Question

From a bag containing $9$ distinct white and $9$ distinct black balls, $9$ balls are drawn at random one by one, the drawn balls being replaced each time. The probability that atleast four balls of each colour is in the draw is

• A. a little less than $\frac{1}{2}$
• B. a little greater than $\frac{1}{2}$
• C. $\frac{1}{2}$
• D. $\frac{1}{3}$

Answer 1

Answer: (A)

a little less than $\frac{1}{2}$

At least $4$ balls of each color can be obtained with either $4$ white and $5$ black balls or $5$ white and $4$ black balls.
Probability of selecting a white ball $=$ Probability of selecting a black ball $=$ $\frac{9}{18}=\frac{1}{2}$
Thus, the required probability

$=$ $\{{\left(\frac{1}{2}\right)}^4\times {\left(\frac{1}{2}\right)}^5\times \frac{9!}{4!\times 5!}\}+\{{\left(\frac{1}{2}\right)}^5\times {\left(\frac{1}{2}\right)}^4\times \frac{9!}{4!\times 5!}\}$

$=\{2\times {\left(\frac{1}{2}\right)}^9\times \frac{9!}{4!\times 5!}\}=\frac{63}{128}$

Hence, (A) is correct.