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------------------------------------------------ For a lot of 10 missiles, 4 are selected at random and fired. If the lot contains 3 defective missiles that will not fire, what is the probability that at least 1 will fire?

A. \(\frac{27}{30}\)

B. \(\frac{28}{30}\)

C. \(\frac{29}{30}\)

D. \(\frac{30}{30}\)

The probability that at least one missile will fire is 1 because the probability that none of the missiles will fire is 0. Therefore, the answer is (d) 30/30.

The complement of "at least 1 missile will fire" is "none of the missiles will fire." So we can find the probability of this happening, and then subtract it from 1 to get the probability that at least 1 missile will fire.

The probability that the first missile selected will not fire is 3/10.

Since the missile is not replaced after being fired, the probability that the second missile selected will not fire is 2/9 (since there are only 9 missiles left in the lot).

Similarly, the probability that the third missile selected will not fire is 1/8.

Finally, the probability that the fourth missile selected will not fire is 0/7 (since there is only 1 missile left in the lot).

Therefore, the probability that none of the missiles will fire is:

(3/10) * (2/9) * (1/8) * (0/7) = 0

So the probability that at least 1 missile will fire is:

1 - 0 = 1

Therefore, the answer is (d) 30/30.

To practice more questions about probability:

https://brainly.com/question/24756209

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