probability of pulling five aces in 10 pulls from deck of cards In the book  bayesian statistics the fun way, there are a number or exercises to complete and I am working my way through them One is:
"what is the probability of pulling five aces in 10 pulls from a deck of cards (remember the card is shuffled back in the deck when it is pulled)?"
It says
"this is the same as B(5; 10, 1/13)"
And the answer:
"As expected, the probability of this is extremely low: about 1/32,000."
But I cannot understand how this was arrived at.
I took k = 5 and n = 10 and calculated that the coefficient is 252
I then multiplied this 252 * (1/13)^5 * (1-1/13)^10-5
But it does not come out to anything near 1/32,000
What am I missing?
 A: Comments (to show R code):
Your way:
choose(10,5)*(1/13)^5*(12/13)^5
[1] 0.0004548553

Using R's built-in binomial PDF dbinom.
dbinom(5, 10, 1/13)
[1] 0.0004548553    # exactly the same

And 1/3200:
1/3200
[1] 0.0003125

The first two (correct) answers are for exactly 5 Aces.
For 5 or more Aces, the probability would be a little larger:
sum(dbinom(5:10, 10, 1/13))
[1] 0.0004879945

The average number of Aces is $10(1/13) = 0.7692,$ so we can expect
lower numbers of Aces will be most probabable.
Here is a bar plot of the distribution $\mathsf{Binom}(10, 1/3).$
You can see that most of the probability is for three or fewer Aces.
k = 0:10;  PDF = dbinom(k, 10, 1/13)
plot(k, PDF, type="h", lwd=3, col="blue")
 abline(h=0, col="green2")


The probability of getting exactly three Aces:
dbinom(3, 10, 1/13)
[1] 0.03119008

The probability of getting exactly three Aces in ten draws, if
cards are not replaced after each draw. (Smaller, roughly because if two aces are already missing, then it's less likely to get a third ace.)
dhyper(3, 4, 48, 10)
[1] 0.01861668
choose(4,3)*choose(48,7)/choose(52,10)
[1] 0.01861668

