Answer: 80.30%
Explanation:
To calculate the probability, we need to determine the total number of possible samples and the number of samples that satisfy the given conditions.
Exactly two white balls and two red balls:
The total number of ways to choose four balls from a bucket of 11 balls is given by the combination formula: C(11, 4) = 11! / (4! * 7!) = 330.
The number of ways to choose exactly two white balls and two red balls can be calculated as follows: C(6, 2) * C(5, 2) = (6! / (2! * 4!)) * (5! / (2! * 3!)) = 15 * 10 = 150.
Therefore, the probability of selecting exactly two white balls and two red balls is 150/330 ≈ 0.4545 or approximately 45.45%.
At least two white balls:
To calculate this probability, we need to consider the following possibilities:
Selecting exactly two white balls and two red balls: We calculated this in the previous question, and the probability is 150/330 ≈ 0.4545 or approximately 45.45%.
Selecting three white balls and one red ball: C(6, 3) * C(5, 1) = (6! / (3! * 3!)) * (5! / (1! * 4!)) = 20 * 5 = 100.
Selecting four white balls and no red ball: C(6, 4) = 6! / (4! * 2!) = 15.
Adding up these possibilities, the total number of favorable outcomes is 150 + 100 + 15 = 265.
Therefore, the probability of selecting at least two white balls is 265/330 ≈ 0.8030 or approximately 80.30%.