Question

An urn A contains $3$ white and $5$ black balls. Another urn $B$ contains $6$ white and $8$ black balls. A ball is picked from $A$ at random and then transferred to $B$. Then, a ball is picked at random from $B$. The probability that it is a white ball is

• A. $\frac{14}{40}$
• B. $\frac{15}{40}$
• C. $\frac{16}{40}$
• D. $\frac{17}{40}$

Answer 1

Answer: (D)

Case Ist Let a white ball be transfered from the first bag to the second bag.
The probability of selecting white ball from the first bag is
$P_{1}=\frac{3}{3+5}=\frac{3}{8}$
Now, the second bag has 7 white and 8 black balls.
The probability of selecting a white ball from the second bag is
$P_{2}=\frac{7}{7+8}=\frac{7}{15}$
The probability that both these events take place simultaneously
$=P_{1} \times P_{2}=\frac{3}{8} \times \frac{7}{15}=\frac{7}{40}$
Case IInd Let a black ball be transfered from the first bag to the second bag.
Its probability is $P_{3}=\frac{5}{3+5}=\frac{5}{8}$
Now, the second bag contains 6 white and 9 black balls.
The probability of drawing a white ball from the second bag is
$P_{4}=\frac{6}{6+9}=\frac{6}{15}$
$\therefore $ The probability of both these events taking place simultaneously
$=P_{3} \times P_{4}=\frac{5}{8} \times \frac{6}{15}=\frac{1}{4}$
$\therefore $ The required probability
$=P_{1} P_{2}+P_{3} P_{4}$
$=\frac{7}{40}+\frac{1}{4}$
$=\frac{7+10}{40}=\frac{17}{40}$