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

The probability that there are at least two calls in a given minute is approximately 0.0902.

Step-by-step explanation:

To find the probability that there are at least two calls in a given minute, we can use the exponential distribution and the Poisson distribution. The average rate of calls is given as 30 per hour, which is equivalent to 0.5 calls per minute. For a Poisson distribution with a rate of 0.5, we can find the probability of having 0 or 1 calls in a minute using the formula P(X=k) = e^(-λ) * (λ^k) / k!, where λ is the rate and k is the number of calls. We can then subtract this probability from 1 to find the probability of having at least two calls in a minute.

P(X=0) = e^(-0.5) * (0.5^0) / 0! = e^(-0.5) ≈ 0.6065

P(X=1) = e^(-0.5) * (0.5^1) / 1! = 0.5 * e^(-0.5) ≈ 0.3033

P(at least 2 calls) = 1 - P(X=0) - P(X=1) ≈ 1 - 0.6065 - 0.3033 ≈ 0.0902

Answer 2

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:} \\ (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)=(e^(-np)*(np)^r)/(r!)`

STEP 4: Find P(X=0)

`\begin{gathered} P(X=r)=(e^(-np)*(np)^r)/(r!) \\ P(X=0)=(e^(-0.5)*(0.5)^0)/(0!)=e^(-0.5)=0.6065 \end{gathered}`

STEP 5: Find P(X=1)

`\begin{gathered} P(X=r)=(e^(-np)*(np)^r)/(r!) \\ P(X=1)=(e^(-0.5)*(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