Question

Let \[A\] and \[B\] are two independent events. The probability that both \[A\] and \[B\] occur together is \[\dfrac{1}{6}\], and the probability that either of them occurs is \[\dfrac{1}{3}\]. Then what is the probability of occurrence of an event \[A\]?
A. 0 or 1
B. \[\dfrac{1}{2}\] or \[\dfrac{1}{3}\]
C. \[\dfrac{1}{2}\] or \[\dfrac{1}{4}\]
D. \[\dfrac{1}{3}\] or \[\dfrac{1}{4}\]

Answer 1

Hint: First, use the product rule of the probability of the independent events and simplify the probabilities of both \[A\] and \[B\] occurring together and either of them occurs. Then use the complement rule of probability and find the simplify the probability equation of either of the event occurring. In the end, solve both equations of probability to get the required answer.

Formula Used:
Product rule: If \[A\] and \[B\] are two independent events, then \[P\left( {A \cap B} \right) = P\left( A \right)P\left( B \right)\].
Complement rule: \[P\left( A \right) + P\left( {A'} \right) = 1\]

Complete step by step solution:
Given:
\[A\] and \[B\] are two independent events
The probability that both \[A\] and \[B\] occur together: \[\dfrac{1}{6}\]
The probability that either of them occurs: \[\dfrac{1}{3}\]
Let \[x\] be the probability of the event \[A\] and \[y\] be the probability of the event \[B\].
From the given information, we get
\[P\left( {A \cap B} \right) = \dfrac{1}{6}\] and \[P\left( {A' \cap B'} \right) = \dfrac{1}{3}\]
Apply the product rule for the above probabilities.
\[P\left( {A \cap B} \right) = P\left( A \right)P\left( B \right)\]
\[ \Rightarrow \]\[\dfrac{1}{6} = xy\] \[.....\left( 1 \right)\]

\[P\left( {A' \cap B'} \right) = P\left( {A'} \right)P\left( {B'} \right)\]
Now apply the complement rule.
\[P\left( {A' \cap B'} \right) = \left( {1 - P\left( A \right)} \right)\left( {1 - P\left( B \right)} \right)\]
Substitute the values in the above equation.
\[\dfrac{1}{3} = \left( {1 - x} \right)\left( {1 - y} \right)\] \[.....\left( 2 \right)\]

Now solve the equations \[\left( 1 \right)\] and \[\left( 2 \right)\].
\[1 - x - y + xy = \dfrac{1}{3}\]
Substitute \[xy = \dfrac{1}{6}\] in the above equation.
\[1 - x - y + \dfrac{1}{6} = \dfrac{1}{3}\]
\[ \Rightarrow \]\[1 - x - y = \dfrac{1}{6}\]
\[ \Rightarrow \]\[x + y = 1 - \dfrac{1}{6}\]
\[ \Rightarrow \]\[x + y = \dfrac{5}{6}\]
Now substitute \[y = \dfrac{1}{{6x}}\] in the above equation.
\[x + \dfrac{1}{{6x}} = \dfrac{5}{6}\]
\[ \Rightarrow \]\[\dfrac{{6{x^2} + 1}}{{6x}} = \dfrac{5}{6}\]
\[ \Rightarrow \]\[\dfrac{{6{x^2} + 1}}{x} = 5\]
\[ \Rightarrow \]\[6{x^2} + 1 = 5x\]
\[ \Rightarrow \]\[6{x^2} - 5x + 1 = 0\]
Factorize the above equation.
\[6{x^2} - 3x - 2x + 1 = 0\]
\[ \Rightarrow \]\[3x\left( {2x - 1} \right) - \left( {2x - 1} \right) = 0\]
\[ \Rightarrow \]\[\left( {2x - 1} \right)\left( {3x - 1} \right) = 0\]
\[ \Rightarrow \]\[2x - 1 = 0\] or \[3x - 1 = 0\]
\[ \Rightarrow \]\[2x = 1\] or \[3x = 1\]
\[ \Rightarrow \]\[x = \dfrac{1}{2}\] or \[x = \dfrac{1}{3}\]
Thus, \[P\left( A \right) = \dfrac{1}{2}\] or \[P\left( A \right) = \dfrac{1}{3}\]

Hence the correct option is B.

Note: Students often get confused with the product rule of the probability.
If \[A\] and \[B\] are two independent events, then \[P\left( {A \cap B} \right) = P\left( A \right)P\left( B \right)\].
If \[A\] and \[B\] are two mutually exclusive events, then \[P\left( {A \cap B} \right) = 0\].