Multiple choice

A bag contains $8$ red and $5$ white balls. Two successive draws of $3$ balls are made without replacement. The probability that the first draw will produce $3$ white balls and the second $3$ red ball is

  1. $\dfrac{5}{223}$
  2. $\dfrac{6}{257}$
  3. $\dfrac{7}{429}$
  4. $\dfrac{3}{548}$
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C Correct answer
AI explanation

Using the multiplication rule of probability for sampling without replacement, the probability of drawing 3 white balls first is 5C3 divided by 13C3, which evaluates to 10/286 or 5/143. Given that 3 white balls are removed, 10 balls remain in the bag, consisting of 8 red and 2 white balls, making the probability of drawing 3 red balls in the second draw 8C3 divided by 10C3, which is 56/120 or 7/15. The combined probability of both events occurring is 5/143 multiplied by 7/15, yielding 35/2145, which simplifies to 7/429. The result is 7/429.

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