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The probability of a man hitting a target is 1/4. How many times must he fire so that the probability of his hitting the target at least once is greater than 2/3 ?

User Holdin
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Final answer:

The man must fire at least 2 times to ensure that the probability of hitting the target at least once is greater than 2/3. To find the number of times the man must fire so that the probability of hitting the target at least once is greater than 2/3, we can use the concept of complementary probability.

Step-by-step explanation:

To find the number of times the man must fire so that the probability of hitting the target at least once is greater than 2/3, we can use the concept of complementary probability. The complementary probability of not hitting the target at least once is equal to 1 minus the probability of hitting the target at least once. Let's represent the number of shots fired as n.

The probability of not hitting the target on a single shot is 1 - 1/4 = 3/4.

The probability of not hitting the target on any of the n shots is (3/4)^n.

The probability of hitting the target at least once is 1 - (3/4)^n.

We need to find the value of n such that the probability of hitting the target at least once is greater than 2/3. So, we have:

1 - (3/4)^n > 2/3

Simplifying the inequality, we get:

(3/4)^n < 1/3

Taking the logarithm of both sides (base 3/4), we have:

n > log((3/4)^n) (base 3/4)

Using logarithmic properties, we can rewrite the inequality as:

n > n * log(3/4) (base 3/4)

Dividing both sides by log(3/4) (base 3/4), we get:

n > 1

Therefore, the man must fire at least 2 times to ensure that the probability of hitting the target at least once is greater than 2/3.

User Donnald Cucharo
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