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So,I have this probability question: In average, 15 clients visit a store in a hour.What is the probability that : a) None of the clients buys b)12 clients buy c)less than 20 clients buy.

But I dont think I have enough data to answer this :/ I thought about using Poisson distribution...

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    $\begingroup$ At the very least, you need to know something about the relation between visiting the store, and actually buying something there. $\endgroup$
    – Gerry Myerson
    Jun 11, 2013 at 12:11

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You do need a bit more information. My instinct is to consider this as a binomial distribution, therefore assuming indepence etc... So what is needed for this approach is the probability that a client purchases something. You could let this be $p$(say) and then calculate the probabilites assuming this value. for example; Let $X$ be the number of clients who make a purchase with probability $p$. Then $X \sim Binomial(15, p)$

a) $P(x=0) = (1-p)^{15}$

b) $P(X =12) = \binom{15}{12} p^{12} (1-p)^{3}$

c) $P(X \le 20) = 1$ Since $20 > 15$.

You could use a Poisson model but then some rate parameter needs to be assumed and a similar method can be used as above using the Poisson distribution instead.

Overall more information is needed in order to explicitly give solutions.

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You should have information about the number of clients that actually buy in the store. If you assume that every client that visits the store buys, you can use Poisson distribution, as you said.

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