Question

The probability that a boy will not pass an M.B.A. examination is $\dfrac{3}{5}$ and that a girl will not pass is $\dfrac{4}{5}$ . Calculate the probability that at least one of them passes the examination.

Answer 1

Hint : In this question, we need to apply the concept of complement of an event, which says that if the probability of an event is say, $p$ , then the probability of the complement of the event, or we can say that the reverse or opposite of the event is $\left( {1 - p} \right)$ . So, we are going to solve this question by considering the opposite of the given event, i.e., the probability that a boy passes the examination and the girl also passes the examination.
Formula Used:
In this question, we are going to apply the concept of union and intersection of two probabilities:
$P\left( {A \cap B} \right) = P\left( A \right) \times P\left( B \right)$
and $P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right)$

Complete step-by-step answer :
We are going to attempt this question in two ways.
Let $P\left( B \right)$ represent the probability that the boy passes and $P\left( G \right)$ that the girl passes.
Given, $P'\left( B \right) = \dfrac{3}{5}$
so, $P\left( B \right) = 1 - \dfrac{3}{5} = \dfrac{2}{5}$
and $P'\left( G \right) = \dfrac{4}{5}$
so, $P\left( G \right) = \dfrac{1}{5}$
1st way
We have been given the probabilities of the boy not passing and the girl not passing.
We are going to multiply the two probabilities which will give us the probability of the event that neither of the two passes.
Then, we are going to apply the formula of complement of an event (subtract from 1) and we are going to get the probability that either of the two passes.
$\Rightarrow P'\left( {B \cap G} \right) = \dfrac{3}{5} \times \dfrac{4}{5} = \dfrac{{12}}{{25}}$
So, the probability that either passes is:
$\Rightarrow P\left( {B \cap G} \right) = 1 - \dfrac{{12}}{{25}} = \dfrac{{13}}{{25}}$
Hence, the required probability is $\dfrac{{13}}{{25}}$ .
2nd way
In this, we are going to find the sum of all possible probabilities of either of the two passing, which is:
boy passes but the girl does not
girl passes but the boy does not
and both pass
$\Rightarrow P\left( {B \cap G'} \right) = \dfrac{2}{5} \times \dfrac{4}{5} = \dfrac{8}{{25}}$
$\Rightarrow P\left( {B' \cap G} \right) = \dfrac{3}{5} \times \dfrac{1}{5} = \dfrac{3}{{25}}$
$\Rightarrow P\left( {B \cap G} \right) = \dfrac{2}{5} \times \dfrac{1}{5} = \dfrac{2}{{25}}$
Thus, total of the probabilities is $\dfrac{8}{{25}} + \dfrac{3}{{25}} + \dfrac{2}{{25}} = \dfrac{{13}}{{25}}$ which gives us the required probability.
So, the correct answer is “$\dfrac{{13}}{{25}}$”.

Note : Here we have multiplied some probabilities and added them at the end. The logic behind that is that we are covering all the cases related to the given problem. In the second case, we took all the cases pertaining to the case that at least one of them will pass – boy passes but the girl does not, girl passes but the boy does not, and both of them pass. We first added the probability of only the boy passing, then we added the probability that only the girl passes and finally we added the probability that both pass.