BrainLab wrote:
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 ?
a) 1/216
b) 75/216
c) 80/216
d) 90/216
e) 100/216
Source: PerfectScores
Good question. +1
For the sake of completeness, you need to mention that the dice is an
unbiased die with numbers
1 to 6. You can not assume these things.
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
C is the correct answer.
I've also set up 2 correct scenarios, but had problenbs calculating them - particularly the second scenario:
2] 4 4 N N N --> 5/6*5/6*5/6 Not4*Not4*Not4, why do we have to choose 1/6 twice after 5/6 ?