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What is the probability of getting a six on either the first or second 09 Jan 2021, 08:51
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Difficulty:

25% (medium)

Question Stats:

68% (01:16) correct 32% (01:05) wrong based on 98 sessions

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What is the probability of getting a six on either the first or second (or on both) roll of a fair six-sided die ?

(A) 1/36
(B) 1/18
(C) 1/6
(D) 11/36
(E) 1/3
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D
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What is the probability of getting a six on either the first or second 09 Jan 2021, 08:58
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Probability of getting a 6 on the first throw or the second throw or both is equivalent to saying getting at least 1 six in 2 throws = 1 - P(not getting a six in either throw)

P(getting a six) = 1/6, therefore P(not getting a six) = 5/6

Required probability = 1 - (5/6 * 5/6) = 1 - 25/36 = 11/36


Option D

Arun Kumar
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What is the probability of getting a six on either the first or second 16 Jan 2021, 18:35
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CrackVerbalGMAT
Probability of getting a 6 on the first throw or the second throw or both is equivalent to saying getting at least 1 six in 2 throws = 1 - P(not getting a six in either throw)

P(getting a six) = 1/6, therefore P(not getting a six) = 5/6

Required probability = 1 - (5/6 * 5/6) = 1 - 25/36 = 11/36


Option D

Arun Kumar
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Good evening Kumar,

Thank you for your explanation. Can you please help me understand why my answer is not right?

I use the the chance to have first “6” plus the chance to have second “6” plus the chance to have both “6”.

1/6 + 1/6 + (1/6 x 1/6) = 13/36......

Can you please help?

Thanks.

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What is the probability of getting a six on either the first or second 16 Jan 2021, 19:43
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Taikiqi
CrackVerbalGMAT
Probability of getting a 6 on the first throw or the second throw or both is equivalent to saying getting at least 1 six in 2 throws = 1 - P(not getting a six in either throw)

P(getting a six) = 1/6, therefore P(not getting a six) = 5/6

Required probability = 1 - (5/6 * 5/6) = 1 - 25/36 = 11/36


Option D

Arun Kumar


Good evening Kumar,

Thank you for your explanation. Can you please help me understand why my answer is not right?

I use the the chance to have first “6” plus the chance to have second “6” plus the chance to have both “6”.

1/6 + 1/6 + (1/6 x 1/6) = 13/36......

Can you please help?

Thanks.

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Hello Taikiqi. You are rolling a dice twice. So if you are getting a 6 on the first throw who's probability is 1/6, then you are not getting a 6 in the 2nd throw. This probability is 5/6. Therefore the combined probability for this event is 1/6 * 5/6 = 5/36

Similarly, when you get a 6 in the 2nd throw, but not on the 1st, then the probability of this event is also 5/6 * 1/6 = 5/36.

When you get a 6 in both throws, as you have done it, it's 1/6 * 1/6 = 1/36


Overall probability = 5/36 + 5/36 + 1/36 = 11/36


Hope this helps.

Arun Kumar
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