If we have a group of 150 people, broken into groups of 3, and only 3 are blue, what is the probability that all 3 blue people are put in the same group?

Is the denominator 150 C 3 = 551,300? For the numerator, my guess was 3!?

If we have a group of 150 people, broken into groups of 3, and only 3 are blue, what is the probability that all 3 blue people are put in the same group?

Is the denominator 150 C 3 = 551,300? For the numerator, my guess was 3!?

I haven't worked this out rigorously, but I think the probability is \(\displaystyle \displaystyle \frac{1}{\binom{149}{2}}\), by reason that: consider blue person #1 (label them arbitrary 1,2,3); blue person #1's two team members are equally likely to be the two other blue people as any other two people. There are \(\displaystyle \displaystyle \binom{149}{2}\) ways to choose two people out of the remaining 149...

The 150 people are divided into 50 groups of 3 people each.

There are: .\(\displaystyle \dfrac{150!}{(3!)^{50}}\) possible groupings.

To have the 3 blue people in one group, there is: .\(\displaystyle {3\choose3} = 1\) way.

The other 147 people are divided into 49 groups of 3: .\(\displaystyle \dfrac{147!}{(3!)^{49}}\) ways. . . Hence, there are: .\(\displaystyle \dfrac{147!}{(3!)^{49}}\) ways to have the 3 blue people in one group.

The probability that all 3 blue people are in one group is:

the probability is \(\displaystyle \displaystyle \frac{1}{\binom{149}{2}}\), by reason that: consider blue person #1 (label them arbitrary 1,2,3); blue person #1's two team members are equally likely to be the two other blue people as any other two people. There are \(\displaystyle \displaystyle \binom{149}{2}\) ways to choose two people out of the remaining 149...