Question

For three events A, B and C, P (Exactly one of A or B occurs) = P (Exactly one of B or C occurs) = P (Exactly one of C or A occurs) = \[\dfrac{1}{4}\] and P (All the three events occurs simultaneously) = \[\dfrac{1}{16}\]. Then the probability that at least one of the events occurs, is.
A. $\dfrac{7}{32}$
B. $\dfrac{7}{16}$
C. $\dfrac{7}{64}$
D. $\dfrac{3}{16}$

Answer 1

Hint: We will be using the concept of probability to solve the problem. We will be using the formulae that if events are not mutually exclusive then the probability of event,
$P\left( A\cup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\cap B \right)$

Complete step-by-step solution -
Now, we have been given that the probability that exactly one of A or B occurs $P\left( A\cup B \right)-P\left( A\cap B \right)=\dfrac{1}{4}$
Now, we know that,
$\begin{align}
  & P\left( A\cup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\cap B \right) \\
 & \Rightarrow P\left( A \right)+P\left( B \right)-2P\left( A\cap B \right)=\dfrac{1}{4}........\left( 1 \right) \\
\end{align}$
Now, we have the probability that exactly one of B or C occurs $=P\left( B\cup C \right)-P\left( B\cap C \right)=\dfrac{1}{4}$.
Also, we know that,
$\begin{align}
  & P\left( B\cup C \right)=P\left( B \right)+P\left( C \right)-P\left( B\cap C \right) \\
 & \Rightarrow P\left( B \right)+P\left( C \right)-2P\left( B\cap C \right)=\dfrac{1}{4}........\left( 2 \right) \\
\end{align}$
Now, we have the probability that exactly one of C or A occurs $=P\left( C\cup A \right)-P\left( C\cap A \right)=\dfrac{1}{4}$.
Also, we know that,
$\begin{align}
  & P\left( C\cup A \right)=P\left( C \right)+P\left( A \right)-P\left( C\cap A \right) \\
 & \Rightarrow P\left( C \right)+P\left( A \right)-2P\left( C\cap A \right)=\dfrac{1}{4}........\left( 3 \right) \\
\end{align}$
Now, we will add (1), (2) and (3). So, we have,
$\begin{align}
  & 2\left( P\left( A \right)+P\left( B \right)+P\left( C \right) \right)-2\left( P\left( A\cap B \right)+P\left( B\cap C \right)+P\left( C\cap A \right) \right)=\dfrac{3}{4} \\
 & \Rightarrow P\left( A \right)+P\left( B \right)+P\left( C \right)-P\left( A\cap B \right)+P\left( B\cap C \right)+P\left( C\cap A \right)=\dfrac{3}{8}..........\left( 4 \right) \\
\end{align}$
Now, we have been given that the probability of all three events occur simultaneously is,
$P\left( A\cap B\cap C \right)=\dfrac{1}{16}............\left( 5 \right)$
Also, we know that the probability of at least one of the events A, B, C occurs is,
$P\left( A\cup B\cup C \right)=P\left( A \right)+P\left( B \right)+P\left( C \right)-P\left( A\cap B \right)-P\left( B\cap C \right)-P\left( C\cap A \right)+P\left( A\cap B\cap C \right)$
Now, we will substitute the values from (4) and (5).
$\begin{align}
  & P\left( A\cup B\cup C \right)=\dfrac{3}{8}+\dfrac{1}{16} \\
 & =\dfrac{6+1}{16} \\
 & =\dfrac{7}{16} \\
\end{align}$
So, the correct option is (C).

Note: To solve these type of questions it is important to remember that if three events A, B, C are events then the probability that at least one of the events occur is,
$P\left( A\cup B\cup C \right)=P\left( A \right)+P\left( B \right)+P\left( C \right)-P\left( A\cap B \right)-P\left( B\cap C \right)-P\left( C\cap A \right)+P\left( A\cap B\cap C \right)$