Question

If the probability of hitting a target by a shooter, in any shot is \[\dfrac{1}{3},\] then the minimum number of independent shots at the target required by him so that the probability of hitting the target at least once is greater than \[\dfrac{5}{6},\] is:
(a) 6
(b) 5
(c) 4
(d) 3

Answer 1

Hint: At first, we will find the probability of not hitting any shot which is \[1-\dfrac{1}{3}\text{ or }\dfrac{2}{3}\] as the total probability is 1 and we can say that the probability of hitting the target at least once is lesser than \[\left( 1-\dfrac{5}{6} \right)\text{ or }\dfrac{1}{6}\] as total probability is 1. So, we can write it as, \[{{\left( \dfrac{2}{3} \right)}^{n}}<\dfrac{1}{6}\] where n is the minimum number of trials.

Complete step-by-step answer:
In this question, we are said that if a probability of hitting a shooter, in any shot is \[\dfrac{1}{3},\] then we have to find the minimum number of shots at the target required by him so that the probability of hitting the target at least once is greater than \[\dfrac{5}{6}.\]
At first, we will define what probability is and understand the basic terms related to the probability to be used in the question. The probability of an event is a measure of the likelihood that the event would occur. If an experiment’s outcome is equally likely to occur, then the probability of an event E is the number of outcomes in the E dividend by the number of outcomes in the sample space. Here, the sample space consists of all the events that can occur possibly.
So, it can be written as, \[P\left( E \right)=\dfrac{n\left( E \right)}{n\left( S \right)}\]
Here, P(E) is the probability of an event or event which is asked, n(E) is the number of favorable events and n(S) is the number of all the events that can occur possibly.
As we know that the probability of hitting the target by a shooter is \[\dfrac{1}{3},\] so the missing target would be \[1-\dfrac{1}{3}\text{ or }\dfrac{2}{3}\] as the total probability is always 1.
Also, the probability of hitting the target at least one is greater than 5 which can also be written as the probability of hitting the target at least one is less than \[\left( 1-\dfrac{5}{6} \right)\text{ or }\dfrac{1}{6}\] as total probability is 1.
Let the number of independent shots be n, so the probability will become \[{{\left( \dfrac{2}{3} \right)}^{n}}.\]
So, according to the question, we are asked to find the value of n such that, \[{{\left( \dfrac{2}{3} \right)}^{n}}<\dfrac{1}{6}\] which can be written as, \[{{\left( 0.66 \right)}^{n}}<0.166\]
For n = 1, the value of \[{{\left( 0.66 \right)}^{n}}\] will be 0.66.
For n = 2, the value of \[{{\left( 0.66 \right)}^{n}}\] will be 0.4356.
For n = 3, the value of \[{{\left( 0.66 \right)}^{n}}\] will be 0.287496.
For n = 4, the value of \[{{\left( 0.66 \right)}^{n}}\] will be 0.18974736.
For n = 5, the value of \[{{\left( 0.66 \right)}^{n}}\] will be 0.1252332576.
So, by observing, we see that only for n = 5, the equation \[{{\left( 0.66 \right)}^{n}}<0.165\] satisfies.
Hence, the correct option is (b).

Note: For this, a formula can be used that if the probability of success of the shot is p and the probability of hitting the target at least greater than one be k, then, \[1-{{\text{ }}^{n}}{{C}_{0}}{{\left( P \right)}^{0}}{{\left( 1-P \right)}^{n}}>k\] and n be the number of events.