The probability of a man hitting a target is 0.25. He shoots 7 times. What is he probability of his hitting at least twice?

Given that, the probability of a man hitting a target is 0.25.

And he shoots 7 times in total.

Let p be the probability of hitting the target.

Then,

p = 0.25

Then, q be the probability of not hitting the target.

And we know that,

p + q = 1

⇒ q = 1 – p

Let X be a random variable representing the number of times the man hits the target out of n shoots.

Then, the probability of hitting the target r times out of n times is given by,

P (X = r) = ^{n}C_{r}p^{r}q^{n-r}

Here, n = 7

Putting the values of n, p, and q in the above equation, we get

…(i)

We need to find the probability of hitting the target at least twice.

This can be expressed as,

Probability = P (X ≥ 2)

Or

P (X ≥ 2) = 1 – P (X < 2)

⇒ P (X ≥ 2) = 1 – [P (X = 0) + P (X = 1)]

So, put r = 0, 1 in equation (i) to get P (X = 0) and P (X = 1) respectively and then, substitute in the above formula.

We get

Thus, the probability of hitting the target at least twice is 4547/8192.

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