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 ? Option 1 : 5/24 Option 2 : 9/160 Option 3 : 1/30 Option 4 : 7/160

Answer 1

Probability of getting 5 = 1/4

Probability of getting 1, 2, 3, 4 or 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

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

= 7/160