Multiple choice

A box contains $20$ identical balls of which $10$ are blue and $10$ are green. The balls are drawn at random from the box one at a time with replacement. The probability that a blue ball is drawn $4th$ time on the $7th$ draw is

  1. $\dfrac {27}{32}$
  2. $\dfrac {5}{64}$
  3. $\dfrac {5}{32}$
  4. $\dfrac {1}{2}$
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C Correct answer
Explanation

This is a negative binomial distribution problem. We want the 4th blue ball on the 7th draw. This means in the first 6 draws, there are 3 blue balls and 3 green balls. The probability is (6C3) * (1/2)^3 * (1/2)^3 * (1/2) = 20 * (1/64) * (1/2) = 20/128 = 5/32.

AI explanation

By the negative binomial distribution, the 7th draw must be a blue ball and exactly 3 of the first 6 draws must be blue. The probability of 3 blue balls in the first 6 draws is found using the binomial probability formula as 6C3 multiplied by (1/2) cubed multiplied by (1/2) cubed, equaling 20/64 or 5/16. Multiplying this by the 1/2 probability that the 7th draw is blue gives a final probability of 5/32.

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