Probability of 3 A's and 2 B's in 5 machine draws

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There is lucky machine with 3 results (letters):
A,B,C
P(A)=0.2 (probability to get A)
P(B)=0.3 "" "" ""
P(C)=0.5
Each round you get just one letter.
You played 5 times.
What the probability that you got three times "A" and two times "B".

i thinked to do that:
$$0.2^{3}$$$$\cdot*$$$$0.3^{2}$$$$\frac{{5}!}{{2}!\left({3}\right)!}$$
$$\frac{{5}!}{{2}!\left({3}\right)!}$$ is the number of options to get that result.
But it doesn't make any sense. because $$\frac{{5}!}{{2}!\left({3}\right)!}$$=10 and if the probability is higher then 0.1 the result will be >1 and it can't happen.
There is any good solution for this problem ? binomial distribution or something?
Thank you.
 

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Hi oriel,

Your calculation looks good to me! You found the probability of one sequence of 3 A's and 2 B's, then multiplied by the number of sequences. Indeed if $(.2^3 \cdot .3^2)$ were greater than .1, then we would have a problem, but it isn't. If you multiply your answer out, we get a valid probability. This isn't proof that the answer is correct, but it isn't outside the range of a probability calculation.
 
I was typing an answer but then I saw Jameson's post and he basically gave the answer. Your calculation is fine! Now, I want to add a comment to your final question: can we use a binomial distribution here? The answer is no. The binomial distribution can only be used in experiments where you have two possible outcomes (a failure and a success). Here you have three. However, you can use the multinomial distribution which is a generalization of the binomial distribution for $k$ different outcomes. If you're interested then I suggest you read the wikipedia page:
https://en.wikipedia.org/wiki/Multinomial_distribution

You'll notice that your calculation is exactly how the probability would be calculated if you had explicitly used the multinomial distribution.

Finally, the probability is indeed small ... do you find the result surprising?
 
Last edited:
Jameson said:
Hi oriel,

Your calculation looks good to me! You found the probability of one sequence of 3 A's and 2 B's, then multiplied by the number of sequences. Indeed if $(.2^3 \cdot .3^2)$ were greater than .1, then we would have a problem, but it isn't. If you multiply your answer out, we get a valid probability. This isn't proof that the answer is correct, but it isn't outside the range of a probability calculation.
Siron said:
I was typing an answer but then I saw Jameson's post and he basically gave the answer. Your calculation is fine! Now, I want to add a comment to your final question: can we use a binomial distribution here? The answer is no. The binomial distribution can only be used in experiments where you have two possible outcomes (a failure and a success). Here you have three. However, you can use the multinomial distribution which is a generalization of the binomial distribution for $k$ different outcomes. If you're interested then I suggest you read the wikipedia page:
https://en.wikipedia.org/wiki/Multinomial_distribution

You'll notice that your calculation is exactly how the probability would be calculated if you had explicitly used the multinomial distribution.

Finally, the probability is indeed small ... do you find the result surprising?
Thank you very much for that clear explanation. I appreciate that.