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

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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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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 [#permalink]
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

Took me 5 minutes to solve. Could someone see if there is anything I can improve with my solving process?

Since there are 12 numbers half will be even, half will be odd. We are choosing 7 numbers which means we need an even amount of odd numbers and an odd number of even numbers.

1 even + 6 odd = even
3 even + 4 odd = even
5 even + 2 odd = even

From here we can find our total combinations which is $$12C7$$

Adding up the combinations from the top we get 6c1 * 6c6 + 6c3 * 6c4 + 6c5 + 6c2 = 396

396/792 = 1/2

The reason this question took so long was I made a calculation mistake when I was adding up all the combinations. Perhaps there is an easier way to do this question
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If Griffin rolls a 12-sided die (numbered 1- 12) 7 times, what is the [#permalink]
kchen1994 wrote:
The reason this question took so long was I made a calculation mistake when I was adding up all the combinations. Perhaps there is an easier way to do this question

One of the primary traps of this problem is to bait you into even doing the calculations at all. This is potentially a 30-second question. See my response above or click on the link below.

https://gmatclub.com/forum/if-griffin-r ... l#p2219488
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Re: If Griffin rolls a 12-sided die (numbered 1- 12) 7 times, what is the [#permalink]
Official Explanation:
This question rewards students who can see through the added complexity of the problem and simplify. This question is asking you not about the sum of the numbers itself, but about whether that sum is even or odd. Remember your number properties here. A sum will be even if both numbers being added are even or if both numbers being added are odd. If you are adding more than two numbers, then the sum will be even as long as there are an even number of odd numbers.

In this case, then, there are four relevant situations: if the sum is even, then Griffin rolled an odd number 0, 2, 4, or 6 times. To find the probability that the sum is even, you then need to compute sum of the possible combinations and divide that by the total number of combinations of even and odd die rolls. Note as you do this that you only need to deal with whether a die roll is even or odd (since the die is numbered 1 - 12, there is a 12
chance of either).

To find the total number of combinations of die rolls, you simply need to use basic counting strategy. Each roll is independent and can happen one of two ways. The total number of possible rolls is therefore 27=128
.

Next you need to find each individual combination. There is only 1 way that Griffin could have rolled 0 odd numbers - he had to have rolled all even numbers.

To compute the number of ways he could have rolled 2 odd numbers and 5 even numbers, you can use the permutation with repeats formula. Simply count the total number of permutations of 5 evens and 2 odds (7!
) and divide it by the number of repeats of each, factorial. This gives you

7!(2!)(5!)=(7)(6)2=21
possible combinations.

You can use the same formula to calculate the number of ways he could have rolled 4 odd numbers and 3 even numbers to get:

7!(4!)(3!)=(7)(6)(5)(3)(2)=35
total possible combinations.

Then you can use it one more time to find the number of ways he could have rolled 6 odds and 1 even to get:

7!6!=7
.

That means that there are a total of 1+21+35+7=64
possible ways that an even sum could have occurred for a total probability of 64128=12
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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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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

Notice that P(even) = 6/12 = ½ and P(odd) = 6/12 = ½.

Recall that odd + odd = even. Thus, if the 7 numbers contain 2 odds, the entire sum will be even. Similarly, if the 7 numbers contain 4 odds, the sum will be even. And if the 7 numbers contain 6 odds, the sum will be even. Thus, the sum will be even if 1) all 7 rolls are even, 2) 5 rolls are even, 3) 3 rolls are even, and 4) 1 roll is even.

Case 1:

P(all 7 rolls are even) = (½)^7 = 1/128

Case 2:

P(5 rolls are even) = P(5 rolls are even and 2 rolls are odd) = (½)^5 x (½)^2 x 7!/(5!2!) = 21/128

Case 3:

P(3 rolls are even) = P(3 rolls are even and 4 rolls are odd) = (½)^3 x (½)^4 x 7!/(3!4!) = 35/128

Case 3:

P(1 roll is even) = P(1 roll is even and 6 rolls are odd) = (½)^1 x (½)^6 x 7!/(1!6!) = 7/128

Therefore, the total probability that the sum of the 7 rolls is even is:

1/128 + 21/128 + 35/128 + 7/128 = 64/128 = 1/2

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If Griffin rolls a 12-sided die (numbered 1- 12) 7 times, what is the [#permalink]
We can approach this question by thinking in the following manner -

1)sum of 7 rolls can be either odd or even
2)We can see that sum of 6 rolls can be odd or even.
3)So in order for the sum of 7 rolls to be even , we should get an even in the 7th roll. (Because even + even = even or odd +odd = even)
4)Hence the probability of getting an even in the 7th roll is - 1/2 (i/e it can either be even or be odd)

cheers ^_^
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If Griffin rolls a 12-sided die (numbered 1- 12) 7 times, what is the [#permalink]
MikeScarn wrote:
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.

Isn't your example of even number of rolls not accounting for the (odd, even) configuration which would be different from the (even, odd) configuration that you have listed? This is given that these are separate rolls and not 2 dice being rolled. So the probability of even sum would still be 1/2 (and that should be the case for any number of rolls).

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

Using symmetry, we can see that probability of getting an even sum is 1/2 and of getting an odd sum is 1/2, just like it works with 6 sided dice rolled 3 times.

We know how the roll of 3 regular dice (1-6) work.
Lowest sum possible = 1 + 1 + 1 = 3 (odd) in 1 way only
Highest sum possible = 6 + 6 + 6 = 18 (even) in 1 way only
Second lowest sum possible (2 + 1 + 1) = 4 (even) in 3 ways
Second highest sum possible (5 + 6 + 6) = 17 (odd) in 3 ways
....
So probability of getting an even sum is 1/2 and of getting an odd sum is 1/2.

This case of 12 sided die rolled 7 times is no different.
Lowest sum possible = 1 + 1 + 1+...+ 1 = 7 (odd) in 1 way only
Highest sum possible = 12 + 12 + 12 +...+12 = 84 (even) in 1 way only
Second lowest sum possible (2 + 1 + 1+...+1) = 8 (even) in 7 ways (7 rolls are distinct)
Second highest sum possible (11 + 12 + 12+ ... + 12) = 83 (odd) in 7 ways (7 rolls are distinct)
...
Same logic here.