawn is non-red (any of the remaining five different colours) and both agreeing that it is red but telling a lie. $E$ is an event that the ball drawn is red and both agree that it is red.