To solve this problem, we can use the formula for probability:
P(event) = number of favorable outcomes / total number of outcomes
First, let's find the total number of outcomes. We are selecting 4 balls from 11 without replacement, so the total number of outcomes is:
11C4 = (11!)/(4!(11-4)!) = 330
where nCr is the number of combinations of n things taken r at a time.
Now let's find the number of favorable outcomes for each part of the problem.
Part 1: Exactly two white balls and two red balls
To find the number of favorable outcomes for this part, we need to select 2 white balls out of 6 and 2 red balls out of 5. The number of ways to do this is:
6C2 * 5C2 = (6!)/(2!(6-2)!) * (5!)/(2!(5-2)!) = 15 * 10 = 150
So the probability of selecting exactly two white balls and two red balls is:
P(2W2R) = 150/330 = 0.45 (rounded to two decimal places)
Part 2: At least two white balls
To find the number of favorable outcomes for this part, we need to consider two cases: selecting 2 white balls and 2 red balls, or selecting 3 white balls and 1 red ball.
The number of ways to select 2 white balls and 2 red balls is the same as the number of favorable outcomes for Part 1, which is 150.
To find the number of ways to select 3 white balls and 1 red ball, we need to select 3 white balls out of 6 and 1 red ball out of 5. The number of ways to do this is:
6C3 * 5C1 = (6!)/(3!(6-3)!) * (5!)/(1!(5-1)!) = 20 * 5 = 100
So the total number of favorable outcomes for selecting at least two white balls is:
150 + 100 = 250
And the probability of selecting at least two white balls is:
P(at least 2W) = 250/330 = 0.76 (rounded to two decimal places)