CA Foundation · Quantitative Aptitude · Theoretical Distributions
A bank branch in Jaipur finds that the number of cheques returned unpaid per day follows a Poisson distribution with mean 1.5. Given e^-1.5 = 0.2231, what is the probability that at least one cheque is returned on a given day?
The probability is 0.7769. The chance of at least one return is one minus the chance of none. With mean 1.5, P(0) = e^-1.5 = 0.2231, so the required probability is 1 − 0.2231 = 0.7769.
- A0.2231
- B0.3347
- C0.7769Correct
- D0.6653
Explanation
P(at least one) = 1 − P(0) = 1 − e^-1.5 = 1 − 0.2231 = 0.7769. Option A is P(0), the probability of no cheque, which is the complement. Option D wrongly computes 1 − P(1) where P(1)=0.3347 giving 0.6653.
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