A bag $P$ contains 5 white marbles and 3 black marbles. Four marbles are drawn at random from $P$ and are…
A bag $P$ contains 5 white marbles and 3 black marbles. Four marbles are drawn at random from $P$ and are put in an empty bag $Q$. If a marble drawn at random from $Q$ is found to be black then the probability that all the three black marbles in $P$ are transfered to the bag $Q$ is.
$\frac{1}{7}$
$\frac{6}{7}$
$\frac{1}{8}$
$\frac{7}{8}$
Solution
Here, $W=$ White, $B=$ Black. Given that bag $P$ contains $5 W$ marbles and $3 B$ marbles. 4 marbles are drawn from the bag $P$ and $A$ black marble is drawn from bag $Q$.
Let Event $E_1: 1 \mathrm{~W}$ and $3 \mathrm{~B}$ marbles are transferred Event $E_2: 2 \mathrm{~W}$ and $2 \mathrm{~B}$ marbles are transferred Event $E_3: 3 \mathrm{~W}$ and $1 \mathrm{~B}$ marbles are transferred Event $E_4: 4 \mathrm{~W}$ and $0 \mathrm{~B}$ marbles are transferred and Event $A$ : a black marble is drawn from bag $Q$.
Then, $P\left(E_1\right)=\frac{{ }^5 C_1 \times{ }^3 C_3}{{ }^8 C_4}=\frac{5}{{ }^8 C_4}$
$
\begin{aligned}
& P\left(E_2\right)=\frac{{ }^5 C_2 \times{ }^3 C_2}{{ }^8 C_4}=\frac{30}{{ }^8 C_4} \\
& P\left(E_3\right)=\frac{{ }^5 C_3 \times{ }^3 C_1}{{ }^8 C_4}=\frac{30}{{ }^8 C_4} \\
& P\left(E_4\right)=\frac{{ }^5 C_4 \times{ }^3 C_0}{{ }^8 C_4}=\frac{5}{{ }^8 C_4}
\end{aligned}
$
$
\begin{aligned}
& P\left(A / E_1\right)=\frac{3}{4} \\
& P\left(A / E_2\right)=\frac{2}{4} \\
& P\left(A / E_3\right)=\frac{1}{4} \\
& P\left(A / E_4\right)=0
\end{aligned}
$
By using Bayes' theorem,
$
\begin{aligned}
P\left(E_1 / A\right)= & \frac{P\left(E_1\right) \cdot P\left(A / E_1\right)}{\left(P\left(E_1\right) \cdot P\left(A / E_1\right)+P\left(E_2\right) \cdot P\left(A / E_2\right)\right.} \\
& +P\left(E_3\right) P\left(A / E_3\right)+P\left(E_4\right) \cdot P\left(A / E_4\right) \\
= & \frac{\frac{5}{{ }^8 C_4} \cdot \frac{3}{4}}{\frac{5}{{ }^8 C_4} \cdot \frac{3}{4}+\frac{30}{{ }^8 C_4} \times \frac{2}{4}+\frac{30}{{ }^8 C_4} \times \frac{1}{4}+0} \\
= & \frac{15}{15+60+30}=\frac{15}{105}=\frac{1}{7}
\end{aligned}
$