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555-605 Level|   Probability|      
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Bunuel
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What is the probability of getting at least one six in a single throw 28 Jan 2019, 04:55
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Difficulty:

35% (medium)

Question Stats:

70% (01:23) correct 30% (01:29) wrong based on 83 sessions

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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
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What is the probability of getting at least one six in a single throw 28 Jan 2019, 06:40
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Bunuel
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

P of not getting a 6 in any of chance : 5/6 * 5/6 * 5/6 = 125/216

gettting a six = 1-125/216 = 91/216
IMO D
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What is the probability of getting at least one six in a single throw 11 Feb 2020, 04:00
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Bunuel
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 ? :? :?
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What is the probability of getting at least one six in a single throw 11 Feb 2020, 04:17
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dave13
Bunuel
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.
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What is the probability of getting at least one six in a single throw 11 Feb 2020, 06:28
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Bunuel
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.

thank you Bunuel for a nice explanation. I got it that we can rearrange in 3 ways, just one question how did you get this 3!/2! ?
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Bunuel
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What is the probability of getting at least one six in a single throw 11 Feb 2020, 06:32
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dave13
Bunuel
dave13
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.

thank you Bunuel for a nice explanation. I got it that we can rearrange in 3 ways, just one question how did you get this 3!/2! ?

It's the number of ways {6, not 6, not 6} can occur, basically the number of permutations of {6NN} (three letters out of which two are the same).
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Bunuel
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