Question

The probability of hitting the target with one shot is 0.96. What is the probability
that: a) hunter will hit the target 2 times making the 11 times shot; b) hunter will
hit the target 18 times making the 18 times shot; c) hunter will hit the target 0 time
making the 17 times shot.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

The probability of hitting the target with one shot is 0.96.

So,

The probability of not hitting the target with one shot is 0.04.

We know that,

The probability of getting r successes out of n independent trials with probability of success p and failure q using binomial distribution is

$\red{ \boxed{ \sf{ \:P(r) \: = \: ^nC_r \: {p}^{r} \: {q}^{n - r} \: }}}$

a) Hunter will hit the target 2 times making the 11 times shot.

So, here

$\rm :\longmapsto\:n = 11$

$\rm :\longmapsto\:p =0.96 = \dfrac{96}{100} = \dfrac{24}{25}$

$\rm :\longmapsto\:q =0.04 = \dfrac{4}{100} = \dfrac{1}{25}$

$\rm :\longmapsto\:r = 2$

So, probability that hunter will hit the target 2 times making the 11 times shot using Binomial distribution is

$\rm :\longmapsto\:P(2) \: = \: ^{11}C_2 {\bigg[\dfrac{24}{25} \bigg]}^{2} {\bigg[\dfrac{1}{25} \bigg]}^{11 - 2}$

$\rm \:  =  \: \dfrac{11 \times 10}{2 \times 1} \times\dfrac{24 \times 24}{ {25}^{2} } \times \dfrac{1}{ {25}^{9} }$

$\rm \:  =  \: \dfrac{55 \times 576}{ {25}^{11} }$

$\rm \:  =  \: \dfrac{11 \times 576}{ {5}^{21} }$

$\rm \:  =  \: \dfrac{6336}{ {5}^{21} }$

b) Hunter will hit the target 18 times making the 18 times shot.

Here,

$\rm :\longmapsto\:n = 18$

$\rm :\longmapsto\:p =0.96 = \dfrac{96}{100} = \dfrac{24}{25}$

$\rm :\longmapsto\:q =0.04 = \dfrac{4}{100} = \dfrac{1}{25}$

$\rm :\longmapsto\:r = 18$

So, probability that hunter will hit the target 18 times making the 18 times shot is

$\rm :\longmapsto\:P(18) \: = \: ^{18}C_{18} {\bigg[\dfrac{24}{25} \bigg]}^{18} {\bigg[\dfrac{1}{25} \bigg]}^{18 - 18}$

$\rm \:  =  \: 1 \times \bigg[\dfrac{24}{25} \bigg]^{18} {\bigg[\dfrac{1}{25} \bigg]}^{0}$

$\rm \:  =  \: \bigg[\dfrac{24}{25} \bigg]^{18}$

c) hunter will hit the target 0 time making the 17 times shot.

Here,

$\rm :\longmapsto\:n = 17$

$\rm :\longmapsto\:p =0.96 = \dfrac{96}{100} = \dfrac{24}{25}$

$\rm :\longmapsto\:q =0.04 = \dfrac{4}{100} = \dfrac{1}{25}$

$\rm :\longmapsto\:r = 0$

So, probability that hunter will hit the target 0 time making the 17 times shot using Binomial distribution is

$\rm :\longmapsto\:P(0) \: = \: ^{17}C_{0} {\bigg[\dfrac{24}{25} \bigg]}^{0} {\bigg[\dfrac{1}{25} \bigg]}^{17 - 0}$

$\rm \:  =  \: \dfrac{1}{ {25}^{17} }$

Additional Information :-

1. Mean of Binomial Distribution is np.

2. Variance of Binomial Distribution is npq.

3. Mean of Binomial Distribution > Variance of Binomial Distribution