Four balls are selected at random without replacement from an urn containing three white balls and five blue balls. Find the probability that two or three of the balls are white. (Round your answer to three decimal places.)
To find the probability that two or three of the balls are white, we need to determine the total number of possible outcomes and the number of favorable outcomes.
First, let's find the total number of possible outcomes. There are 8 balls in total, so the first ball can be chosen in 8 ways, the second in 7 ways (since one ball has already been chosen), the third in 6 ways, and the fourth in 5 ways. Therefore, the total number of possible outcomes is 8 * 7 * 6 * 5 = 1,680.
Now, let's find the number of favorable outcomes. We need to consider two cases:
Case 1: Two white balls and two blue balls
The number of ways to choose 2 white balls out of 3 is given by the combination formula: C(3, 2) = 3.
The number of ways to choose 2 blue balls out of 5 is given by the combination formula: C(5, 2) = 10.
Therefore, the number of favorable outcomes for this case is 3 * 10 = 30.
Case 2: Three white balls and one blue ball
The number of ways to choose 3 white balls out of 3 is given by the combination formula: C(3, 3) = 1.
The number of ways to choose 1 blue ball out of 5 is given by the combination formula: C(5, 1) = 5.
Therefore, the number of favorable outcomes for this case is 1 * 5 = 5.
Now, let's calculate the total number of favorable outcomes: 30 + 5 = 35.
Finally, we can calculate the probability by dividing the number of favorable outcomes by the total number of possible outcomes: P = 35 / 1680 = 0.0208 (rounded to three decimal places).
Therefore, the probability that two or three of the balls are white is approximately 0.021.