dave13 wrote:
Bunuel wrote:
What is the probability of getting at least one six in a single throw of three unbiased dice?
(A) 1/36
(B) 1/6
(C) 81/216
(D) 91/216
(E) 125/216
hi
Bunuel, how is your gmatclub life
can you please explain why my solution is wrong ?
At least ONE six means we can get one SIX or more than ONE SIX right ? So
1/6*5/6*5/6 = 25/216 probability of getting one SIX
1/6 * 1/6*5/6 = 5/216 probability of getting TWO SIX 1/6 *1/6*1/6 = 1/216 probability of getting SIX in each dice
All combined = 31/216
so whats wrong with it ?
The red parts are not correct.
P(one six) = 1/6*5/6*5/6*3!/2!. We multiply by 3!/2! because {6, not 6, not 6} ({6NN}) can occur in 3!/2! ways (6NN, N6N, NN6). For example: first die = 6, second die = not 6, third die = not 6 is different from first die = not 6, second die = 6, third die = not 6.
P(two sixes) = 1/6*1/6*5/6*3!/2!. We multiply by 3!/2! because {6, 6, not 6} ({66N}) can occur in 3!/2! ways (66N, 6N6, N66).
P(three sixes) = 1/6*1/6*1/6. We do not multiply this by anything because {666} can occur only in one way.
The sum of the above is 91/216.
for a nice explanation. I got it that we can rearrange in 3 ways, just one question how did you get this 3!/2! ?