Probability question?

Question

A bag holds 8 chips of 4 different colors (2 chips per color).

What is the probability that when four chips are chosen from the bag WITH REPLACEMENT, there is:

a. at least three different colors?

b. all four colors?

Please show solution steps!

Answer 1

We might as well consider a bag with only one chip of each colour; since we are using replacement, each chip has a 1/4 chance of being any particular colour.

P(all four colours used) = 1 (3/4) (2/4) (1/4) = 3/32 : the first chip can be any colour, the second any of the three unused colours, and so on.

P(three colours used): one colour must be used twice, and another not at all. There are 12 ways to specify the colours. Then there are (4C2) = 6 ways to specify which two chips will be the colour that is repeated. Once these are specified, we have (1/4) (1/4) probability of getting those two chips the right colour, and (1/2) (1/4) probability of getting the other two chips the right colour (two possibilities for the first one, only one for the second). So the overall probability for this case is 12(6)(1/16)(1/8) = 9/16.

Similarly we can work out P(two colours used) = 21/64
and P(1 colour used) = 1/64 - that's easy because we don't care what the first chip is and all the others must be the same, so we get (1/4)^3 = 1/64. You can check that these add up to 1.

This is a somewhat trickier problem when you try it without replacement.