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

Answer Verified Verified
Hint: Poisson distribution with an average of phone calls using Euler’s numbers. First we convert $k!$ in expanded form, then we put the values in the formula mentioned below:

Formula used:
$P(k$events in an interval) $ = \dfrac{{{e^{ - \lambda }} \times {\lambda ^k}}}{{k!}}$


Complete step by step solution:
Probability of an event for a Poisson distribution is given by the formula,
\[P(k\]events in an interval) $ = \dfrac{{{e^ - }^\lambda \times {\lambda ^k}}}{{k!}}$
Where \[\lambda = \]average number of events per interval, $e$ is the number with value \[2.71828\]…..[Euler’s number where k takes value \[0,{\text{ }}1,{\text{ }}2, \ldots ..\]]
\[k! = k(k - 1) \times (k - 2)....... \times 3 \times 2 \times 1,\] is the factorial of $k$.
In the problem \[\lambda = 5\] (given)
Probability of getting at most one call is,
$P(k \leqslant 1) = p(k = 0) + p(k = 1)$
$ = \dfrac{{{e^{ - \lambda }}}}{{0!}} + {e^{ - \lambda }}\dfrac{\lambda }{{1!}}$ …………… (i)
Now, we solve the above expression by putting the value of $\lambda $ that is equal to $5$. We will separately calculate the value of $p(k = 0)$ and $p(k = 1)$
For $p(k = 0)$
$p(k = 0) = \dfrac{{5^\circ \times {e^{ - 5}}}}{{0!}}$
Here ${{\text{e}}^{{\text{ - 5}}}} = \dfrac{1}{{{{\text{e}}^{\text{5}}}}}$(because when we put the value of power from numerator to denominator its sign gets changed)
$ = \dfrac{{1 \times 1}}{{{e^5}}}$ $(\therefore \,{5^0} = 1)$
$ = \dfrac{1}{{{e^5}}}$ ………… (ii)
For $p(k = 1)$
$P\left( {k = 1} \right) = \dfrac{{{5^1} \times {e^{ - 5}}}}{{1!}}$
$ \Rightarrow P\left( {k = 1} \right) = 1 \times \dfrac{5}{{{e^5}}}$ $(\therefore {5^1} = 5)$
$ \Rightarrow P\left( {k = 1} \right) = \dfrac{5}{{{e^5}}}$ …………….. (iii)
By putting equation\[\left( {ii} \right){\text{ }}and{\text{ }}\left( {iii} \right)\;\], in $P\left( {k \leqslant 1} \right) = P\left( {k{\text{ }} = {\text{ }}0} \right){\text{ }} + p\left( {k{\text{ }} = 1} \right)$ we have,
$P\left( {k \leqslant 1} \right) = P\left( {k{\text{ }} = {\text{ }}0} \right){\text{ }} + p\left( {k{\text{ }} = 1} \right)$
\[P\left( {k \leqslant 1} \right) = \dfrac{1}{{{e^5}}} + \dfrac{5}{{{e^5}}}\]
\[
   = \dfrac{{1 + 5}}{{{e^5}}} \\
   = \dfrac{6}{{{e^5}}} \\
 \]
Thus, $\dfrac{6}{{{e^5}}}$.
Hence, the correct option is C.


Note: A Poisson distribution is a tool that helps to predict the probability of certain events from happening, when you know how often the event has occurred. Poisson distribution can be applied only when there is mean/average involved.It is not applicable in all the cases