Poisson

**dwsmith** · May 3rd 2011, 10:44 AM

During business hours, the number of calls passing through a particular cellular relay system averages 5 per minute.

(a)
Find the probability that no call will pass through the relay system during a given minute.
$P(X=0)=p(0)=\frac{5^0}{0!}e^{-5}=.0067$

(b)
Find the probability that no call will pass through the relay system during a given 2 minute period.
$[P(X=0)]^2=[p(0)]^2=\frac{5^0}{0!}e^{-10}=.0000454$
Not so sure about this one.

(c)
Find the probability that 3 calls will pass through the relay system during a given minute.
$P(X=3)=p(3)=\frac{5^3}{3!}e^{-5}=.1404$

Correct?

**SpringFan25** · May 3rd 2011, 10:54 AM

all fine

For 2, you can also note that the number of calls in a t minute period is poisson( \lamda * t) . (you end up with exactly the same calculation)

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