Question

A bowl contains 10 balls, of which 4 are red and 6 are white. Balls are randomly selected with replacement from the bowl until 4 red balls have been selected. Let X be the number of white balls drawn before the fourth red ball is selected. Calculate the probability that X is 6.

Answer 1

Answer:

The probability that X is 6 = 0.2508

Step-by-step explanation:

Probability of picking a red ball = 4 / 10 = 0.4

Probability of picking a white ball = 6 / 10 = 0.6

Let's first state that the last ball selected in this scenario will be red. This implies that we selected 6 white balls and 3 red balls in any manner before we selected the last red ball. So we need to find the probability of taking out 6 white balls and 3 red balls, and use permutation to account for the order they get drawn in:

Total Probability = Probability of 6 white balls * Probability of 3 red balls * number of ways this can happen

Probability of 6 white balls = 0.6 ^ 6

Probability of 3 red balls = 0.4 ^ 3

Number of ways this can happen (permutation with similar objects) = $\frac{9!}{6!*3!}$

Number of ways this can happen = 84

Total probability = $0.6^6 * 0.4^3 * 84$

Total probability = 0.2508