Probability of selecting a combination of coloured balls

Question

Good day all,

I'm no mathematical genius but I'm trying to learn.

My problem is as follows:

I have 7 blue balls, 3 red balls and 5 green balls. I wish to know the probability, if I had to randomly pick 11 balls without replacement, of selecting exactly 5 blue balls, at least 3 green balls (so 3,4 or 5) and the combination of exactly 6 red and green balls.

I have done the following in Excel for calculating the probability of selecting exactly 5/3/3 of the respective colours but it needs tweaking for the above scenario:

=combin(7,5)*combin(3,3)*combin(5,3)/combin(15,11)

Then my understanding is that you will square the answer to work out the probability of this combination occurring consecutively?

If anyone can assist it would be great!

Answer 1

You are rather close to the solution, you just need to add the cases of 4 and 5 green balls (and thus, 2 and 1 red balls, respectively), so:

$$ \LARGE \frac{ \color{blue}{\binom{7}{5}} \color{red}{\binom{3}{3}} \color{green}{\binom{5}{3}} + \color{blue}{\binom{7}{5}} \color{red}{\binom{3}{2}} \color{green}{\binom{5}{4}} + \color{blue}{\binom{7}{5}} \color{red}{\binom{3}{1}} \color{green}{\binom{5}{5}} }{\binom{15}{11} } $$

Or, equivalently, adapting you excel formulas:

=( combin(7,5)*combin(3,3)*combin(5,3) + combin(7,5)*combin(3,2)*combin(5,4) + combin(7,5)*combin(3,1)*combin(5,5) ) /combin(15,11)