Question

The two events A and B have probabilities $0.25$ and $0.50$ respectively. The probability that both A and B occur simultaneously is $0.14$. Then the probability that neither A nor B occurs is
A. $0.39$
B. $0.25$
C. $0.904$
D. none of these

Answer 1

Hint: We first express the probabilities in the mathematical form. We use the theorem of complementary events as $p\left( {{A}^{c}} \right)=1-p\left( A \right)$ and $p\left( {{A}^{c}}\cap {{B}^{c}} \right)=p\left[ {{\left( A\cup B \right)}^{c}} \right]$. With the sue of inclusion theorem $p\left( A\cup B \right)=p\left( A \right)+p\left( B \right)-p\left( A\cap B \right)$, we get the value of $p\left( A\cup B \right)$. Putting the values, we get the final solution.

Complete step by step answer:
The two events A and B have probabilities $0.25$ and $0.50$ respectively. The probability that both A and B occur simultaneously is $0.14$.
In mathematical form we can write that $p\left( A \right)=0.25$, $p\left( B \right)=0.5$ and $p\left( A\cap B \right)=0.14$.
We have to find the probability that neither A nor B occurs, which is $p\left( {{A}^{c}}\cap {{B}^{c}} \right)$.
We know the theorem of complementary events which tells us $p\left( {{A}^{c}} \right)=1-p\left( A \right)$ and $p\left( {{A}^{c}}\cap {{B}^{c}} \right)=p\left[ {{\left( A\cup B \right)}^{c}} \right]$.
Therefore, $p\left( {{A}^{c}}\cap {{B}^{c}} \right)=p\left[ {{\left( A\cup B \right)}^{c}} \right]=1-p\left( A\cup B \right)$.
We also know the inclusion theorem gives $p\left( A\cup B \right)=p\left( A \right)+p\left( B \right)-p\left( A\cap B \right)$.
Putting the values, we get $p\left( A\cup B \right)=0.25+0.5-0.14=0.61$.
Again, the putting the value of $p\left( A\cup B \right)=0.61$ in $p\left( {{A}^{c}}\cap {{B}^{c}} \right)=1-p\left( A\cup B \right)$, we get
$p\left( {{A}^{c}}\cap {{B}^{c}} \right)=1-0.61=0.39$.
Therefore, the probability that neither A nor B occurs is $0.39$.

So, the correct answer is “Option A”.

Note: We need to remember that the universal set is similar for all the given probabilities like \[P\left( A \right),P\left( B \right),P\left( A\cap B \right),P\left( {{A}^{'}}\cap B \right)\]. That’s why we didn’t use the concept of number of points in a set and instead we directly used the probability form to find the solution.