An urn $A$ contains 3 white and 5 black balls. Another urn $B$ contains 6 white and 8 black balls. A ball is…
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
$\frac{14}{40}$
$\frac{15}{40}$
$\frac{16}{40}$
$\frac{17}{40}$
Solution
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 $=P_1 P_2+P_3 P_4$
$=\frac{7}{40}+\frac{1}{4}$
$=\frac{7+10}{40}=\frac{17}{40}$