Question

A biased die has a probability of 1/4 of showing a 5, while the probability of any of 1, 2, 3, 4, or 6 turning up is the same . If three such dice are rolled, what is the probability of getting a sum of atleast 14 without getting a 6 on any die ?

Answer 1

Solution:-
Probability of getting 5 = 1/4
Probability of getting 1, 2, 3, 4 and 6 = 1 - 1/4 = 3/4
Therefore,
Probability of getting 1 = (3/4)/5 = 3/20 
Probability of getting 2 = (3/4)/5 = 3/20
Probability of getting 3 = (3/4)/5 = 3/20
Probability of getting 4 = (3/4)/5 = 3/20
Probability of getting 6 = (3/4)/5 = 3/20
Event: 3 dice rolled and their sum should be at least 14 without getting a 6 on any die.
i.e. (5,5,5), (4,5,5), (5,4,5), (5,5,4)
Probability of getting (5,5,5) = 1/4*1/4*1/4 = 1/64
Probability of getting (4,5,5) = 3/20*1/4*1/4 = 3/320
Probability of getting (5,4,5) = 1/4*3/20*1/4 = 3/320
Probability of getting (5,5,4) = 1/4*1/4*3/20 = 3/320
Probability of getting a sum of at least 14
⇒ 1/64 + 3/320 + 3/320 + 3/320
⇒ 14/320 
= 7/160   Answer