Question

A,B and C fire one shot each at a target. The probability that A will hit the target is $\dfrac{1}{5}$ , The probability that B will hit the target is $\dfrac{1}{4}$, and the probability that C will hit the target is $\dfrac{1}{3}$. If they fire together, calculate the probability that:
(a) all three shots hit the target
(b) only C’s shot hits the target
(c) at least one shot hits the target
(d) given that only one shot hits the target, it is the shot by C.

Answer 1

Hint: In this given problem, we are given the probabilities of A,B and C hitting the target. And we are to find the given cases to find the solutions. We can find the solutions by using simple properties of probability and simplifying them to get the result.

Complete step by step solution:
According to the question, we are given A,B and C fire one shot each at a target.
We are also given, The probability that A will hit the target is $\dfrac{1}{5}$ , The probability that B will hit the target is $\dfrac{1}{4}$, and the probability that C will hit the target is $\dfrac{1}{3}$.
Now, we have to consider the case when they are firing together.
So, to start with, $P\left( A\,hits \right)=\dfrac{1}{5}\,\And \,P\left( A\,miss \right)=\dfrac{4}{5}$
And, $P\left( B\,hits \right)=\dfrac{1}{4}\,\And \,P\left( B\,miss \right)=\dfrac{3}{4}$.
And again, $P\left( C\,hits \right)=\dfrac{1}{3}\,\And \,P\left( C\,miss \right)=\dfrac{2}{3}$.

(a) Now, if they fire together, calculating the probability that, all three shots hitting the target,
$P\left( A,B,C\,hit \right)=\dfrac{1}{5}.\dfrac{1}{4}.\dfrac{1}{3}=\dfrac{1}{60}$

(b) C’s shot only hits the target:
$P\left( A\,miss,B\,miss,C\,hit \right)=\dfrac{4}{5}.\dfrac{3}{4}.\dfrac{1}{3}=\dfrac{1}{5}$

(c) at least one shot hits the target:
$P\left( at\,least\,one\,hit \right)=1-P\left( no\,hits \right)=1-\dfrac{4}{5}.\dfrac{3}{4}.\dfrac{2}{3}=1-\dfrac{2}{5}=\dfrac{3}{5}$

(d) exactly two shots hit the target:
Considering the cases,
$P\left( A\,hit,B\,hit,C\,miss \right)=\dfrac{1}{5}.\dfrac{1}{4}.\dfrac{2}{3}=\dfrac{1}{30}$
$P\left( A\,hit,B\,miss,C\,hit \right)=\dfrac{1}{5}.\dfrac{3}{4}.\dfrac{1}{3}=\dfrac{1}{20}$
$P\left( A\,miss,B\,hit,C\,hit \right)=\dfrac{4}{5}.\dfrac{1}{4}.\dfrac{1}{3}=\dfrac{1}{15}$
Therefore, $P\left( exactly\,two\,hits \right)=\dfrac{1}{30}+\dfrac{1}{20}+\dfrac{1}{15}=\dfrac{2+3+4}{60}=\dfrac{9}{60}$
After more simplification, dividing both numerator and denominator with 3, we will now get, \[\dfrac{3}{20}\] .

Note: In this solution, we are finding different solutions as a part of the solution. We need to remember that, Either an event will occur for sure, or not occur at all. Or there are possibilities to different degrees the event may occur.
An event that occurs for sure is called a certain event and its probability is 1. An event that doesn’t occur at all is called an impossible event and its probability is 0. This means that all other possibilities of an event occurrence lie between 0 and 1.