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Math Expert V
Joined: 02 Sep 2009
Posts: 56303
If Griffin rolls a 12-sided die (numbered 1- 12) 7 times, what is the  [#permalink]

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12 00:00

Difficulty:   35% (medium)

Question Stats: 74% (01:54) correct 26% (02:35) wrong based on 105 sessions

### HideShow timer Statistics 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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Math Expert V
Joined: 02 Aug 2009
Posts: 7764
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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If Griffin rolls a 12-sided die (numbered 1- 12) 7 times, what is the  [#permalink]

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

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3
Don't fall for the trap here! This problem is NOT about the raw math. Many explanations in this forum focus rather blindly on the math, but they fail to realize that the GMAT is a critical-thinking test, not a "let's-see-if-we-can-do-the-math-the-long-way-around" test. Oftentimes, the shape and structure of a problem can point to leverage that allows you to avoid a lot of messy mathematical gymnastics. This problem is no exception.

First of all, the problem asks you about the probability that a sum of a set of numbers is "even." This is MASSIVE leverage. It means you don't have to care about the numbers; you only need to worry about whether the numbers are odd or even. Since a 12-sided die has 6 evens ($$2$$,$$4$$,$$6$$,$$8$$,$$10$$) and 6 odds ($$1$$,$$3$$,$$5$$,$$7$$,$$9$$,$$11$$), it basically can be treated as a coin toss where one side is even and the other side is odd. You have the same chance of getting either.

Since odd+odd = even, the sum of $$7$$ integers will be odd if there are an odd number of odds! (Pairs of odd values will effectively cancel each other out.) There is no reason to do any complicated math here, and certainly no reason to pull out combinatorics equations. Just visualize what is happening. For every combination of odds/evens, there is its "mirror" that reverses the odds and evens. (In other words, for every $$O+O+O+O+O+E+E$$ with $$5$$ odds and $$2$$ evens, there is an $$E+E+E+E+E+O+O$$ with 5 evens and 2 odds.)

You don't need to spend time messily calculating the number of combinations. Since every "odd" combination can be paired with an "even" combination, then the chance of getting an odd sum will be 50%. The answer is B.

Now, let’s look back at this problem through the lens of strategy. Your job as you study for the GMAT isn't to memorize the solutions to specific questions; it is to internalize strategic patterns that allow you to solve large numbers of questions. This problem can teach us patterns seen throughout the GMAT. This solution is an application of a strategy I call in my classes "Number Crunchers." The idea is simple: Watch for patterns of numbers, especially when it asks for "positive/negative" issues or "odd/even" issues. You just need to look at how the problem is structured and determine what would limit or define the value you are looking for. Determine what those limits are, and you have your answer. Patterns turn "inefficient" math into great critical-thinking opportunities. And that is how you think like the GMAT.
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Originally posted by AaronPond on 02 Feb 2019, 13:22.
Last edited by AaronPond on 13 Feb 2019, 14:34, edited 1 time 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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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]

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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]

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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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