Question 167214
<font size = 7 color = "red"><b>Edwin's solution:</b></font>

I have 27 balls. 14 are red and 13 blue. I decided to group the balls into a group of 3s the balls will be choosen randomly.
Then one of the groups of three was chosen at random.
What is the probablity that the group chosen will have 2 red and 1 blue.

<pre><font size = 4 color = "indigo"><b>

We can pick the two reds 14C2 ways and the one blue 13C1 ways.

So the numerator of the probability is (14C2)(13C1)  

The denominator is 27C3.

So the probability is {{{((14C2)(13C1))/(27C3) = 91/225}}}

2)-the probability that the group will have all reds or all blue

P(all reds or all blue) = P(all reds) + P(all blue)

P(all reds) = {{{(14C3)/(27C3)}}}

P(all blues) = {{{(13C3)/(27C3)}}}

P(all reds or all blue) = P(all reds) + P(all blue) =

{{{(14C3)/(27C3)+(13C3)/(27C3)}}} =  {{{(14C3+13C3)/(27C3)}}} = {{{2/9}}}.

Edwin</pre>