A bag contains 6 white and 4 black balls. A die is rolled once and the number of balls equal to the number obtained on the die are drawn from the bag at random. The probability that all the balls drawn are white is :

Solution

Maths sir · 3 steps

Split by what the die shows, find the all-white chance in each case, then average over the six faces.

1. Total probability. Let the die show k, where k = 1, 2, …, 6, each with probability 1/6. Drawing k balls, all of them are white with probability C(6, k)/C(10, k). This is the total-probability sum that forms the denominator of Bayes' theorem.

   P(all white) = 1/6Σₖ₌₁⁶ C(6, k)/C(10, k)

   Die shows k	C(6, k) ÷ C(10, k)	Value	Over 210
   1	6 ÷ 10	3/5	126/210
   2	15 ÷ 45	1/3	70/210
   3	20 ÷ 120	1/6	35/210
   4	15 ÷ 210	1/14	15/210
   5	6 ÷ 252	1/42	5/210
   6	1 ÷ 210	1/210	1/210

2. Add up. The six terms add to (126 + 70 + 35 + 15 + 5 + 1)/210 = 252/210 = 6/5.
3. So. P(all white) = 1/6 × 6/5 = 1/5, which is option (c).

The idea to keep. When a first random step decides how the second experiment runs, condition on each outcome of the first step, then add (probability of case) × (success in that case).

Answer from the NTA final answer key. Solution by PrepWiser.