Vtyjkyuk

The ice-cream guy sells 25 different kind of ice cream. 5 of those are vegan. Mother buys 6 Ice Cream Caps for her son (he is vegan) at random. How many are vegan?

I try this.

So we have all posibilities 25 right ? The picks she gets are(N=normal, V=vegan):

1. 6N, 0V;


2. 5N, 1V;


3. 4N, 2V;


4. 3N, 3V;


5. 2N, 4V;


6. 1N, 5V;


7. 0N, 6V.

I dont understand the question of this example? Does it mean that at least one is vegan ? Than it is 6/25 ? Is this correct ?

Assuming that the mother buys six different kinds of ice cream at random, we can assume that the number $X$ of vegan cups purchased has a hypergeometric distribution. For example,
$$P(X = 0) = frac5choose 020 choose 625 choose 6 = 0.2189.$$

So the probability she buys at least one vegan cup is
$P(X ge 1) = 1 - P(X = 0) = 1 - 0.2189 = 0.7811.$

You can express the binomial coefficients in terms of factorials to compute the answer. For example,
$20 choose 6 = frac20!6! cdot 14!= 38,760.$

Alternatively, you can use the hypergeometric PDF function dhyper in R statistical software to obtain the answer (or a hypergeometric PDF from other software).

dhyper(0, 5, 20, 6)
[1] 0.2188594
1 - dhyper(0, 5, 20, 6)
[1] 0.7811406

choose(20,6)
[1] 38760

The following R code gives the six probabilities of getting 0 through 5 vegan cups out of six. [Notice that it is impossible to get 6 vegan cups because there are not 6 different ones available.]

x = 0:5; pdf = dhyper(x, 5, 20, 6); cbind(x, pdf)
 x pdf
[1,] 0 0.2188594015
[2,] 1 0.4377188029
[3,] 2 0.2735742518
[4,] 3 0.0643704122
[5,] 4 0.0053642010
[6,] 5 0.0001129305

If you might have to compute such probabilities on an ordinary calculator during an exam, maybe you should try computing a couple of these using the general formula below. (The job can be simplified somewhat by noticing that a lot of factors get
canceled.)

$$P(X = k) = frac5choose k20 choose 6-k25 choose 6,$$
for $k = 0, 1, dots, 5.$

Note: If the mother buys six cups out of many, of which $1/5$ are vegan, without making sure she buys all of different kinds, then the number $Y$ of vegan cups purchased has the distribution
$Y sim mathsfBinom(n = 6,, p=1/5),$ which has
$$P(Y = k) = 6choose kleft(frac15right)^kleft(frac45right)^6-k,$$
where $k = 0, 1, dots, 6.$ For example,
$P(Y = 0) = (4/5)^6 = 0.2621.$

Addendum prompted by @Bungo's comments:

The figure below compares the PDF's of the
hypergeometric and binomial distributions mentioned above.

R code below also makes the figure and shows mean and variances of the two distributions.

x = 0:6; pdf.h = dhyper(x, 5, 20, 6)
pdf.b = dbinom(x, 6, 1/5)
plot(x-.05, pdf.h, type="h", col="blue", lwd=2, ylab="PDF", 
 xlab ="x", main="Hypergeometric (blue) and Binomial PDFs")
lines(x+.05, pdf.b, type="h", col="red", lwd=2)
abline(h=-.001, col="green2")

mu.h=sum(x*pdf.h); mu.h; var.h=sum(x^2*pdf.h) - mu.h^2; var.h
[1] 1.2 # E(X)
[1] 0.76 # Var(X)
mu.b=sum(x*pdf.b); mu.b; var.b=sum(x^2*pdf.b) - mu.b^2; var.b
[1] 1.2 # E(Y)
[1] 0.96 # Var(Y)