Multiple choice

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

  1. a little less than $\dfrac {1}{2}$
  2. a little greater than $\dfrac {1}{2}$
  3. $\dfrac {1}{2}$
  4. $\dfrac {1}{3}$
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A Correct answer
Explanation

With 9 white and 9 black balls, the probability of drawing a white ball is 1/2. Drawing 9 balls with replacement follows a binomial distribution B(9, 0.5). The probability of at least four of each color is 1 - P(0, 1, 2, or 3 of one color). Due to symmetry, this is 1 - 2 * (P(0) + P(1) + P(2) + P(3)). Calculating this gives 1 - 2 * (1/512 + 9/512 + 36/512 + 84/512) = 1 - 2 * (130/512) = 1 - 260/512 = 252/512, which is slightly less than 1/2.

AI explanation

Because the balls are replaced, each draw is an independent Bernoulli trial with a 1/2 chance of drawing a white ball and a 1/2 chance of drawing a black ball. By the binomial distribution, the probability of drawing 4 white and 5 black balls is the combination of 9 choose 4 times (1/2)^9, which is 126/512. The probability of 5 white and 4 black balls is identical, making the combined probability 252/512, which equals 0.4921875. This value is a little less than 1/2.

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