Six balls are selected at random without replacement from an urn containing three white balls and five blue balls. find the probability of the given event. (round your answer to three decimal places.) two or three of the balls are white.

25/28 is the probability of picking two or three white balls. Given that there are only 8 balls in total and of those 8, 5 of them are blue, this problem can be turned around to "What's the probability of selecting 5 blue balls?" And then subtracting that result from 1. There are a total of 6 ways of selecting all 5 blue balls. The only difference is when the single white ball is selected. The 6 ways are W, WB, ..., W. So let's calculate the probability of each of those 6. 5/8 * 4/7 * 3/6 * 2/5 * 1/4 * 3/3 + 5/8 * 4/7 * 3/6 * 2/5 * 3/4 * 1/3 + 5/8 * 4/7 * 3/6 * 3/5 * 2/4 * 1/3 + 5/8 * 4/7 * 3/6 * 3/5 * 2/4 * 1/3 + 5/8 * 3/7 * 4/6 * 3/5 * 2/4 * 1/3 + 3/8 * 5/7 * 4/6 * 3/5 * 2/4 * 1/3 = ? If you look closely at each of the 6 lines, you'll realize that the numerator will always be the product of 5!3 and the denominator will be 3*4*5*6*7*8. So, let's simplify to 6*5!3/(3*4*5*6*7*8) = 6*5*4*3*2*3/(3*4*5*6*7*8) = 2*3/(7*8) = 3/(7*4) = 3/28 Now we just need to calculate 1 - 3/28 = 28/28 - 3/28 = 25/28

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