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

Complete Solution:

The number of balls drawn (n) can range from 1 to 6 (as the die shows numbers 1 to 6). For each n, the probability of selecting n balls such that all are white is:

P(all white for a given n) = C(6, n) / C(10, n)

where:

• C(6, n): Number of ways to choose n white balls from 6 white balls.
• C(10, n): Number of ways to choose n balls from a total of 10 balls.

Since the die is fair, the probability of rolling any number (n) is \( 1/6 \). The overall probability of drawing n balls where all are white is:

P(all white) = (1/6) Σ [C(6, n) / C(10, n)] for n = 1 to 6.

We calculate \( C(6, n) / C(10, n) \) for each value of \( n \):

• For n = 1: (6, 1) = 6, (10, 1) = 10 so P = 6/10 = 0.6
• For n = 2: (6, 2) = 15, C(10, 2) = 45 , so P = 15/45 = 0.333
• For n = 3: (6, 3) = 20, C(10, 3) = 120, so ( P = 20/120 = 0.167
• For n = 4: (6, 4) = 15, C(10, 4) = 210 , so ( P = 15/210 = 0.0714
• For n = 5: (6, 5) = 6, C(10, 5) = 252 , so ( P = 6/252 = 0.0238
• For n = 6: (6, 6) = 1, C(10, 6) = 210, so P = 1/210 = 0.00476 

The overall probability is:

P(all white) = (1/6) × (0.6 + 0.333 + 0.167 + 0.0714 + 0.0238 + 0.00476

Simplifying:

P(all white) = (1/6) × 1.19996 ≈ 0.2

Final Answer:

The probability that all the balls drawn are white is: 0.2 or 1/5.