Mbawarrior01 wrote:
A certain dice game can only be won if, when a player throws
two fair six-sided dice, the number showing on one of the dice
is a multiple of the number showing on the other. What is the
probability that a player wins this game?
(A) 11/18
(B) 11/36
(C) 1/4
(D) 2/9
(E) 7/36
My approach is as follows :
For first dice, with 1, there will be 6 ways, (1,1),(1,2),(2,1),(1,3),(3,1),(1,4),(4,1),(1,5),(5,1),(1,6),(6,1) = 11 ways
Similarly for 2 on first dice, there will be 3 ways (2,4,6) (2,2), (2,4),(4,2), (2,6),(6,2) = 5 ways
For 3, on first dice, there will be 2 ways (3,6) (3,3),(3,6),(6,3) = 3 ways
For 4, 5 ,6 , on first dice, 1 way for each. = 3 ways
Total ways = 11 + 5 + 3 + 3 = 22 ways
Probability = 22/36 = 11/18
The answer doesn't seem to be right. Am I missing out on any possibility?
: (
Mbawarrior01My approach is as follows :
For first dice, with 1, there will be 11 ways, (1,1),(1,2),(2,1),(1,3),(3,1),(1,4),(4,1),(1,5),(5,1),(1,6),(6,1) = 11 ways
Similarly for 2 on first dice, there will be 5 ways (2,4,6) (2,2), (2,4),(4,2), (2,6),(6,2) = 5 ways
For 3, on first dice, there will be 3 ways (3,6) (3,3),(3,6),(6,3) = 3 ways
For 4, 5 ,6 , on first dice, 1 way for each. = 3 ways
Total ways = 11 + 5 + 3 + 3 = 22 ways
Probability = 22/36 = 11/18
IMO A
This is correction in your approach which works.
Hope it helps
_________________
Kinshook Chaturvedi
Email: kinshook.chaturvedi@gmail.com