# 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.

Thank you for your answer.

• 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
• As you are new to the site: people here tend to not respond well (or at all) to questions like this that look like ordinary homework problems and which show no effort. What have you tried? Where are you getting stuck? – lulu Jun 3 '18 at 15:08
• 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
• Perfectly correct. Do you see the answer now? – lulu Jun 3 '18 at 15:18
• 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

## 1 Answer

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 $\binom{7}{2} \binom{5}{3}0.6^20.3^30.1^2$.