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a toy manufacturer inspects boxes of toys before shipment. each box contains 9 toys. the inspection procedure consists of randomly selecting three toys from the box. if one or more of the toys are defective, the box is not shipped. suppose that a given box has two defective toys. what is the probability that it will be shipped?

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Final answer:

To find the probability that a box with two defective toys will be shipped, we can use the hypergeometric distribution. The probability is approximately 0.4167.

Step-by-step explanation:

To find the probability that a box with two defective toys will be shipped, we need to use the hypergeometric distribution.

The hypergeometric distribution is used when sampling without replacement, which is the case here as the toys are not put back in the box after being selected.

The formula for the hypergeometric distribution is:
P(X ≤ k) = (C(k, 0) * C(N - k, n)) / C(N, n)

where N is the total number of toys in the box, k is the number of defective toys in the box, n is the number of toys selected for inspection, and C(a, b) represents the number of combinations of a items taken b at a time.

In this case, N = 9 (total toys in the box), k = 2 (number of defective toys), and n = 3 (toys selected for inspection).

Using the formula, we can calculate the probability:
P(X ≤ 2) = (C(2, 0) * C(9 - 2, 3)) / C(9, 3)

Plugging in the values, we get:
P(X ≤ 2) = (1 * C(7, 3)) / C(9, 3)

Simplifying further:
P(X ≤ 2) = (1 * 35) / 84

= 0.4167

Therefore, the probability that the box will be shipped is approximately 0.4167.

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