The question is as follows:
"Each meal at a fast food restaurant comes with a prize. There are three types of prizes, $A$, $B$, and $C$. Each meal comes with prize $A$ with probability $0.5$, prize $B$ with probability $0.4$, and prize $C$ with probability $0.1$, independently of previous meals. You buy four meals. What’s the probability that you get a prize of each type?"
This is a very basic question, but I can't figure it out. My attempt was to do $4$ choose $3$ multiplied by each of the probabilites, $0.1, 0.4, 0.5$, but $0.8$ is incorrect. Not sure how to approach this problem. Intution behind solution would be appreciated
You need to account for the fact that, if you get all three prizes, one of them comes twice. So you could get $AABC$, $CBCA$, etc. The number of ways to choose a permutation of two $A$'s, one $B$, and a $C$ is $\binom{4}{2} \times 2 \times 1 = 12$ (the number of ways to count the other similar permutations is the same). The probability of any such permutation occurring is $(0.5)^2(0.4)(0.1) = 0.01$. Similarly, the probability of a permutation where $B$ is repeated is $(0.5)(0.4)^2(0.1) = 0.008$ and the probability of a permutation where $C$ is repeated is $(0.5)(0.4)(0.1)^2 = 0.002$. Adding over all such permutations gives a $0.24$ probability of getting all three prizes.