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Question: In a shooting competition, the probability that Rahul will hit the target is \( \dfrac{5}{7} \) and ...

In a shooting competition, the probability that Rahul will hit the target is 57\dfrac{5}{7} and the probability that Sheela will hit the target is 34\dfrac{3}{4} . So, what is the probability that both of them will hit the target?

Explanation

Solution

Hint : We first define the events for the persons hitting the target. We find the relation between the events. We use the formula of P(AB)=P(A)P(B)P\left( A\cap B \right)=P\left( A \right)P\left( B \right) to find the final solution.

Complete step-by-step answer :
In a shooting competition, the probability that Rahul will hit the target is 57\dfrac{5}{7} and the probability that Sheela will hit the target is 34\dfrac{3}{4} .
We denote the events of them hitting them the target as event A and event B respectively.
The complement events will be missing the targets.
Therefore, P(A)=57P\left( A \right)=\dfrac{5}{7} which gives P(Ac)=157=27P\left( {{A}^{c}} \right)=1-\dfrac{5}{7}=\dfrac{2}{7} .
Similarly, P(B)=34P\left( B \right)=\dfrac{3}{4} which gives P(Bc)=134=14P\left( {{B}^{c}} \right)=1-\dfrac{3}{4}=\dfrac{1}{4} .
Now we need to find the probability that both of them will hit the target which can be denoted by P(AB)P\left( A\cap B \right) .
The probabilities of them hitting the target are independent of one another.
Therefore, P(A)P\left( A \right) and P(B)P\left( B \right) are independent events.
So, P(AB)=P(A)P(B)=57×34=1528P\left( A\cap B \right)=P\left( A \right)P\left( B \right)=\dfrac{5}{7}\times \dfrac{3}{4}=\dfrac{15}{28} .
The probability that both of them will hit the target is 1528\dfrac{15}{28} .
So, the correct answer is “ 1528\dfrac{15}{28} .”.

Note : Dependent events is when two events are dependent when the outcome of the first event influences the outcome of the second event. The probability of two dependent events is the product of the probability of X and the probability of Y after X occurs.