Question

(Urn Poker) An urn contains 8 red balls numbered 1 through 8, 8 yellow balls numbered 1 through 8, 8 green balls numbered 1 through 8, and 8 black balls numbered 1 through 8. If 4 balls are randomly selected, find the probability of getting:

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, i.e., a pair of one denomination and a second pair of a different denomination.)

Answer 1

8 red balls numbered 1 through 8, 8 yellow balls numbered 1 through 8, 8 green balls numbered 1 through 8, and 8 black balls numbered 1 through 8

total balls = 8*4 = 32

a)

there is 8 possible cases of one denomination

from each set(4 cards), 3 cards are drawn in 4C3 ways
for second denomination, 8C1 ways are there          
now, picking 1 cards from 7 cards , and have 4 identical each          
so, total possible outcome= 8C1*4C3 * 7C1 * 4 = 896

required probability = 896/32C4 = 0.024916574

  
b)

for 2 pair,there is 8C2 ways to choose two denomination.
from each set(4 cards), 2 cards are drawn in 4C2*4C2 ways

required probability = 8C2*4C2*4C2 / 32C4 = 0.028031146