A bag contains 6 balls. If 4 balls are drawn at a time and all of them are found to be red, then the…
A bag contains 6 balls. If 4 balls are drawn at a time and all of them are found to be red, then the probability that exactly 5 of the balls in the bag are red is
$\frac{10}{19}$
$\frac{5}{21}$
$\frac{1}{21}$
$\frac{5}{7}$
Solution
Let us define the following events
$E=$ drawn balls are red. $\quad A=4$ red balls in bag.
$B=5$ red balls in bag.
$C=6$ red balls in bag.
Then, $P(A)=P(B)=P(C)$
$
\text { and } \begin{aligned}
P\left(\frac{E}{A}\right) & =\frac{{ }^4 C_4}{{ }^6 C_4}=\frac{1}{15} \\
P\left(\frac{E}{B}\right) & =\frac{{ }^5 C_4}{{ }^6 C_4}=\frac{5}{15} \text { and } P\left(\frac{E}{C}\right)=\frac{{ }^6 C_4}{{ }^6 C_4}=\frac{15}{15}
\end{aligned}
$
By applying Baye's theorem,
$
\begin{gathered}
P\left(\frac{B}{A}\right)=\frac{P(B) \cdot P\left(\frac{E}{B}\right)}{P(A) P\left(\frac{E}{A}\right)+P(B) P\left(\frac{E}{B}\right)+P(C) P\left(\frac{E}{C}\right)} \\
=\frac{\frac{5}{15}}{\frac{1}{15}+\frac{5}{15}+\frac{15}{15}}=\frac{5}{21}
\end{gathered}
$