Question

(Um Poker) An urn contains 11 red balls numbered 1 through 11, 11 yellow balls numbered 1 through 11, 11 green balls numbered 1 through 11, and 11 black balls numbered 1 through 11. If 4 balls are randomly selected find the probability of getting (a) a flush (ie., all balls the same color) (b) three of a kind. (Three of a kind is 3 balls of one denomination and a fourth ball of a different denomination. e.g., 5,5,5,2) (c) two pairs. (A pair is two balls of the same denomination. e.g., 3,3. Two pairs is something like 3,3,5,5, ie., a pair of one denomination and a second pair of a different denomination.)

Answer 1

a)

total balls = 11+11+11+11 = 44      
P(selecting 4 balls) = 44C4      

total possible ways of flush = 4 * 11C4

P(flsuh) = 4 * 11C4 / 44C4= 0.009723685

b)

P(three of a kind) = 11C3*33C1*4 /44C4 = 0.160440807

c)

for 2 pair,there is 11C2 ways to choose two denomination.      
from each set(4 cards), 2 cards are drawn in 4C2*4C2 ways      

P(two pairs) = 11C2*4C2*4C2/ 44C4 = 0.014585528