CA Foundation · Quantitative Aptitude · Theoretical Distributions
A Poisson variable has mean 3. Using e^-3 = 0.0498, what is the probability that X is at most 1, i.e. P(X ≤ 1)?
P(X ≤ 1) is 0.1992. It is the sum of P(0) = 0.0498 and P(1) = 3 × 0.0498 = 0.1494. Using only P(1) would ignore the zero-event case and give the incorrect value 0.1494.
- A0.1494
- B0.1992Correct
- C0.0498
- D0.2490
Explanation
P(0) = e^-3 = 0.0498. P(1) = e^-3 * 3 = 0.1494. Sum = 0.0498 + 0.1494 = 0.1992. The option 0.1494 gives only P(1), missing P(0). The option 0.2490 wrongly adds P(2)-type extra term of 0.0498 times 5.
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