Question

At a checkout counter customers arrive at an average of 1.5 per minute. Find the probabilities that ( i ) at most 4 will arrive in any given minute; ( ii ) one customer will arrive in the first one minute and two customers will arrive in the next one minute.

Answer 1

Poisson distribution

$P(X)= \frac{e^{-λ}λ^x}{x!} \\ λ=1.5$

(i) P(X≤4) = P(X=0) + P(X=1) + P(X=2) + P(X=3)+ P(X=4)

$P(X=0) = \frac{e^{-1.5}1.5^0}{0!} = 0.2231 \\ P(X=1) = \frac{e^{-1.5}1.5^1}{1!} = 0.3347 \\ P(X=2) = \frac{e^{-1.5}1.5^2}{2!} = 0.2510 \\ P(X=3) = \frac{e^{-1.5}1.5^3}{3!} = 0.1255 \\ P(X=4) = \frac{e^{-1.5}1.5^4}{4!} = 0.0470 \\ P(X≤4) =0.2231+0.3347+0.2510+0.1255+0.0470 = 0.9813$

(ii) Both are independent events.

$P=P(X=1) \times P(X=2) \\ = 0.3347 \times 0.2510 = 0.0840$