# Probability that Davis got 3 Souvenirs

Davis likes baseball games. Suppose there are a 100 souvenirs in one of the games and 4936 fans that attend. Calculate the probability that Davis receives 3 souvenirs.

What I did was I used a Poisson model. To find $\lambda=\dfrac{100}{4936}$ which will give me the mean of one person getting a souvenir.

Plugging this into the Poisson distribution I get $1.36 \times 10^{-6}$. The books answer is $8 \times 10^{-6}$ is there more to this problem I am not seeing? Are there other ways to determine $\lambda$?

• Did the book intend you get the prob dave got 3 or more? Commented Dec 20, 2013 at 8:31
• no it just specified only 3. I dont know if they are counting the number of ways of getting 3 from 100. But than it wouldn't be a poisson distribution would it?
Commented Dec 20, 2013 at 8:37
• I get $\binom{100}{3}(\frac1{4936})^3(1-\frac1{4936})^{100-3}=1.31841\times10^{-6}$ - so closer to, but not exactly, your answer. Commented Dec 20, 2013 at 9:13
• Interesting so for large values of $n$ the poisson and the binomial distributions are nearly the same. Perhaps the book has a mistake I'll just move on to another problem
Commented Dec 20, 2013 at 9:20

As mentioned by @Greg Martin, you can use the binomial distribution here. Treat each time they give out a souvenir as a Bernoulli trial, where a success is Davis getting a souvenir (the probability being $p = \frac{1}{4936}$) and a failure is not getting a souvenir (the probability being $1-p = 1-\frac{1}{4936}$).

The question asks what the probability of getting 3 successes, which plugging into the Bernoulli distribution is \begin{align}P\left(k=3\right)=&\dbinom{n}{k}p^k\left(1-p\right)^{n-k} \\ =&\left(\frac{100}{3}\right)\left(\frac{1}{4936}\right)^3\left(1-\frac{1}{4936}\right)^{100-3} \\ =& 1.3184059\ldots \times10^{-6} \end{align}

If you don't need the exact answer, then you can use the Poisson distribution as an approximation.

The Poisson distribution can be derived as a limiting case to the binomial distribution as the number of trials goes to infinity and the expected number of successes remains fixed — see law of rare events below. Therefore it can be used as an approximation of the binomial distribution if n is sufficiently large and p is sufficiently small. $$F_{binomial}\left(k;n,p\right) \approx F_{Poisson}\left(k; \lambda = np\right)$$

(ref)

For your case, $\lambda=\left(100\right)\left(\frac{1}{4936}\right)=\frac{100}{4936}$, which you seem to have figured out already. Plugging this into the Poisson equation gives \begin{align}P\left(k=3\right)=& \frac{\lambda^k}{k!}e^{-\lambda}\\ =& \frac{\left(\frac{100}{4936}\right)^3}{3!} e^{-\frac{100}{4936}} \\ \approx& 1.4 \times10^{-6} \end{align}

So, you were indeed correct and your book appears to be wrong.