Question

If the probability of X to fail in the examination is 0.3 and that for Y is 0.2, then the probability that either X or Y fail in the examination is?
(a) 0.5
(b) 0.44
(c) 0.6
(d) None of these

Answer 1

Hint:Assume the probability of X to fail as P (X) and that of Y to fail as P (Y). To find the probability that either X or Y fail in the exam use the formula $P\left( X\cup Y \right)=P\left( X \right)+P\left( Y \right)-P\left( X\cap Y \right)$, where $\cup $ is the symbol of union and $\cap $ is the symbol of intersection. Use the fact that the two events are independent of each other and apply the formula $P\left( X\cap Y \right)=P\left( X \right)\times P\left( Y \right)$ to get the answer.

Complete step-by-step solution:
Here we have been provided with the probability of X of Y to fail in an examination as 0.3 and 0.5 respectively. We have been asked to find the probability that either X or Y fail in the exam.
Let us assume the probability of X to fail as P (X) and that of Y to fail as P (Y), so we have $P\left( X \right)=0.3$ and $P\left( Y \right)=0.2$. Since we have to find the probability of either X or Y to fail, that means we have to consider the union of the two events. So we have,
$\Rightarrow P\left( X\cup Y \right)=P\left( X \right)+P\left( Y \right)-P\left( X\cap Y \right)$ …… (1)
Here $P\left( X\cup Y \right)$ is the probability of either X or Y fail and $P\left( X\cap Y \right)$ is the probability of both X and Y fail simultaneously. We have $\cup $ is the symbol of union and $\cap $ is the symbol of intersection. Since both the events are independent of each other because if X fail then it will have no effect on Y, so we have the relation $P\left( X\cap Y \right)=P\left( X \right)\times P\left( Y \right)$.
$\begin{align}
  & \Rightarrow P\left( X\cap Y \right)=0.3\times 0.2 \\
 & \Rightarrow P\left( X\cap Y \right)=0.06 \\
\end{align}$
Substituting the values in relation (1) we get,
$\begin{align}
  & \Rightarrow P\left( X\cup Y \right)=0.3+0.2-0.06 \\
 & \Rightarrow P\left( X\cup Y \right)=0.5-0.06 \\
 & \therefore P\left( X\cup Y \right)=0.44 \\
\end{align}$
Hence, option (b) is the correct answer.

Note:You must remember the condition for the probability of two independent events to occur simultaneously. In case events are not independent then we have to find the common favourable outcomes for them to find the probability. You may relate the above formula of the probability with the formula that we use in sets to find the number of elements.