CA Foundation · Quantitative Aptitude · Theoretical Distributions
A call centre in Pune receives faults reports at an average of 3 per hour, and the number of reports in an hour follows a Poisson distribution. What is the probability that no report is received in a given hour?
The probability of no report is e^(-3). In a Poisson distribution P(X = 0) equals e^(-m) because m to the power zero and zero factorial both equal one. With a mean of 3 reports per hour, the result is e^(-3).
- A3e^(-3)
- Be^(-3)Correct
- C1 - e^(-3)
- De^(3)
Explanation
For a Poisson variable, P(X = x) = e^(-m) m^x / x!. With m = 3 and x = 0, m^0 = 1 and 0! = 1, so P(X = 0) = e^(-3). The option 3e^(-3) is P(X = 1), which is a different event.
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