Answer:
Explanation:
To solve this problem, we can use the formula for probability:
P(event) = number of favorable outcomes / total number of outcomes
There are a couple of ways to approach this problem, but one common method is to use combinations. The number of ways to choose k objects from a set of n objects is given by the combination formula:
C(n, k) = n! / (k!(n-k)!)
where n! represents n factorial (the product of all positive integers up to n).
a) To find the probability of selecting exactly two white balls and two red balls, we need to count the number of ways to choose 2 white balls and 2 red balls out of 6 white balls and 5 red balls, respectively. We can then divide this number by the total number of ways to choose any 4 balls out of the 11.
Number of favorable outcomes:
C(6, 2) * C(5, 2) = 15 * 10 = 150
Total number of outcomes:
C(11, 4) = 330
Probability of selecting exactly two white balls and two red balls:
P(2 white and 2 red) = 150/330 = 5/11
Therefore, the probability of selecting exactly two white balls and two red balls is 5/11.
b) To find the probability of selecting at least two white balls, we can count the number of ways to choose 2, 3, or 4 white balls, and then add up these probabilities. Alternatively, we can subtract the probability of selecting 0 or 1 white balls from 1 (the total probability).
Number of favorable outcomes:
C(6, 2) * C(5, 2) + C(6, 3) * C(5, 1) + C(6, 4) * C(5, 0) = 150 + 120 + 15 = 285
Total number of outcomes:
C(11, 4) = 330
Probability of selecting at least two white balls:
P(at least 2 white) = 1 - P(0 white) - P(1 white)
To find P(0 white), we count the number of ways to choose 4 red balls out of 5, and divide by the total number of outcomes:
C(5, 4) / C(11, 4) = 5/33
To find P(1 white), we count the number of ways to choose 1 white ball and 3 red balls, and multiply by the number of ways to arrange these balls (4!):
C(6, 1) * C(5, 3) * 4! / C(11, 4) = 120/33
Therefore, we have:
P(at least 2 white) = 1 - 5/33 - 120/33 = 208/33
So the probability of selecting at least two white balls is 208/33, or about 0.63 (rounded to two decimal places).