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If Griffin rolls a 12-sided die (numbered 1- 12) 7 times, what is the

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If Griffin rolls a 12-sided die (numbered 1- 12) 7 times, what is the  [#permalink]

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New post 28 Aug 2018, 05:25
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If Griffin rolls a 12-sided die (numbered 1- 12) 7 times, what is the probability that the sum of the results of the 7 rolls will be even?

A. 11/32

B. 1/2

C. 5/8

D. 11/16

E. 13/16

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If Griffin rolls a 12-sided die (numbered 1- 12) 7 times, what is the  [#permalink]

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New post Updated on: 28 Aug 2018, 08:33
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Bunuel wrote:
If Griffin rolls a 12-sided die (numbered 1- 12) 7 times, what is the probability that the sum of the results of the 7 rolls will be even?

A. 11/32

B. 1/2

C. 5/8

D. 11/16

E. 13/16


I used logic and reasoning and came to Answer: B

My thought process:

We have an odd number of of rolls (7). The likelihood that the result of the dice is odd vs even is equal. 6 even options and 6 odd options (1-12)

Let's use small numbers. Imagine we had only 3 rolls (still odd number). Possible results would be:
3 even | 0 odd
2 even | 1 odd
1 even | 2 odd
0 even | 3 odd

even + even + even = even
e + e + o = odd
e + o + o = even
odd + odd + odd = odd

50% if we roll an odd amount of dice. It would be different if he rolled an even amount of dice.

IF he rolled an even amount of dice.... say 2.

e + e = e
e + o = o
o +o = e

Then we would have a \(\frac{2}{3}\) chance of the sum being an even number.
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Originally posted by MikeScarn on 28 Aug 2018, 08:30.
Last edited by MikeScarn on 28 Aug 2018, 08:33, edited 2 times in total.
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Re: If Griffin rolls a 12-sided die (numbered 1- 12) 7 times, what is the  [#permalink]

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New post 28 Aug 2018, 08:33
Bunuel wrote:
If Griffin rolls a 12-sided die (numbered 1- 12) 7 times, what is the probability that the sum of the results of the 7 rolls will be even?

A. 11/32

B. 1/2

C. 5/8

D. 11/16

E. 13/16



There are only two possibilities - even or odd..
So straightway it should be 1/2 for each
B

Otherwise, if odd number, say n, of 7 rolls give even outcome, ans will be even..
n*even + (7-n)*odd=even
So 7C1+7C3+7C5+7C7=7+35+21+1=64

Total possibilities = 7C1+7C2+7C3+7C4+7C5+7C6+7C7=2^7=128

Probability of even = 64/128=1/2

B
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Re: If Griffin rolls a 12-sided die (numbered 1- 12) 7 times, what is the &nbs [#permalink] 28 Aug 2018, 08:33
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