In the game of Yahtzee, a "full house" occurs when a player rolls five fair dice and the result is two dice showing one number and the other three all showing another (three of a kind). What is the probability that a player will roll a full house on one roll of each of the three remaining dice if that player has already rolled a pair on the first two? Choose the most appropriate option:  A) 1/3  B) 5/54  C) 5/72  D) 5/216  E) 1/1296
We have that the player rolled XX on the first two dice. To get a "full house" the next three dice must show YYY or XYY in any combination.   The probability of YYY is 5/6*1/6*1/6 = 5/216.   The probability of XYY is 3*1/6*5/6*1/6 = 15/216 (multiply by 3 because XYY can occur in three ways: XYY, YXY, YYX).   The overall probability is 5/216 + 15/216 = 20/216 = 5/54.   Answer: 5/54.
<result>B</result>
In the game of Yahtzee, a "full house" occurs when a player rolls five fair dice and the result is two dice showing one number and the other three all showing another (three of a kind). What is the probability that a player will roll a full house on one roll of each of the three remaining dice if that player has already rolled a pair on the first two?