Question

A person is known to hit the target $\dfrac{3}{4}$ shots and another person is known to hit the target $\dfrac{2}{3}$. Find the probability that the target is hit when they both try.

Answer 1

In any random test or experiment, the sum of probabilities of happening and not happening an event is always equal to $1$, i.e., $P\left( E \right) + P\left( {{E'}} \right) = 1$. We will use this formula to get the desired answer.

Complete step-by-step answer:
As given in question,
Probability that first person hit the target is
$P\left( A \right) = \dfrac{3}{4}$
Probability that the first person can’t hit the target is
$P\left( {{A'}} \right) = 1 - P\left( A \right) = 1 - \dfrac{3}{4} = \dfrac{1}{4}$
The probability that first person hit the target is
$P\left( B \right) = \dfrac{2}{3}$
Probability that the first person can’t hit the target is
$P\left( {{B'}} \right) = 1 - P\left( B \right) = 1 - \dfrac{2}{3} = \dfrac{1}{3}$
Since, $A$ and $B$ are independent events. Therefore,
Probability that the target can’t be hit when they both try=$P\left( {{A'} \cap B'} \right)$
$= P\left( {{A'}} \right) \times P\left( {{B'}} \right)$
$= \dfrac{1}{4} \times \dfrac{1}{3}$
$=$ $\dfrac{1}{{12}}$
Probability that the target will be hit=$1 - P\left( {{A'}} \right) \times P\left( {{B'}} \right)$
$= 1 - \dfrac{1}{{12}}$
$=$ $\dfrac{{11}}{{12}}$

Hence, the probability that the target is hit when they both try will be $\dfrac{{11}}{{12}}$.

Note: If the two events $A$ and $B$ are independent, then we can write it as $P\left( {A \cap B} \right) = P\left( A \right) \times P\left( B \right)$ or $P\left( {A' \cap B'} \right) = P\left( {A'} \right) \times P\left( {B'} \right)$.