Daniel is playing a dice game in which a dice is rolled 5 times. If a number turns up exactly 3 times then the game is won. Daniel has thrown the dice two times and got number 4 both times. What is the probability that Daniel will win the game ?
Note: the dice is an unbiased die with numbers 1 to 6	Pick
A) 1/216  B) 75/216  C) 80/216  D) 90/216  E) 100/216
Probability of any number from 1 to 6 = 1/6.   Given that you already have 2 4s , leaving you with 3 more slots to determine if you will win or not.   2 scenarios for a win:   Scenario 1: 4 4 T N N , with T= 4 , N= not 4   Probability of this scenario = 1*1*(1/6)*(5/6)*(5*6)*3!/2!=75/216 .... (3!/2! to show the arrangement of TNN in 3 slots)   Scenario 2: 4 4 N N N , with N = any particular number but 4 and in this case for a 'win' all 'N's are the same number.   Probability of this scenario = 1*1*(5/6)*(1/6)*(1/6) = 5/216 ...(with 1/6 shows the probability of choosing that same number out of 6 possible)   Thus the final probability = 75/216 + 5/216 = 80/216   80/216 is the correct answer.
<result>C</result>
Daniel is playing a dice game in which a dice is rolled 5 times. If a number turns up exactly 3 times then the game is won. Daniel has thrown the dice two times and got number 4 both times. What is the probability that Daniel will win the game ?
Note: the dice is an unbiased die with numbers 1 to 6