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.

• A. $\frac{24}{25}$
• B. $\frac{40}{41}$
• C. $\frac{1}{25}$
• D. $\frac{1}{41}$

Answer 1

Answer: (B)

$\frac{40}{41}$

There are two hypotheses (i) their coincident testimony is true (ii) it is false.

$\therefore P_1=\frac{1}{6}$(red ball), $P_2=\frac{5}{6}$(no red ball)

$p_1=\frac{2}{3}\times \frac{4}{5},P_2=(\frac{1}{5}\times \frac{1}{5})\times \frac{1}{3}\times \frac{1}{5}$

For in estimating $p_2$ we must take into account the chance that $A$ and $B$ will both select the red ball, when it has not been drawn.

Thus, $P_1{p}_1:P_2{p}_2=8:\frac{1}{5}=40:1$

Hence, the probability that the statement is true is $\frac{40}{41}.$