Multiple choice

A bag contains 12 white balls and 3 black balls .Balls are drawn one by one without replacement till all the black balls are drawn . The probability that the procedure will come to an end at the $7^{th}$ draw is

  1. $\dfrac{27}{91}$
  2. $\dfrac{9}{91}$
  3. $\dfrac{3}{91}$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For the process to end at the 7th draw, the 7th draw must be the 3rd black ball, and exactly 2 black balls must have been drawn in the first 6 draws. Ways = (6C2 * 3C2 * 12C10) / 15C7. Calculation: (15 * 3 * 66) / 6435 = 2970 / 6435 = 3/91.

AI explanation

For the procedure to end at the 7th draw, the first six draws must contain exactly two black balls and four white balls, and the 7th draw must be the final black ball. The probability of drawing two black and four white balls in the first six draws is found using combinations, calculated as (3C2 times 12C4) divided by 15C6, which equals 3 times 495 divided by 5005, simplifying to 1485/5005. The probability of drawing the last black ball on the 7th turn from the remaining 9 balls is 1/9. Multiplying these probabilities gives (1485/5005) times (1/9), which simplifies to 165/5005 and reduces to 3/91.

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