Question

At a telephone enquiry system, the number of phone calls regarding relevant enquiry follows a Poisson distribution with an average of $5$ phone calls during $10$-minute time intervals. The probability that there is at the most one phone call during a $10$-minute time period is-
A)$\dfrac{6}{5^e}$ B) $\dfrac{5}{6}$ C) $\dfrac{6}{{55}}$ D) $\dfrac{6}{{{{\text{e}}^5}}}$

Answer 1

Hint:We can solve the given question using Poisson distribution formula \[{\text{P}}\left( {{\text{X = r}}} \right) = \dfrac{{{{\text{e}}^{ - {\text{m}}}}{{\text{m}}^{\text{r}}}}}{{{\text{r!}}}}\] where ‘m’ is the expected number of occurrences or the average number of events in a given time interval and ‘X’ is the number of events observed over a given time period .Put the given values in the formula and simplify it. Here, e is the base of natural logarithm (also called Euler’s number)

Complete step-by-step answer:
Given, an average number of phone calls during $10$-minute time intervals(m)=$5$.We know that the number of phone calls at the telephone enquiry system follows the Poisson distribution. We have to find the probability that there is at most one phone call during a $10$-minute time interval. So here X≤$1$.Now, according to Poisson distribution, \[{\text{P}}\left( {{\text{X = r}}} \right) = \dfrac{{{{\text{e}}^{ - {\text{m}}}}{{\text{m}}^{\text{r}}}}}{{{\text{r!}}}}\] where, ‘m’ is the expected number of occurrences or the average number of events in a given time interval and ‘X’ is the number of events observed over a given time period. Here, e is the base of the natural logarithm (also called Euler’s number). Now, for there to be at the most one phone call to happen during the given interval, the value of X must be equal to or greater than one.
∴${\text{P}}\left( {{\text{X}} \leqslant {\text{1}}} \right) = {\text{P}}\left( {{\text{X = 0}}} \right) + {\text{P}}\left( {{\text{X}} = 1} \right)$
On putting the values in the formula we get, \[{\text{P}}\left( {{\text{X}} \leqslant {\text{1}}} \right) = \dfrac{{{{\text{e}}^{ - {\text{m}}}}{{\text{m}}^0}}}{{{\text{0!}}}} + \dfrac{{{{\text{e}}^{ - {\text{m}}}}{{\text{m}}^1}}}{{{\text{1!}}}} = {{\text{e}}^{ - 5}} + (5 \times {{\text{e}}^{ - 5}})\]
Om simplifying it, we get,
$ \Rightarrow {{\text{e}}^{ - 5}}\left( {1 + 5} \right) = \dfrac{6}{{{{\text{e}}^5}}}$

Hence the answer is ‘D’.
Note: Poisson distribution is used to predict the probability of certain events from happening when you know how often the event has occurred. The condition for Poisson distribution is-
1. An event can occur any number of times during a time period.
2. Events occur independently. Example-for the number of phone calls an office would receive, there is no reason to expect a caller to affect the chances of another person calling.
3. The rate of occurrence is constant and not based on time.
4. The probability of an event occurring is proportional to the length of the time period.