Question

In a series of n independent trials for an event of constant probability p, the most probable number r of successes is given by $(n+1)p-1<r<(n+1)p$. Hence, the most probable number of successes is the integral part of $(n+1)p$. But if $(n+1)p$ is an integer, the chance of r successes is equal to that of $r+1$ successes and both $r,r+1$ are most probable numbers of successes.
A bag contains 2 white balls and 1 black ball. A ball is drawn at random and returned to the bag.

The experiment is done 10 times. The probability that a white ball is drawn exactly 5 times is

• A. $\frac{10!}{5!5!}{\left(\frac{2}{3}\right)}^5$
• B. $\frac{10!}{5!5!}{\left(\frac{1}{3}\right)}^5$
• C. $\frac{10!}{{(5!)}^2}{\left(\frac{2}{9}\right)}^5$
• D. $\frac{10!}{5!}{\left(\frac{2}{9}\right)}^5$

Answer 1

The correct option is C $\frac{10!}{{(5!)}^2}{\left(\frac{2}{9}\right)}^5$

We need to select which 5 of the 10 are white. Also, probability of a white is $\frac{2}{3}$ and that of a black is $\frac{1}{3}$
Probability of choosing a white ball 5 times out of 10 is ${}_5^{10}C\ast ({\frac{2}{3})}^5\ast {\left(\frac{1}{3}\right)}^5=\frac{10!}{{(5!)}^2}\ast {\left(\frac{2}{9}\right)}^5$
Hence, (c) is correct.