Probability of 3 people ending up on same team?

[steve212]

There are 20 people and we are forming 2 teams of 10 people. 3 of the people (3/20) are women, 17 (17/20) are men. Names are drawn from random, in alternating order for teams. What is the probability that all 3 women end up on the same team?

Initially I thought it was 25%. 100% probability for 1st woman and 50% for each of remaining 2, so .5*.5 = .25

Then I thought it was 17 choose 10 / 20 choose 10 = 0.105
or possibility 2*0.105 (accounting for 0 women on same team as equal probability)

any help!?

 

[MarkFL]

I think your second method is correct. Suppose we call the teams $X$ and $Y$...and now we need only look at one team, so let's look at team $X$. If there are to be 3 women on the same team, which we'll call event $A$, then that will happen if team $X$ gets zero women (in which case team $Y$ has three women) OR if team $X$ gets three women. So, I would state:

$$P(A)=\frac{{17 \choose 10}{3 \choose 0}}{{20 \choose 10}}+\frac{{17 \choose 7}{3 \choose 3}}{{20 \choose 10}}=\frac{4}{19}$$

 

[steve212]

Thanks! If I change the size of the team to 1000 (keeping the ratio of W:M the same) the result is 0.245 - very close to the 0.25 I also thought - is there a rule that explains why this is for large numbers?

I think your second method is correct. Suppose we call the teams $X$ and $Y$...and now we need only look at one team, so let's look at team $X$. If there are to be 3 women on the same team, which we'll call event $A$, then that will happen if team $X$ gets zero women (in which case team $Y$ has three women) OR if team $X$ gets three women. So, I would state:

$$P(A)=\frac{{17 \choose 10}{3 \choose 0}}{{20 \choose 10}}+\frac{{17 \choose 7}{3 \choose 3}}{{20 \choose 10}}=\frac{4}{19}$$

 

[MarkFL]

Thanks! If I change the size of the team to 1000 (keeping the ratio of W:M the same) the result is 0.245 - very close to the 0.25 I also thought - is there a rule that explains why this is for large numbers?

If we have 2000 people, 300 of which are women, then the probability that all 300 women will be on the same team is given by:

$$P(A)=\frac{{1700 \choose 1000}{300 \choose 0}}{{2000 \choose 1000}}+\frac{{1700 \choose 700}{300 \choose 300}}{{2000 \choose 1000}}\approx 3.00417429967\,\times\,10^{-102}$$

 

[Greg]

I think your second method is correct.

I concur.

$$P(A)=2\cdot\frac{10\choose3}{20\choose3}=\frac{4}{19}$$

This method also gives the same result for your second example.

 

[steve212]

Sorry that wasn't exactly clear - I meant 2000 people of which 3 are women and 1997 are men, p~=0.25

If we have 2000 people, 300 of which are women, then the probability that all 300 women will be on the same team is given by:

$$P(A)=\frac{{1700 \choose 1000}{300 \choose 0}}{{2000 \choose 1000}}+\frac{{1700 \choose 700}{300 \choose 300}}{{2000 \choose 1000}}\approx 3.00417429967\,\times\,10^{-102}$$