Question

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.

• A. $\frac{1}{7}$
• B. $\frac{6}{7}$
• C. $\frac{1}{8}$
• D. $\frac{7}{8}$

Answer 1

Answer: (A)

Let $E_{1}$ be the event that $1\, W$ and $3\, B$ marbles are transferred.
$E_{2}$ be the event that $2\, W$ and $2\, B$ marbles are transferred.
$E_{3}$ be the event that $3\, W$ and $1\, B$ marbles are transferred.
$E_{4}$ be the event that $4\, W$ and $0\, B$ marbles are transferred
and $A$ be the event that a black marble is drawn from bag $Q$.
Then,
$P\left(E_{1}\right)=\frac{{ }^{5} C_{1} \times{ }^{3} C_{3}}{{ }^{8} C_{4}}=\frac{5}{{ }^{8} C_{4}}$
$P\left(E_{2}\right)=\frac{{ }^{5} C_{2} \times{ }^{3} C_{2}}{{ }^{8} C_{4}}=\frac{30}{{ }^{8} C_{4}}$
$P\left(E_{3}\right)=\frac{{ }^{5} C_{3} \times{ }^{3} C_{1}}{{ }^{8} C_{4}}=\frac{30}{{ }^{8} C_{4}}$
$P\left(E_{4}\right)=\frac{{ }^{5} C_{4} \times{ }^{3} C_{0}}{{ }^{8} C_{4}}=\frac{5}{{ }^{8} C_{4}} $
$P\left(A / E_{1}\right)=\frac{3}{4}$
$P\left(A / E_{2}\right)=\frac{2}{4}$
$P\left(A / E_{3}\right)=\frac{1}{4}$
$P\left(A / E_{4}\right)=0$
Required probability $=P\left(E_{1} / A\right)=$
$\left(P\left(E_{1}\right) \cdot P\left(A / E_{1}\right)+P\left(E_{2}\right) \cdot P\left(A E_{2}\right)\right.$
$+P\left(E_{3}\right) P\left(A / E_{3}\right)+P\left(E_{4}\right) \cdot P\left(A / E_{4}\right)$
$=\frac{\frac{5}{{ }^{8} C_{4}} \cdot \frac{3}{4}}{\frac{5}{{ }^{8} C_{4}} \cdot \frac{3}{4}+\frac{30}{{ }^{8} C_{4}} \times \frac{2}{4}+\frac{30}{{ }^{8} C_{4}} \times \frac{1}{4}+0}$
$= \frac{15}{15+60+30}=\frac{15}{105}=\frac{1}{7}$