Probability distribution question

**Glitch** · February 1st 2011, 07:35 PM

Note: Sorry if this is considered a uni question, I'm not sure, so you'll have to tell me.

The question
A busy switchboard receives 150 calls an hour on average. Assume that the probability, $p_k$ , of getting k calls in a given minute is:

$p_n = e^{-\lambda}\frac{\lambda^{k}}{k!}$

where $\lambda = s$ the average number of calls per minute.

a) Find the probability of getting exactly 3 calls in a given minute.
b) Find th probability of getting at least 2 calls in a given minute.

I'm not sure how to attempt this. I'm sure I'm missing something obvious, but there's no examples in this text to guide me. Any assistance would be great.

**pickslides** · February 1st 2011, 07:46 PM

Originally Posted by Glitch

a) Find the probability of getting exactly 3 calls in a given minute.

$\displaystyle P(k=3) = e^{-150}\frac{150^{3}}{3!}$

Originally Posted by Glitch

b) Find th probability of getting at least 2 calls in a given minute.

$\displaystyle P(k\geq 2) = 1-P(k=1)-P(k=0) = 1-e^{-150}\frac{150^{1}}{1!}-e^{-150}\frac{150^{0}}{0!}$

Recall $0!=1$

**Glitch** · February 1st 2011, 10:25 PM

Your answer to a) doesn't match the textbook solution of 0.214

Even by intuition (as bad as it tends to be for probability) it looks much too small.

I'm yet to find a solution myself. :/

**pickslides** · February 2nd 2011, 11:52 AM

Ok, reading it a second time, you 150 calls per hour, which is the same as 2.5 per minute....

So

$\displaystyle P(k=3) = e^{-2.5}\frac{2.5^{3}}{3!} = 0.214$

Sorry for the confusion! Do you follow?

Same applies to part b)

**Glitch** · February 2nd 2011, 05:56 PM

Ahh, that's better.

With part b, I got this far:

$e^{-2.5}\sum\limits_{k = 2}^{\infty} \frac{(2.5)^k}{k!}$

I know this converges by the ratio test, but I can't remember how to find the value it converges to. :/

**pickslides** · February 2nd 2011, 06:18 PM

Why are you wanting to know what this converges to?

Just find your calculator!

$\displaystyle P(k\geq K) = 1-P(k<K)$

so

$\displaystyle P(k\geq 2) = 1-P(k=1)-P(k=0) = 1-e^{-2.5}\frac{2.5^{1}}{1!}-e^{-2.5}\frac{2.5^{0}}{0!}$

**Glitch** · February 2nd 2011, 06:20 PM

It's confirmed, my brain is officially fried in this heat.

Thanks. :P

**pickslides** · February 2nd 2011, 06:21 PM

Originally Posted by Glitch

my brain is officially fried in this heat.

You in Melbourne as well?

**Glitch** · February 2nd 2011, 06:25 PM

Originally Posted by pickslides

You in Melbourne as well?

Sydney. It's a cloudy day, but this house has an uncanny ability to absorb and amplify heat. :P

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