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 $1W$ and $3B$ marbles are transferred.
$E_2$ be the event that $2W$ and $2B$ marbles are transferred.
$E_3$ be the event that $3W$ and $1B$ marbles are transferred.
$E_4$ be the event that $4W$ and $0B$ 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 $=P\left(E_1/A\right)=$
$(P\left(E_1\right)\cdot P\left(A/E_1\right)+P\left(E_2\right)\cdot P\left(A{E}_2\right)$
$+P\left(E_3\right)P\left(A/E_3\right)+P\left(E_4\right)\cdot P\left(A/E_4\right)$
$=\frac{\frac{5}{{}^8{C}_4}\cdot \frac{3}{4}}{\frac{5}{{}^8{C}_4}\cdot \frac{3}{4}+\frac{30}{{}^8{C}_4}\times \frac{2}{4}+\frac{30}{{}^8{C}_4}\times \frac{1}{4}+0}$
$=\frac{15}{15+60+30}=\frac{15}{105}=\frac{1}{7}$