Question

The probability of a man hitting a target is \[\dfrac{2}{5}\]. He fires at the target \[k\] times (\[k\], a given number). Then the minimum \[k\], so that the probability of hitting the target at least once is more than \[\dfrac{7}{{10}}\], is

Answer 1

Hint: Here we have to use the concept of the probability to find the minimum value of \[k\]. So firstly we will find what will be the probability of hitting the target in one fire. Then we find the probability of hitting the target in two fires. Then similarly we will find it for three fires and so on up to the \[k\] times. Then we will add all these probabilities to get the probability of hitting the target at least once and by solving that we will get the value of \[k\].

Complete step-by-step answer:
It is given that the probability of a man hitting a target is \[\dfrac{2}{5}\].
So we can say that the probability of a man not hitting a target is \[ = 1 - \dfrac{2}{5} = \dfrac{3}{5}\].
So firstly we will find the probability of the hitting the target in one fire \[ = \dfrac{2}{5}\]
Now we will find the probability of the hitting the target in two fires\[ = \dfrac{3}{5} \times \dfrac{2}{5}\], so we can say that it missed in first fire and then hit the target at second fire.
Now we will find the probability of the hitting the target in three fires\[ = \dfrac{3}{5} \times \dfrac{3}{5} \times \dfrac{2}{5}\], so we can say that it missed in first and second fire and then hit the target at third fire. So, similarly it goes on up to the \[k\] times.
We have to find the probability of hitting the target at least once which will be equal to the addition of the probabilities of hitting the target with equal to or more than one time. Therefore, we get
Probability of hitting the target at least once \[ = \dfrac{2}{5} + \dfrac{3}{5} \times \dfrac{2}{5} + \dfrac{3}{5} \times \dfrac{3}{5} \times \dfrac{2}{5} + ..... + k{\rm{ \,times}}\]
Now we have to simplify the equation to get the answer. We can take\[\dfrac{2}{5}\] common from the equation we get
Probability of hitting the target at least once\[ = \dfrac{2}{5}\left( {1 + \dfrac{3}{5} + \dfrac{3}{5} \times \dfrac{3}{5} + ..... + k{\rm{\, times}}} \right)\]
We can clearly see that the term inside the bracket follows the G.P. and we know that the sum of n number of terms is equal to \[\dfrac{{1 - {r^n}}}{{1 - r}}\]where, \[r\]is the common ratio and \[n\]is the number of terms. Therefore, we get
Probability of hitting the target at least once \[ = \dfrac{2}{5}\left( {\dfrac{{1 - {{\left( {\dfrac{3}{5}} \right)}^k}}}{{1 - \dfrac{3}{5}}}} \right)\]
So by simplifying the equation we get
Probability of hitting the target at least once \[ = \dfrac{2}{5}\left( {\dfrac{{1 - {{\left( {\dfrac{3}{5}} \right)}^k}}}{{\dfrac{2}{5}}}} \right) = 1 - {\left( {\dfrac{3}{5}} \right)^k}\]
It is given in the equation that the probability of hitting the target at least once is more than\[\dfrac{7}{{10}}\]. Therefore, we get
Probability of hitting the target at least once \[ = 1 - {\left( {\dfrac{3}{5}} \right)^k} \ge \dfrac{7}{{10}}\]
Now we have to solve this equation to get the value of\[k\].
\[ \Rightarrow 1 - {\left( {\dfrac{3}{5}} \right)^k} \ge \dfrac{7}{{10}}\]
\[ \Rightarrow {\left( {\dfrac{3}{5}} \right)^k} \ge \dfrac{3}{{10}}\]
Now by solving this equation we will get the value of\[k\]. So, we get
\[ \Rightarrow k \ge 3\]
Hence, the minimum numbers of fires at the target to get the probability of hitting the target at least once is more than\[\dfrac{7}{{10}}\] is 3.

Note: Here we have to keep in mind that the probability of occurrence is always less than or equal to one. If the probability of the event is one then that event is known as the true event. We should also note that the sum of the probability of occurrence of an event and sum of the probability of not occurrence of an event is always equal to one. We used the same concept to find our required probability. Probability is the subject which helps us a lot in the prediction of the events. We use this probability concept in our day to day life also like in weather forecasting etc.
\[P({\rm{occurrence\, of\, event}}) + P({\rm{not\, occurrence\, of\, event}}) = 1\]
We should know that the addition of n terms of G.P. series is equal to \[\dfrac{{1 - {r^n}}}{{1 - r}}\] where, \[r\] is the common ratio and \[n\] is the number of terms.