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

Question

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.

Thank you in advance.

Answer 1

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 ?

Answer 2

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.