The probability of a shooter hitting a target is 3/4 . How many minimu

1.	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?
Answer» Let the shooter fire n times. Obviously, n fires are n Bernoulli trials. In each trial, p = probability of hitting the target = 3/4 and q = probability of not hitting the target = 1/4 . Therefore P(X = x) = ⁿC_xq^{n - x}p^x, x = 0, 1, 2, ...., n = ⁿC_x(1/4)^{n - x}(3/4)^x = ⁿC_x(3^x/4ⁿ) Now, given that, P(hitting the target at least once) > 0.99 i.e. P(x ≥ 1) > 0.99 Therefore, 1 – P (x = 0) > 0.99 ⇒1 - ⁿC₀(1/4ⁿ) > 0.99 ⇒ ⁿC₀(1/4ⁿ) < 0.01 ⇒ 1/4ⁿ < 0.01 ⇒ 4ⁿ > 1/0.01 = 100 The minimum value of n to satisfy the inequality (1) is 4. Thus, the shooter must fire 4 times.

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?

Answer»

Let the shooter fire n times. Obviously, n fires are n Bernoulli trials. In each trial, p = probability of hitting the target = 3/4 and q = probability of not hitting the target = 1/4 .

Therefore P(X = x) = ⁿC_xq^{n - x}p^x, x = 0, 1, 2, ...., n

= ⁿC_x(1/4)^{n - x}(3/4)^x = ⁿC_x(3^x/4ⁿ)

Now, given that, P(hitting the target at least once) > 0.99

i.e. P(x ≥ 1) > 0.99

Therefore, 1 – P (x = 0) > 0.99

⇒1 - ⁿC₀(1/4ⁿ) > 0.99 ⇒ ⁿC₀(1/4ⁿ) < 0.01

⇒ 1/4ⁿ < 0.01 ⇒ 4ⁿ > 1/0.01 = 100

The minimum value of n to satisfy the inequality (1) is 4.

Thus, the shooter must fire 4 times.

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