Multiple choice

In a series of n independent trials for an event of constant probability p, the most probable number r of successes is given by $\left ( n+1 \right )p-1< r< \left ( n+1 \right )p$. Hence, the most probable number of successes is the integral part of $\left ( n+1 \right )p$. But if $\left ( n+1 \right )p$ 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

  1. $\dfrac{10!}{5!5!}\left ( \dfrac{2}{3} \right )^5$
  2. $\dfrac{10!}{5!5!}\left ( \dfrac{1}{3} \right )^5$
  3. $\dfrac{10!}{\left(5! \right)^2}\left(\dfrac{2}{9} \right )^5$
  4. $\dfrac{10!}{5!}\left(\dfrac{2}{9} \right )^5$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

This is a binomial distribution problem with n=10, p=2/3 (probability of white ball), and q=1/3. The probability of exactly 5 successes is C(10, 5) * (2/3)^5 * (1/3)^5 = (10! / (5!5!)) * (2^5 / 3^10). Option C is (10! / (5!5!)) * (2^5 / 9^5) = (10! / (5!5!)) * (2^5 / 3^10), which matches.

AI explanation

The probability of drawing a white ball is 2/3 and the probability of drawing a black ball is 1/3. Using the binomial distribution formula, the probability of exactly 5 successes in 10 trials is 10C5 multiplied by the success probability to the 5th power and the failure probability to the 5th power. Substituting these probabilities gives 10! / (5! times 5!) multiplied by (2/3)^5 and (1/3)^5, which simplifies to 10! / (5!)^2 multiplied by (2/9)^5.

Keep practicing — related questions