We can use the multinomial probability formula to solve this problem. The multinomial probability formula is:
P(x1, x2, ..., xn) = n! / (x1! * x2! * ... * xn!) * p1^x1 * p2^x2 * ... * pn^xn
where:
n = total number of trials (in this case, 12 uniforms)
x1, x2, ..., xn are the number of trials resulting in each outcome (5, 4, and 3 uniforms)
p1, p2, ..., pn are the probabilities of each outcome (0.45, 0.35, and 0.20)
First, we need to calculate the factorials:
12! = 12 * 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = 479001600
5! = 5 * 4 * 3 * 2 * 1 = 120
4! = 4 * 3 * 2 * 1 = 24
3! = 3 * 2 * 1 = 6
Now we can plug these values into the multinomial probability formula:
P(5, 4, 3) = (479001600) / (120 * 24 * 6) * 0.45^5 * 0.35^4 * 0.20^3
P(5, 4, 3) = (479001600) / (165888) * 0.45^5 * 0.35^4 * 0.20^3
Now we can calculate the powers:
0.45^5 ≈ 0.018
0.35^4 ≈ 0.015
0.20^3 ≈ 0.008
We can now multiply these values together:
P(5, 4, 3) = (479001600) / (165888) * 0.018 * 0.015 * 0.008
P(5, 4, 3) ≈ 2881.45 * 0.018 * 0.015 * 0.008
P(5, 4, 3) ≈ 0.006
Therefore, the probability that 5 uniforms contain only 1 color, 4 uniforms contain 2 colors, and 3 uniforms contain 3 or more colors is approximately 0.006.
In a bowling league, the probability that the uniforms are all 1 color is 0,45, that 2 colors are used is 0.35, and that 3 or more colors are use is 0.20. if a sample of 12 uniforms is selected, find the probability that 5 contain only 1 color, 4 contain 2 colors, and 3 contain 3 or more colors.
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