Question

A can hit a target $4$ times in $5$ shots. ${\text{B 3}}$ times in $4$ shots and $C$ twice in $3$ shots. They fire a volley. What is the probability that two shots hit the target?
A.$\dfrac{{13}}{{30}}$
B.$\dfrac{{17}}{{30}}$
C.$\dfrac{{11}}{{30}}$
D.None of these

Answer 1

Hint: We have given the $A$ can hit the largest $4$ times in $5$ shots, $B$ can hit target $3$ times in $4$ shots and $C$ hit tangent $2$ times in $3$ shots. We have to find the probability that two shots hit the tangent. So firstly we calculate the probability of hitting target and not hitting target by each. There we have different combinations in which two shots hit the target. First is $A$ hit target and $C$ does not . Second is $A{\text{ and }}B$ hit the target but $C$ does not and third is $B{\text{ and }}C$ hit the target but $A$ does not. We find the probability of each combination and add them. This will give the required result.

Complete step-by-step answer:
We have given that $A$ hit the target $4$ times in $5$ shots. So probability of hitting target by Probability of not hitting the target by $A = P(A') = \dfrac{4}{5}$ probability of not hitting the target by $A = P(A'){\text{ }} = 1 - P(A)$
$ = 1 - \dfrac{4}{5}{\text{ }} = {\text{ }}\dfrac{1}{5}$
Probability of not hitting the target by \[B = P(B) = \dfrac{3}{4}\] . Probability of not hitting the target by $A = P(B'){\text{ }} = 1 - P(B)$
$ = 1 - \dfrac{3}{4}{\text{ }} = {\text{ }}\dfrac{{4 - 3}}{4}{\text{ }} = {\text{ }}\dfrac{1}{4}$
Probability of hitting target by $C = P(C) = \dfrac{2}{3}$
Probability of not hitting target by $C = P(C) = 1 - \dfrac{2}{3}$
$ = \dfrac{{3 - 2}}{3}{\text{ }} = {\text{ }}\dfrac{1}{3}$
Now Probability of hitting target by $A{\text{ and }}B$ but not $C = P(ABC')$ .
As hitting the target is an independent event so
$P(ABC') = P(A)P(B)P(C')$
$ = \dfrac{4}{5} \times \dfrac{3}{4} \times \dfrac{1}{3} = \dfrac{1}{5}$
Probability of hitting target by $A{\text{ and }}C$ but not \[B = P(ACB'){\text{ }} = {\text{ }}P(A)P(C)P(B')\]
$ = \dfrac{4}{5} \times \dfrac{2}{3} \times \dfrac{1}{4} = \dfrac{2}{{15}}$
Probability of hitting target by $B{\text{ and }}C$ but not $A$
$ = P(A'BC){\text{ }} = {\text{ }}P(A')P(B)P(C)$
\[ = \dfrac{1}{5} \times \dfrac{3}{4} \times \dfrac{2}{3} = \dfrac{1}{{10}}\]
Thus probability that two shots hit the target
$ = P(ABC'){\text{ + }}P(ACB') + P(A'BC)$
$ = \dfrac{1}{5} + \dfrac{2}{{15}} + \dfrac{1}{{10}}{\text{ }} = {\text{ }}\dfrac{{6 + 4 + 3}}{{30}}{\text{ }} = {\text{ }}\dfrac{{13}}{{30}}$
So option $A$ is correct.

Note: Probability is a branch of mathematics that deals with numerical descriptions of how likely an event occurs or not. That is it tells about the chance of occurrence of an event. The probability of the event always lies between $0{\text{ to }}1$. The formula of probability as given as
${\text{Probability = }}\dfrac{{{\text{Number favourable outcome}}}}{{{\text{Total number of outcomes}}}}$