card probability?

Question

A new card game consists of 8 red cards, 7 yellow cards, 6 green cards and 5 blue cards. If you have to draw 8 cards, what is the probability that you would:

a) draw 2 of each color?

b) draw 3 red cards, 3 blue cards and 2 yellow cards?

c) draw 4 green cards and 4 cards of another color?

d) draw no blue cards?

e) draw at least 1 blue card?

Answer 1

All parts of the question can be done by using a denominator of ₂₆C₈ , which represents the number of distinct outcomes, i.e. the sample space, because it represents the number of distinct 8-element subsets in a group of 26 (the total number of cards).

For the numerator of each fraction, we use the number of distinct ways to meet the description. For example,

a) The # of ways we can choose 2 of each color is ₈C₂·₇C₂·₆C₂·₅C₂.

similarly, ...

c) P(exactly 4 green cards) = ₆C₄·₂₀C₄ / ₂₆C₈

Lastly, for part e) , we can just do 1 - the answer for d) , since those 2 events are complements.

Answer 2

C(x,y) = combination of x items taken y at a time:

x!

-----

y! (x-y)!

a) C(8,2) * C(7,2) * C(6,2) * C(5,2) / C(26,8) = 28 * 21 * 15 * 10 / 1562275 = 0.056

b) C(8,3) * C(7,2)* C(6,0) * C(5,3) /1562275

c) C(6,4) * C(20,4)/1562275

d) C(21,8) * C(5,0)/1562275

e) 1 - probability computed in d)