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!?
Any thoughts?
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!?
Any thoughts?
I haven't worked this out rigorously, but I think the probability is , 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 ways to choose two people out of the remaining 149...
I'm not sure. I found this thread from back in the day which is a similar question...
http://www.mathhelpforum.com/math-he...mutations.html
Hello, mathceleb!
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?
The 150 people are divided into 50 groups of 3 people each.
There are: . possible groupings.
To have the 3 blue people in one group, there is: . way.
The other 147 people are divided into 49 groups of 3: . ways.
. . Hence, there are: . ways to have the 3 blue people in one group.
The probability that all 3 blue people are in one group is:
. .
Hmm since Soroban's answer differs from mine I decided to do a little more research.
I believe this should be because the groups of 3 can be permuted.
Likewise I believe this should be .
Leading to:
See here (a PDF file) for some more info; under subheading "Partitioning".
Edit: Plato answered a lot quicker than I did, didn't refresh the page.