JEE MAIN - Mathematics (2024 - 1st February Evening Shift - No. 8)
Explanation
We are given that the probability of Ajay not appearing in the JEE exam is $\mathrm{p}=\frac{2}{7}$, and the probability that both Ajay and Vijay will appear in the exam is $\mathrm{q}=\frac{1}{5}$.
We are asked to find the probability that Ajay will appear in the exam and Vijay will not. Let's denote this probability as $\mathrm{r}$.
To find $\mathrm{r}$, we need to use the concept of complementary events. The probability that Ajay will appear in the exam is the complement of the probability that he will not appear. So,
$$ P(\text{Ajay appears}) = 1 - P(\text{Ajay does not appear}) = 1 - \mathrm{p} = 1 - \frac{2}{7} = \frac{5}{7}. $$
The event that both Ajay and Vijay appear in the exam is independent of the event that only Ajay appears (and Vijay does not). Therefore, we can express the probability that only Ajay will appear (and Vijay will not) as the difference of Ajay appearing minus both Ajay and Vijay appearing, because the probability of both appearing ($\mathrm{q}$) is included in the probability of Ajay appearing:
$$ \mathrm{r} = P(\text{Ajay appears}) - P(\text{Both Ajay and Vijay appear}) = \frac{5}{7} - \frac{1}{5}. $$
To subtract these two fractions, we need a common denominator, which would be $35$ in this case. So,
$$ \mathrm{r} = \frac{5}{7} \cdot \frac{5}{5} - \frac{1}{5} \cdot \frac{7}{7} = \frac{25}{35} - \frac{7}{35} = \frac{25 - 7}{35} = \frac{18}{35}. $$
Therefore, the probability that Ajay will appear in the exam and Vijay will not appear is $\mathrm{r} = \frac{18}{35}$, which corresponds to Option D.
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