Suppose that an ice cream truck has $3$ different flavors of ice cream (Vanilla Chocolate, Strawberry).
A group of three kids come along and purchase one ice cream cone each. Ice cream flavors come in $2$ scoops each. Currently, the truck has $17$ vanilla double-scoops, $14$ chocolate double-scoops, and $9$ strawberry double-scoops in stock. In addition, the ice cream stand also has combo flavors in stock, there are $10$ vanilla/chocolate combo scoops, and $6$ vanilla/strawberry combo scoops.
What is the probability that each flavor of ice cream is randomly chosen by at least one of the kids (each flavor is present in at least one scoop of ice cream among the kids)?
Here is my approach:
The total number of possibilities of choosing an ice cream selection among the three kids is
$$
{{\text{Number of ice creams in stock}}\choose{3}} = {{17+14+9+10+6}\choose{3}} = {{56}\choose{3}} = 27720
$$
The tricky part is finding the number of possible selections where each flavor is present in at least one scoop of ice cream:
Let $V$ denote the set of all vanilla ice creams available, $C$ for chocolate, and $S$ for strawberry.
$|V| = 17 + 10 + 6 = 33$
$|C| = 14 + 10 = 24$
$|S| = 9 + 6 = 15$
So if each kid gets a different flavor of ice cream, should the total number of selections be
$33*24*15 = 11880$?
Then the probability is $11880/27720 = 3/7$.
My textbook says the answer is wrong, so I'm not sure where I messed up? Is there something I forgot to do?
Thanks for any hints, much appreciated!
Best Answer
Going with the interpretation that all $\binom{56}{3}$ selections of icecream packages are equally likely (which I still don't like... what icecream salesman doesn't keep their supply organized?) we approach by way of inclusion-exclusion over the events that a particular flavor was missing. Let $X$ be the event that there was no vanilla, $Y$ that there was no chocolate, and $Z$ that there was no strawberry.
You are tasked with finding $\Pr(X^c\cap Y^c\cap Z^c)$ which can be rearranged and expanded into:
$$=1-\Pr(X\cup Y\cup Z) = 1 - \Pr(X)-\Pr(Y)-\Pr(Z)+\Pr(X\cap Y)+\Pr(X\cap Z)+\Pr(Y\cap Z)-\Pr(X\cap Y\cap Z)$$
The probability that no chocolate appeared... well, that is found simply by excluding all cones that have chocolate from our count and selecting from what remains. There are $56$ cones originally, $24$ have chocolate... so $\Pr(Y) = \dfrac{\binom{32}{3}}{\binom{56}{3}}$. $\Pr(X)$ and $\Pr(Z)$ is found similarly.
As for finding $\Pr(X\cap Y)$, here we exclude all cones which have chocolate and/or vanilla in any capacity... leaving only those which are purely strawberry. Here that would be $\Pr(X\cap Y)=\dfrac{\binom{9}{3}}{\binom{56}{3}}$. We can similarly find the others.
For the intersection of all three, well... that is impossible to have avoided all flavors given the supply the ice cream man has with him.
The final probability is then:
$$1 - \left(\dfrac{\binom{23}{3}+\binom{32}{3}+\binom{41}{3}-\binom{9}{3}-\binom{14}{3}-\binom{7}{3}}{\binom{56}{3}}\right)$$
This solution is easily generalized to a larger number of children as well (just replace $3$ at the bottom of each binomial coefficient with the number of kids), something which would have been extremely difficult to do with your original plan.