# 100 tickets given to guests each having p probability of winning? what is the expected number of guests who win the prize?

A host gives a ticket to each of the $$100$$ guests at his party. Each ticket has the probability $$p$$ of winning some prize (independent of the others).

(a) What is the probability that exactly $$10$$ guests win a prize?

(b) What is the expected number of guests who win a prize?

(c) What is the probability that at least $$3$$ guests win prizes?

Hi so so far I have attempted (a), my assumption is that since each ticket has the probability $$p$$ and its independent. I can say $$p^{10}$$chance that exactly $$10$$ guests win a prize.

With (b) the expected value will be $$(1p) + (2p^2) + (3p^3)$$ and so on up to $$100$$. I believe there should be a formula or something where I can relate this question to so I can plug in the numbers to understand the question better.

Would anyone be able to let me know what formula this question relates to.

• Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. – José Carlos Santos May 14 at 10:26
• It usually helps to check answers on some small cases or extreme parameters. For example, $p = 1$ (everyone wins) implies probability of exactly $10$ guests winning is $0$, but your answer gives $1$. – mihaild May 14 at 10:31
• @mihaild ahh thank you. I think with part B it should be E[x] = p given that if p is 0.50 then its expected half or 50 guests will win. Is that the right way to think about it? – Grammer May 14 at 10:48
• What is $x$? If it's number of winners, then it should depend on number of guests... – mihaild May 14 at 10:59
• you are right so it should be 100p given that there is a hundred guests? – Grammer May 14 at 11:16

Let $$X_i=0$$ if the i-th guest does not win and $$1$$ if he wins. Let $$X=\sum X_i$$. Then $$X$$ is the number of guests who win. $$X_i$$'s are Bernoulli rv's and $$X$$ has Binomial distribution with parameters $$100$$ and $$p$$. Hence $$P(X=k)=\binom {100} {k} p^{k} (1-p)^{100-k}$$. Can you solve the problem using this ?

The independence together with the fact that all guest have the same chance to win a prize indicates that you are dealing here with binomial distribution having $$n=100$$ and $$p$$ as parameters.

You can speak of $$100$$ independent events that can end up in success or failure where the probability on a success (win a prize) is the same for each experiment.

If $$X$$ denotes the number of persons that win a prize then to be found are:

a) $$P(X=10)$$

b) $$\mathbb EX$$

c) $$P(X\geq3)=1-P(X=0)-P(X=1)-P(X=2)$$

I leave that to you.