Question

Calls for a particular center occur at an average rate of 30 per hour. Find the probability that there are at least two calls in a given minute, correct to 2 decimal places.

Answer 1

SOLUTION

Given the question in the image, the following are the solution steps to answer the question.

STEP 1: Get the probability of calls within a minute

$\begin{gathered} 30\text{ calls = 1 hour} \\ 30\text{ calls = 60 minutes} \\ \text{Converting to calls per minute will be:} \\ \frac{30}{60}=0.5 \end{gathered}$

STEP 2: Write the formula for getting the probability that there are at least two calls in a given minute.

$At\text{ least 2=}P(X\ge2)=1-(P(X=0)+P(X=1))$

STEP 3: Write the formula for P(X=r)

$P(X=r)=\frac{e^{-np}\times(np)^r}{r!}$

STEP 4: Find P(X=0)

$\begin{gathered} P(X=r)=\frac{e^{-np}\times(np)^r}{r!} \\ P(X=0)=\frac{e^{-0.5}\times(0.5)^0}{0!}=e^{-0.5}=0.6065 \end{gathered}$

STEP 5: Find P(X=1)

$\begin{gathered} P(X=r)=\frac{e^{-np}\times(np)^r}{r!} \\ P(X=1)=\frac{e^{-0.5}\times(0.5)^1}{1!}=0.3033 \end{gathered}$

STEP 6: Find P(X>=2)

$\begin{gathered} P(X\ge2)=1-(0.6065+0.3033) \\ P(X\ge2)=1-0.9098=0.0902 \end{gathered}$

Hence, the probability that there are at least two calls in a given minute is 0.0902