Answer 1

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,...

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)^2)/(2!) + ("e"^-4 (4)^1)/(1!) + 0.1464`   ...[From (i)]

= 0.0183 + 4 x 0.0183 + 0.1464
= 0.2379