What is the probability that exactly 3 one hundred dollar bills are chosen? STATEMENT At a casino, one lucky player is blindfolded and allowed to pick $5$ bills from a bowl containing $25$ bills. The bowl contains $6$ one hundred dollar bills, $3$ fifty dollar bills, and the rest of the bills of various other denominations.
QUESTION:
a. What is the probability that exactly 3 one hundred dollar bills are chosen?
b. Find the variance for the number of one hundred dollar bills that can be chosen?
MY WORKING:
Let $X$ denote the number of one hundred dollar bills chosen by blindfolded player then $X$ is binomial random variable with $p$ denoting the probability of success (which is to chose one hundred dollar bill), with $p=\frac{6}{25}=0.24$. Then:
a. By using binomial distribution we have: $P(X=3)={25 \choose 3}(0.24)^3(0.76)^{25-3}=2300\times0.013824\times0.002387=0.075$
b. since $Var(X)=np(1-p)=25\times0.24\times0.76=4.56$
I am preparing for my upcoming exams by solving past papers. I don't have the key to the questions. I need surety if my calculations are correct. Guidance will be appreciated, in case there is a mistake. Thanks
 A: By using the binomial distribution, you have considered the case with replacement i.e. after you choose a bill, you put it back, so there are still 25 options.
The question seems like it wants the case without replacement, however. To pick three \$100 bills out of five, you need 3 bills from the pile of 6 \$100 bills and 2 bills from the pile of 19 not-\$100 bills. The number of ways to choose this, where the order matters, is:
$$ 6\times5\times4\times19\times18 $$
But the order doesn't matter, so we divide through by symmetries to get:
$$ \frac{6\times5\times4\times19\times18}{3\times 2\times 1\times 2\times 1}=5\times2\times19\times18 $$
And how many total ways are there to choose $5$ bills out of $25$, dividing by symmetries?
$$ \frac{25\times 24\times 23\times 22\times 21}{5\times 4\times 3\times 2\times 1} $$
So the probability is the ratio of the two:
$$ \frac{5\times2\times19\times18\times5\times 4\times 3\times 2\times 1}{25\times 24\times 23\times 22\times 21} = \frac{19\times6}{23\times11\times7}=\frac{114}{1771} $$
If you know the hypergeometric distribution, you could skip this derivation of its pmf from first principles and jump straight to: $$\frac{{6\choose 3}\times{19\choose 2}}{25\choose 5}$$
To find the variance, you could have learned a formula for the variance of a hypergeometric distribution. If not, maybe you know that $\text{Var}(X)=\mathbb{E}(X^2)-\mathbb{E}(X)^2$. The possible values of $X$ here are $\{0,1,2,3,4,5\}$ and generalising the above calculation (for $X=2$) tells you how to calculate the probability that $X$ takes each value.
