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The number of complaints which a bank manager receives per day is a Poisson random variable with parameter m = 4. Find the probability that the manager will receive -

(a) only two complaints on any given day.

(b) at most two complaints on any given day

[Use e^{-4} =0.0183]

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#### Solution

`P(X=k)=e^(-m) xx m^k/(k!)`

(a) only two complaints on any given day.

k=2, m=4

`P(X=2)=e^(-4) xx 4^2/2`

`=0.0183 xx 8 `

`=0.1464`

Probability that the manager will recieve only two complaints on any given day = 0.1464

(b) at most two complaints on any given day

`=P(x<=2)`

`=P(x=0)+P(x=1)+P(x=2)`

`= e^(-4) + 4e^(-4) + 4^2 (e^(-4))/(2!)`

`= e^(-4) ( 1 + 4 + 8)`

`= 13 x e^(-4)`

= 13 x 0.0183

= 0.2379

Probab ility that the manager will receive · at most two complaints on any given day = 0.2379.

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