The probability of a shooter hitting a target is 3/4. How many minimum number of times must he/she fire so that the probability of hitting the target at least once is more than 0.99?

Given that, the probability of a shooter hitting a target is 3/4.

Let p be the probability of hitting a target and q be the probability of not hitting a target.

Then,

But, we know p + q = 1

⇒ q= 1 – p

Let the shooter shoot n times in total.

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

Then, the probability of hitting the target r times out of total n times is given by Binomial distribution as,

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

Substitute the value of p and q in the above formula, we get

…(i)

We need to find the minimum number of times the shooter must fire so that the probability of hitting the target at least once is more than 0.99.

It can be represented as,

P(hitting the target atleast once) > 0.99

⇒ P (X ≥ 1) > 0.99

⇒ 1 – P (X < 1) > 0.99

⇒ 1 – P (X = 0) > 0.99

Put r = 0 in equation (i) and then substitute in the above equation, we have

⇒ 4^{n} > 100

We need to find the minimum value of n to satisfy this inequality.

Take n = 0.

⇒ 4^{0} > 100

⇒ 1 > 100

But 1 ≯ 100.

Take n = 1.

⇒ 4^{1} > 100

⇒ 4 > 100

But 4 ≯ 100.

Take n = 2.

⇒ 4^{2} > 100

⇒ 16 > 100

But 16 ≯ 100.

Take n = 3.

⇒ 4^{3}> 100

⇒ 64 > 100

But 64 ≯ 100.

Take n = 4.

⇒ 4^{4} > 100

⇒ 256 > 100

It is true.

Hence, the minimum value of n to satisfy the inequality is 4.

∴, the shooter must fire 4 times.

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