CA Foundation · Quantitative Aptitude · Theoretical Distributions
Defective bulbs in a box from a Nashik supplier are Poisson distributed with mean 1.5 bulbs per box. What is the probability expression for a box having at most one defective bulb, if e^(-1.5) = 0.2231?
The probability is 0.5578. At most one defective means zero or one, so add P(0) = 0.2231 and P(1) = 0.2231 × 1.5 = 0.3347, giving 0.5578, which equals e^(-1.5) multiplied by 2.5.
- A0.2231
- B0.3347
- C0.5578Correct
- D0.4463
Explanation
P(X≤1) = P(0)+P(1) = e^(-1.5)(1 + 1.5) = 0.2231 × 2.5 = 0.5578. The value 0.2231 counts only P(0), and 0.3347 counts only P(1), so both miss one term.
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