Question

The probability of a man hitting a target is$1/2$. How many times must he fire so that the probability of hitting the target at least once is more than$90\%$.

Answer 1

Use Binomial Distribution Formula, ${\text{P(X = }}x{\text{)}}{{\text{ = }}^{\text{n}}}{{\text{C}}_x}{{\text{p}}^x}{{\text{q}}^{{\text{n}} - x}}$
Probability of man hitting the target, ${\text{p = }}\dfrac{1}{2}$
Probability of man not hitting the target ,${\text{q = 1 - p = 1 - }}\dfrac{1}{2} = \dfrac{1}{2}$

Complete step-by-step answer:
Using Binomial Distribution:
${\text{P(X = }}x{\text{)}}{{\text{ = }}^{\text{n}}}{{\text{C}}_x}{{\text{p}}^x}{{\text{q}}^{{\text{n}} - x}}$
In this question we have to find the value of n, that is the number of times he fire so that the probability of hitting the target at least once is more than $90\%$.
$\Rightarrow {\text{P(X}} \geqslant {\text{1) > 0}}{\text{.9}}$ (given)
$\Rightarrow 1 - {\text{P(X = 0) > 0}}{\text{.9}}$
$\Rightarrow 1{ - ^{\text{n}}}{{\text{C}}_0}{(\dfrac{1}{2})^0}{(\dfrac{1}{2})^{{\text{n - 0}}}} > 0.9$
$\Rightarrow 1 - \dfrac{1}{{{2^{\text{n}}}}} > 0.9$
$\Rightarrow 1 - 0.9 > \dfrac{1}{{{2^{\text{n}}}}}$
$\Rightarrow 0.1 > \dfrac{1}{{{2^{\text{n}}}}}$
$\Rightarrow {2^{\text{n}}} > \dfrac{1}{{0.1}}$
$\Rightarrow {2^{\text{n}}} > 10$
$\Rightarrow {\text{n = 4,5,}}......$
$\therefore$the smallest value ,${\text{n = 4}}$

Note: The binomial distribution is frequently used to model the number of successes in a sample of size n drawn with replacement from a population of size N. In these types of problems use Binomial Distribution Formula, ${\text{P(X = }}x{\text{)}}{{\text{ = }}^{\text{n}}}{{\text{C}}_x}{{\text{p}}^x}{{\text{q}}^{{\text{n}} - x}}$.