Basic Probability Question

jc2008

This should be easy, but I'm not too sure.

4 people choose random integers between 1 and 12 inclusive.
(a) Probability that 3 (but not 4) people choose the same integer
(b) Probability exactly two different are chosen.

What I have:
(a)
Pr(3 pick same) = (12/12) * (1/12) * (11/12) * (1/12) = 0.00637
Is this correct?

(b)
Pr(2 integers exactly) = (12/12) * (11/12) * (2/12) * (2/12) = 0.0255
I don't think this is right though - an alternative method

Pr(2 integers exactly) = (12 C 2)/(12 C 4) = 0.1333

Giving me drastically different answers, neither of which are right I think. Do you know where I'm going wrong?

Yakuza

The first one can be done as a Bernoulli trial with the probabilty of success of 1/12. What are the chances of getting 3 sucesses in 4 trials?

The second one doesn't fit that pattern, but can be done from first principles easily enough I think.
There are 12 options for the first number, and 11 options for the second number. Lastly, how many ways can you arrange 4 things when there are two pairs of equal items?

4!/(2! * 2!)

If you multiply all those together and divide by 12^4 (the total number of possibilites of 4 people picking from 12 numbers) should give you the right answer.