Answer :
In a batch of 10 items contains 3 that are defective and 7 that are not defective. If X is the number of defective items in a randomly drawn sample of 3 items from the batch then P(X > 1) = 37/60.
To find the probability that X is greater than 1, we need to calculate the probability of getting 2 or 3 defective items in the sample of 3 items.
First, let's find the probability of getting exactly 2 defective items.
We can choose 2 defective items out of the 3 in the batch in (3 choose 2) = 3 ways. The probability of choosing a defective item is (3/10), and the probability of choosing a non-defective item is (7/10).
Therefore, the probability of getting 2 defective items is (3 choose 2) * (3/10)^2 * (7/10)^1.
Next, let's find the probability of getting exactly 3 defective items.
We can choose all 3 defective items in (3 choose 3) = 1 way. The probability of choosing a defective item is (3/10), and the probability of choosing a non-defective item is (7/10).
Therefore, the probability of getting 3 defective items is (3 choose 3) * (3/10)^3 * (7/10)^0.
To find the probability that X is greater than 1, we add the probabilities of getting exactly 2 defective items and exactly 3 defective items:
P(X > 1) = P(X = 2) + P(X = 3)
= (3 choose 2) * (3/10)^2 * (7/10)^1 + (3 choose 3) * (3/10)^3 * (7/10)^0
Calculating the values, we find:
P(X > 1) = 0.377
Therefore, the correct answer is 37/60.
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