Question

A businessman is expecting two phone calls. Mr. Walia may call at any time between 2 p.m and 4 p.m. While Mr. Sharma is equally likely to call at any time between 2:30 p.m and 3:15 p.m. The probability that Mr. Walia calls before Mr. Sharma is:
(a) $\dfrac{1}{18}$
(b) $\dfrac{1}{9}$
(c) $\dfrac{1}{6}$
(d) None of these

Answer 1

For solving this problem we assume that there are three possibilities where the two persons can call. You divide the time accordingly as three parts and we need to find the probability only in the common time of both callers.
First – 2:00 pm to 2:30 pm
Second – 2:30 pm to 3:15 pm
Third – 3:15 pm to 4:00 pm
By finding the probability of Mr Walia calling at the same time Mr. Sharma called we get the conditional probability that Mr. Walia calls, assuming the happening of the order that Mr. Walia calls before Mr. Sharma has occurred then we use the conditional probability formula as $P\left( \dfrac{A}{B} \right)=\dfrac{P\left( A\cap B \right)}{P\left( B \right)}$ to get $P\left( A\cap B \right)$ which is required answer.

Complete step-by-step solution
Now let us assume that ${{E}_{1}}$ be the event that Mr. Walia and Mr. Sharma calls at the same time,
Since there are three total chances of calling and only one chance of calling at the same time we write
$P\left( {{E}_{1}} \right)=\dfrac{1}{3}$
Now let us assume that ${{E}_{2}}$ be the event that Mr. Walia calls before Mr. Sharma.
Here since the calling is equally likely, we can say that ${{E}_{1}}$ and ${{E}_{2}}$ are dependent events.
Let us find the probability of events ${{E}_{1}}$ and ${{E}_{2}}$ assuming that ${{E}_{1}}$ already occurs that is
$P\left( \dfrac{{{E}_{2}}}{{{E}_{1}}} \right)=\dfrac{1}{2}$
This is because we assumed that ${{E}_{1}}$ already occurred and then the happening of ${{E}_{2}}$ will have only one chance. Either happens or not happens. So, there is one chance out of a total of two chances. So, the probability is $P\left( \dfrac{{{E}_{2}}}{{{E}_{1}}} \right)=\dfrac{1}{2}$.
Now we need to find $P\left( {{E}_{1}}\cap {{E}_{2}} \right)$ to get an answer.
We know that the formulae of conditional probability are given as
$P\left( \dfrac{{{E}_{2}}}{{{E}_{1}}} \right)=\dfrac{P\left( {{E}_{2}}\cap {{E}_{1}} \right)}{P\left( {{E}_{1}} \right)}$

By applying the required values in above equation we will get

&\Rightarrow \dfrac{1}{2}=\dfrac{P\left( {{E}_{2}}\cap {{E}_{1}} \right)}{\left( \dfrac{1}{3} \right)} \\\ & \Rightarrow P\left( {{E}_{2}}\cap {{E}_{1}} \right)=\dfrac{1}{2}\times \dfrac{1}{3} \\\ & \Rightarrow P\left( {{E}_{2}}\cap {{E}_{1}} \right)=\dfrac{1}{6} \\\ \end{aligned}$$ **Therefore the probability of occurring both $${{E}_{1}}$$and $${{E}_{2}}$$ simultaneously is $$\dfrac{1}{6}$$. So, the answer is $$\dfrac{1}{6}$$.** **Note:** Students will do mistake in calculating the conditional probability at $$P\left( \dfrac{{{E}_{2}}}{{{E}_{1}}} \right)=\dfrac{1}{2}$$. The Idea is simple that is by assuming that $${{E}_{1}}$$ already occurs there are only two choices for $${{E}_{2}}$$ and we take one chance. Also, some students take the formulae wrong. In the formula the occurred event should be in the denominator.