At a telephone enquiry system the number of phone cells regarding relevant enquiry follow 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 :

Probability PYQ — JEE Main 2006 · Shift 0 · Q64 with Solution

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Correct answer: D

Let the number of phone calls in a $10$ -minute interval be $X$ .

Since the calls follow a Poisson distribution with average $5$ calls per $10$ minutes, $X \sim \text{Poisson}(\lambda=5).$
For a Poisson random variable, $P(X=k)=\frac{e^{-\lambda}\lambda^k}{k!}.$
We need the probability of at most one phone call: $P(X\le 1)=P(X=0)+P(X=1).$
Compute each term:

$P(X=0)=\frac{e^{-5}5^0}{0!}=e^{-5}$

$P(X=1)=\frac{e^{-5}5^1}{1!}=5e^{-5}$
Add them: $P(X\le 1)=e^{-5}+5e^{-5}=6e^{-5}.$
Rewrite: $6e^{-5}=\frac{6}{e^5}.$
Comparing with the options:
- A: $\dfrac{6}{5^e}$ — incorrect
- B: $\dfrac{5}{6}$ — incorrect
- C: $\dfrac{6}{55}$ — incorrect
- D: $\dfrac{6}{e^5}$ — correct

Therefore, the required probability is $\boxed{\frac{6}{e^5}}.$