Three unbiased dice are thrown simultaneously. What is the probability...
When three unbiased dice are thrown, there are 63 = 216 combinations possible.
The least possible sum is 3 and the highest possible sum is 18.
The multiples of 4 in this range are 4, 8, 12 and 16
Sum = 4:
Obtained for the combination (1,1, 2).
This combination can be made in 3!/2! = 3 ways
Sum = 8:
Obtained for the combinations(1, 1,6)- possible in 3!/2! = 3 ways (1,2, 5) - possible in 3! = 6 ways (1, 3, 4) - possible in 3! = 6 ways (2, 2, 4) - possible in 3!/2! = 3 ways (2, 3, 3) - possible in 3!/2! = 3 ways
Total number of ways = 3 + 6 + 6 + 3 + 3 = 21
Sum = 12:
Obtained for the combinations
(1, 5, 6) - possible in 3! = 6 ways
(2, 4, 6) - possible in 3! = 6 ways
(2, 5, 5) - possible in 3!/2! = 3 ways (3, 3, 6) - possible in 3!/2! = 3 ways
(3, 4, 5) - possible in 3! = 6 ways
(4, 4, 4) - possible in 1 way
Total number of ways = 6 + 6 + 3 + 3 + 6 + 1 =25
Sum = 16:
Obtained for the combinations
(4, 6, 6) - possible in 3!/2! = 3 ways
(5, 5, 6) - possible in 3!/2! = 3 ways Total number of ways = 3 + 3 = 6 Overall total unmber of ways = 3 + 21 + 25 + 6 = 55
Required probability = 55/216 = 0.254 Hence, option 1.