Question

The probability of hitting a target is 1/10 then find the minimum number of trials so that the probability of at least one success is greater than 1/4 is:

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

Bernoulli Trials

Under same condition if an experiment is repeated n times. Then the probability of r success is

$\displaystyle \: = \large^{n}C_r \: \times {p}^{r} \times {(1 - p)}^{n - r}$

where p = probability of success in a single trial

EVALUATION

The probability of hitting a target is

$\displaystyle \: \frac{1}{10}$

So the probability of success

$= \displaystyle \: \frac{1}{10}$

Probability of failure

$= \displaystyle \:1 - \frac{1}{10} = \frac{9}{10}$

Let the minimum number of trials required = n

So the probability of at least one success

$= 1 - P(0)$

$\displaystyle \: = 1 - \large^{n}C_0 \: { \bigg( \frac{1}{10} \bigg)}^{0} \times { \bigg( \frac{9}{10} \bigg)}^{n - 0}$

$\displaystyle \: = 1 - { \bigg( \frac{9}{10} \bigg)}^{n }$

By the given condition

$\displaystyle \: 1 - { \bigg( \frac{9}{10} \bigg)}^{n } \geqslant \frac{1}{4}$

$\implies \: \displaystyle \: { \bigg( \frac{9}{10} \bigg)}^{n } \leqslant \: 1 - \frac{1}{4}$

$\implies \: \displaystyle \: { \bigg( \frac{9}{10} \bigg)}^{n } \leqslant \: \frac{3}{4}$

$\implies \: \displaystyle \: n \: log { \bigg( \frac{9}{10} \bigg)} \leqslant \: log \bigg( \frac{3}{4} \bigg)$

$\implies \: \displaystyle \: n(log \: 9 - log \: 10) \leqslant (log3 - log4)$

$\implies \: \displaystyle \: n(0.9542 - 1) \leqslant (0.4771 - 0.6021)$

$\implies \: \displaystyle \: n \times 0.0458 \geqslant 0.125$

$\implies \: \displaystyle \: n \geqslant 2.729$

Hence the minimum number of trials = 3