If the probability that A and B will die within a year are p and q respectively, then probability that only one of them will be alive at the end of the year is:
(a)$p+q$
(b)$p+q-2pq$
(c)$p+q-pq$
(d)$p+q+pq$

Hint: It is given that the probability of A and B will die within a year is p and q respectively so the probability that A and B will be alive within a year is $\left( 1-p \right)\And \left( 1-q \right)$ respectively. The probability of one of them will be alive at the end of the year is by adding the probability when A is alive and B is dead to the probability when B is alive and A is dead.

Complete step-by-step answer:
It is given that:
Probability of A will die within a year is equal to p.
Probability of B will die within a year is equal to q.
If the probability of A’s death is p then the probability of A’s life is $\left( 1-p \right)$and if the probability of B’s death is q then the probability of B’s life is $\left( 1-q \right)$.
We are asked to find the probability that only one of them will be alive at the end of the year.
The probability that one of them will be alive at the end of the year is equal to the combined probability of when A is alive and B is dead to when B is alive and A is dead.
The probability of when A is alive and B is dead is equal to:
$\left( 1-p \right)q$……. Eq. (1)
The probability of when B is alive and A is dead is equal to:
$p\left( 1-q \right)$………. Eq. (2)
The combined probability is the sum of eq. (1) and eq. (2).
$\begin{align}
  & \left( 1-p \right)q+p\left( 1-q \right) \\
 & =q-pq+p+pq \\
\end{align}$
In the above expression, $pq$ will be cancelled out and we get,
$p+q$
Hence, the correct option is (a).

Note: You might think why we multiply the probability when A is alive to the probability when B is dead is because the life and death of both A and B are independent of each other or are independent events.
And we know that when two events are independent to each other then we multiply both of them.
Similarly, we have multiplied the probabilities when B is alive and A is dead.