Multiple choice

A bag contains 10 white and 3 black balls. Balls are drawn one by one without replacement til all the black balls are drawn. The probability that the test will come to an end at the 7th draw is

  1. $\frac{^{10}c_{4}\times ^{3}c_{2}}{^{13}c_{6}}$
  2. $\frac{^{10}c_{4}\times ^{3}c_{2}}{^{13}c_{6}}\times \frac{2}{7}$
  3. $\frac{^{10}c_{4}\times ^{3}c_{2}}{^{13}c_{7}}$
  4. $\frac{^{10}c_{5}\times ^{3}c_{2}}{^{13}c_{7}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

To end at the 7th draw, the 3rd black ball must be drawn on the 7th attempt. This means in the first 6 draws, exactly 2 black balls and 4 white balls were drawn. The probability is (ways to pick 2 black from 3 AND 4 white from 10) / (total ways to pick 6 from 13) * (probability that the 7th is the remaining black ball). The formula provided in A correctly represents the selection of the first 6 balls.

AI explanation

Using the hypergeometric probability method, we calculate the total ways to choose 6 balls from 13, which is 13C6. For the 7th draw to end the test, the first 6 draws must contain exactly 2 black balls and 4 white balls, which can be done in 3C2 * 10C4 ways. The probability of this specific combination in the first six draws is exactly (10C4 * 3C2) divided by 13C6.

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