Question

From an urn containing six balls, $3$ white and $3$ black ones, a person selects at random an even number of balls (all the different ways of drawing an even number of balls are considered equally probable, irrespective of their number).

Then the probability that there will be the same number of black and white balls among them is?

• A. $4/5$
• B. $11/15$
• C. $11/30$
• D. $2/5$

Answer 1

The correct option is B $11/15$

Probability of selecting 2,4, or 6 balls $=\frac{1}{3}$ each

Total ways of selecting 2 balls from 6 balls$=(\frac{6}{2})$

Ways of selecting 1 white and 1 black ball$=(\frac{3}{1})\times (\frac{3}{1})$

Probability that 2 balls are picked having same no. of black and white balls$=P_1=\frac{1}{3}\times \frac{(\frac{3}{1})\times (\frac{3}{1})}{(\frac{6}{2})}$

$\Rightarrow \frac{3\times 3\times 2}{3\times 6\times 5}$

$\Rightarrow \frac{1}{5}$

Total ways of selecting 4 balls from 6 balls$=(\frac{6}{4})$

Ways of selecting 2 white and 2 black balls$=(\frac{3}{2})\times (\frac{3}{2})$

Probability that 4 balls are picked having same no. of black and white balls$=P_2=\frac{1}{3}\times \frac{(\frac{3}{2})\times (\frac{3}{2})}{(\frac{6}{4})}$

$\Rightarrow \frac{3\times 3\times 2}{3\times 6\times 5}$

$\Rightarrow \frac{1}{5}$

Total ways of selecting 6 balls from 6 balls$=(\frac{6}{6})$

Ways of selecting 3 white and 3 black balls$=(\frac{3}{3})\times (\frac{3}{3})$

Probability that 6 balls are picked having same no. of black and white balls$=P_3=\frac{1}{3}\times \frac{(\frac{3}{3})\times (\frac{3}{3})}{(\frac{6}{6})}$

$\Rightarrow \frac{1\times 1}{3\times 1}$

$\Rightarrow \frac{1}{3}$

Total probability$=P_1+P_2+P_3$

$\Rightarrow \frac{1}{5}+\frac{1}{5}+\frac{1}{3}$

$\Rightarrow \frac{11}{15}$

Hence, answer is option $B$