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प्रश्न
The number of complaints which a bank manager receives per day follows a Poisson distribution with parameter m = 4. Find the probability that the manager receives a) only two complaints on a given day, b) at most two complaints on a given day. Use e−4 = 0.0183.
उत्तर
Let X denote the number of complaints which a bank manager receives per day.
Given, m = 4 and e–4 = 0.0183
∴ X ∼ P(m) = X ∼ p(4)
The p.m.f. of X is given by
P(X = x) = `("e"^-"m" "m"^x)/(x!)`
∴ P(X = x) `("e"^-4(4)^x)/(x!), x` = 0, 1, 2,...
a) P(only two complaints on a given day)
= P(X = 2)
= `("e"^-4 (4)^2)/(2!)`
= `(0.0183 xx 16)/(2)`
= 0.1464
b) P(at most two complaints)
= P(X ≤ 2)
= P(X = 0 or X = 1 or X = 2)
= P(X = 0) + P(X = 1) + P(X = 2)
= `("e"^-4 (4)^0)/(0!) + ("e"^-4 (4)^1)/(1!) + 0.1464` .....[From (i)]
= 0.0183 + 4 × 0.0183 + 0.1464
= 0.2379
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