# probability of a 3 outcomes event

I struggle with this simple probability exercise. In a lake, there are 3 types of fishes A, B and C. The probability of catching one fish of type A is $p_A=0.6$, the probability of catching a fish of type B is $p_B=0.3$ and the probability of catching a fish of type C is $p_C=0.1$. We catch 7 fishes, what is the probability of the having 2 fishes of type A, 3 fishes of type B and 2 fishes of type C.

• To get started: what's the probability of catching the exact sequence $AABBBCC$? How many such patterns are there? – lulu Jun 3 '18 at 14:52
• First, thank you for your answer. It is indeed the last question on my exercice sheet. If there were only two types of fishes, it would be a simple binomial law but here I wasn't sure on how to start. I saw your previous post and I think that the probability of the sequence $AABBBCC$ should be $0.6^20.3^30.1^2$. And Now I would guess that the number of such sequences is $\binom{7}{2} \binom{5}{3}$. Is that true ? – Alain Jun 3 '18 at 15:17
• I would guess that it is simply $\binom{7}{2} \binom{5}{3} 0.6^20.3^30.1^2$ – Alain Jun 3 '18 at 15:19
The probability of the sequence $AABBBCC$ is simply $0.6^20.3^30.1^2$. Now the number of such sequences is $\binom{7}{2} \binom{5}{3}$. Hence the desired probability is given by $\binom{7}{2} \binom{5}{3}0.6^20.3^30.1^2$.