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# Two number cubes with faces numbered 1 to 6 are rolled. What is the

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Joined: 23 Feb 2014
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GMAT 1: 560 Q41 V26
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Two number cubes with faces numbered 1 to 6 are rolled. What is the  [#permalink]

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Updated on: 17 Dec 2018, 02:41
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25% (medium)

Question Stats:

73% (01:04) correct 27% (01:22) wrong based on 187 sessions

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Two number cubes with faces numbered 1 to 6 are rolled. What is the probability that that sum of the rolls is 8.

A. 1/12
B. 1/11
C. 1/9
D. 5/36
E. 1/6

I am struggling to understand a concept - This seems like an error in MGMAT book - But i want to get it clarified and get a better understanding.

The questions is: Two number cubes with faces numbered 1 to 6 are rolled. What is the probability that that sum of the rolls is 8.

The Concern/Error: The successful outcomes is being counted as 2-6, 3-5,4-4, 5-3, 6-2.
I think there needs to be another 4-4 that should be counted.

Because 4 in Dice 1 is different from 4 in dice 2.

Can someone please explain? Anyone facing same issue?

Originally posted by vingmat001 on 02 Mar 2014, 20:20.
Last edited by Bunuel on 17 Dec 2018, 02:41, edited 3 times in total.
Renamed the topic and edited the question.
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Re: Two number cubes with faces numbered 1 to 6 are rolled. What is the  [#permalink]

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02 Mar 2014, 22:22
3
vingmat001 wrote:
I am struggling to understand a concept - This seems like an error in MGMAT book - But i want to get it clarified and get a better understanding.

The questions is: Two number cubes with faces numbered 1 to 6 are rolled. What is the probability that that sum of the rolls is 8.

The Concern/Error: The successful outcomes is being counted as 2-6, 3-5,4-4, 5-3, 6-2.
I think there needs to be another 4-4 that should be counted.

Because 4 in Dice 1 is different from 4 in dice 2.

Can someone please explain? Anyone facing same issue?

Say one die is red in color and the other is yellow. Why do we take two cases (2, 6) and (6, 2)?
Because a 2 on the red one and 6 on the yellow one is different from 2 on the yellow one and 6 on the red one.

In the case of (4, 4), you have 4 on the red one and 4 on the yellow one. How can you have another (4, 4) case? The other one will also be 4 on the red one and 4 on the yellow one.
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Karishma
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Re: Two number cubes with faces numbered 1 to 6 are rolled. What is the  [#permalink]

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03 Mar 2014, 16:30
Wow !! Thanks for the explanation Karishma!- Just add color to the dice and it all becomes clear.
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Re: Two number cubes with faces numbered 1 to 6 are rolled. What is the  [#permalink]

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03 Mar 2014, 22:47
2
vingmat001 wrote:
I am struggling to understand a concept - This seems like an error in MGMAT book - But i want to get it clarified and get a better understanding.

The questions is: Two number cubes with faces numbered 1 to 6 are rolled. What is the probability that that sum of the rolls is 8.

The Concern/Error: The successful outcomes is being counted as 2-6, 3-5,4-4, 5-3, 6-2.
I think there needs to be another 4-4 that should be counted.

Because 4 in Dice 1 is different from 4 in dice 2.

Can someone please explain? Anyone facing same issue?

Another way to visualize this problem is as shown in the attached diagram. In the test, you do not need to list all possible sums. Just look for the one you are interested in and count them.
Attachments

File comment: Probability of getting a defined sum from the sum of the numbers on two faces of a randomly rolled cube.
2 rolled cubes numbered 1 to 6.docx [58.01 KiB]

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Schools: CBS '20 (S)
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Re: Two number cubes with faces numbered 1 to 6 are rolled. What is the  [#permalink]

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08 Jul 2017, 19:01
"Two number cubes with faces numbered 1 to 6 are rolled. What is the probability that the sum of the rolls is 8?" (MGMAT example question Guide 5 Chapter 4)

We all know the right way solve this question, which is - Total outcomes 36, Desirable outcomes 5. So, Probability is 5/36.

My question - why is the # of total outcomes not 11? Why are we considering the total outcomes of die-rolls, and not the total outcomes of the sum. If I consider the total outcomes of the sum = 2 to 12 then my probability is 1/11

Again, I completely understand that the right answer takes into effect the higher number of individual outcomes as below.

Sum # outcomes
2 1
3 2
4 3
.
.
....
11 2
12 1

My question is what makes the solution of 1/11 wrong? Does it depend on what we define as an "outcome"?
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Posts: 998
Re: Two number cubes with faces numbered 1 to 6 are rolled. What is the  [#permalink]

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08 Jul 2017, 20:41
1
rbramkumar wrote:
"Two number cubes with faces numbered 1 to 6 are rolled. What is the probability that the sum of the rolls is 8?" (MGMAT example question Guide 5 Chapter 4)

We all know the right way solve this question, which is - Total outcomes 36, Desirable outcomes 5. So, Probability is 5/36.

My question - why is the # of total outcomes not 11? Why are we considering the total outcomes of die-rolls, and not the total outcomes of the sum. If I consider the total outcomes of the sum = 2 to 12 then my probability is 1/11

Again, I completely understand that the right answer takes into effect the higher number of individual outcomes as below.

Sum # outcomes
2 1
3 2
4 3
.
.
....
11 2
12 1

My question is what makes the solution of 1/11 wrong? Does it depend on what we define as an "outcome"?

number of ways u get 8
2,6
6,2
3,5
5,3
4,4
= 5ways

and total possible outcomes of two dice = 6*6=36

thus probability = 5/36....
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Re: Two number cubes with faces numbered 1 to 6 are rolled. What is the  [#permalink]

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09 Jul 2017, 00:55
Sorry, I don't think you read my question right. Anyway, thanks for your input.
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Re: Two number cubes with faces numbered 1 to 6 are rolled. What is the  [#permalink]

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09 Jul 2017, 01:22
1
rbramkumar wrote:
"Two number cubes with faces numbered 1 to 6 are rolled. What is the probability that the sum of the rolls is 8?" (MGMAT example question Guide 5 Chapter 4)

We all know the right way solve this question, which is - Total outcomes 36, Desirable outcomes 5. So, Probability is 5/36.

My question - why is the # of total outcomes not 11? Why are we considering the total outcomes of die-rolls, and not the total outcomes of the sum. If I consider the total outcomes of the sum = 2 to 12 then my probability is 1/11

Again, I completely understand that the right answer takes into effect the higher number of individual outcomes as below.

Sum # outcomes
2 1
3 2
4 3
.
.
....
11 2
12 1

My question is what makes the solution of 1/11 wrong? Does it depend on what we define as an "outcome"?

Since there are 1 outcome for getting sum of 2 and this pattern increases till 7(where the number of outcomes is 6)
The maximum outcomes happen, when the sum if 7 {(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)} and then the outcomes reduce till we reach 12 as sum,
where the number of outcomes is 1(6,6)

The probability of an event happening = $$\frac{P(A)}{P(T)}$$
where P(A) = Possibility_of_an_event_happening
and P(T) = Total_number_of_outcomes

Here the Possibility of getting 8 is 5 {(2,6),(3,5),(4,4),(5,3),(6,2)}
Total possibilities are 1+2+3+4+5+6+5+4+3+2+1 = 36.

Hope that helps!
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Re: Two number cubes with faces numbered 1 to 6 are rolled. What is the  [#permalink]

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09 Jul 2017, 02:46
rbramkumar wrote:
"Two number cubes with faces numbered 1 to 6 are rolled. What is the probability that the sum of the rolls is 8?" (MGMAT example question Guide 5 Chapter 4)

We all know the right way solve this question, which is - Total outcomes 36, Desirable outcomes 5. So, Probability is 5/36.

My question - why is the # of total outcomes not 11? Why are we considering the total outcomes of die-rolls, and not the total outcomes of the sum. If I consider the total outcomes of the sum = 2 to 12 then my probability is 1/11

Again, I completely understand that the right answer takes into effect the higher number of individual outcomes as below.

Sum # outcomes
2 1
3 2
4 3
.
.
....
11 2
12 1

My question is what makes the solution of 1/11 wrong? Does it depend on what we define as an "outcome"?

Merging topics.

I think pushpitkc answers your question but just to make sure you understand. Not all sums out of 11 have the same number of occurrences. The sum of 2 can occur in 1 way - (1, 1) but the sum of two can occur in 2 way - (1, 2) and (2, 1).

Hope it helps.
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Re: Two number cubes with faces numbered 1 to 6 are rolled. What is the  [#permalink]

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09 Jul 2017, 05:19
Thank you, pushpitkc and bunuel.

I understand the concept of "# on dice 1, # on dice 2" as an outcome, but was merely speculating how the answer looked like when we look at "SUM" as an outcome.

Let me elaborate my thoughts:

Former: Outcome (# on dice1, # on dice 2)
Total outcomes 36

Latter: Outcome is defined as the "Sum" of the #s in 2 dices
Total outcomes 11

I was getting derailed by the fact that the # Total outcomes needed to be 11, but I think the latter definition of "outcome" changes the basic assumption in Probability questions, i.e., equally likely outcomes (no bias). I missed the fact that this is a probability distribution where we (sum the area of desired/sum the total area under histogram) to get to the required probability. This seems to result in the original answer of 5/36.

There's really no need to think in the latter manner, just trying out different ways. Thanks and all the best!
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Re: Two number cubes with faces numbered 1 to 6 are rolled. What is the  [#permalink]

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17 Oct 2017, 05:20
VeritasPrepKarishma wrote:
vingmat001 wrote:
I am struggling to understand a concept - This seems like an error in MGMAT book - But i want to get it clarified and get a better understanding.

The questions is: Two number cubes with faces numbered 1 to 6 are rolled. What is the probability that that sum of the rolls is 8.

The Concern/Error: The successful outcomes is being counted as 2-6, 3-5,4-4, 5-3, 6-2.
I think there needs to be another 4-4 that should be counted.

Because 4 in Dice 1 is different from 4 in dice 2.

Can someone please explain? Anyone facing same issue?

Say one die is red in color and the other is yellow. Why do we take two cases (2, 6) and (6, 2)?
Because a 2 on the red one and 6 on the yellow one is different from 2 on the yellow one and 6 on the red one.

In the case of (4, 4), you have 4 on the red one and 4 on the yellow one. How can you have another (4, 4) case? The other one will also be 4 on the red one and 4 on the yellow one.

Hi, I can understand why in the "total number of desired outcome" we count only 1 combination {4;4}

However, in the "total number of possible outcome" we put 36 = 6*6. This number include all the duplicated {1;1}, {2,2}.... {6,6}; Otherwise we don't have 36, we only have 30 = 6*5 possible outcome.

Thank you!
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Re: Two number cubes with faces numbered 1 to 6 are rolled. What is the  [#permalink]

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17 Oct 2017, 21:39
1
phanthibaovien wrote:
VeritasPrepKarishma wrote:
vingmat001 wrote:
I am struggling to understand a concept - This seems like an error in MGMAT book - But i want to get it clarified and get a better understanding.

The questions is: Two number cubes with faces numbered 1 to 6 are rolled. What is the probability that that sum of the rolls is 8.

The Concern/Error: The successful outcomes is being counted as 2-6, 3-5,4-4, 5-3, 6-2.
I think there needs to be another 4-4 that should be counted.

Because 4 in Dice 1 is different from 4 in dice 2.

Can someone please explain? Anyone facing same issue?

Say one die is red in color and the other is yellow. Why do we take two cases (2, 6) and (6, 2)?
Because a 2 on the red one and 6 on the yellow one is different from 2 on the yellow one and 6 on the red one.

In the case of (4, 4), you have 4 on the red one and 4 on the yellow one. How can you have another (4, 4) case? The other one will also be 4 on the red one and 4 on the yellow one.

Hi, I can understand why in the "total number of desired outcome" we count only 1 combination {4;4}

However, in the "total number of possible outcome" we put 36 = 6*6. This number include all the duplicated {1;1}, {2,2}.... {6,6}; Otherwise we don't have 36, we only have 30 = 6*5 possible outcome.

Thank you!

These are the 36 cases:

{1, 1}, {1, 2}, {1, 3}, {1, 4}, {1, 5}, {1, 6}
{2, 1}, {2, 2}, {2, 3}, {2, 4}, {2, 5}, {2, 6}
{3, 1}, {3, 2}, {3, 3}, {3, 4}, {3, 5}, {3, 6}
{4, 1}, {4, 2}, {4, 3}, {4, 4}, {4, 5}, {4, 6}
{5, 1}, {5, 2}, {5, 3}, {5, 4}, {5, 5}, {5, 6}
{6, 1}, {6, 2}, {6, 3}, {6, 4}, {6, 5}, {6, 6}
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Re: Two number cubes with faces numbered 1 to 6 are rolled. What is the  [#permalink]

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27 Jan 2018, 17:30
1
vingmat001 wrote:
Two number cubes with faces numbered 1 to 6 are rolled. What is the probability that that sum of the rolls is 8.

A. 1/12
B. 1/11
C. 1/9
D. 5/36
E. 1/6

I am struggling to understand a concept - This seems like an error in MGMAT book - But i want to get it clarified and get a better understanding.

The questions is: Two number cubes with faces numbered 1 to 6 are rolled. What is the probability that that sum of the rolls is 8.

The Concern/Error: The successful outcomes is being counted as 2-6, 3-5,4-4, 5-3, 6-2.
I think there needs to be another 4-4 that should be counted.

Because 4 in Dice 1 is different from 4 in dice 2.

Can someone please explain? Anyone facing same issue?

Here is an alternate approach.

We choose 1 dice and individually calculate the possibility that it will make a sum of 8 when combined with a number in the 2nd dice.

If the first dice rolls a...

1: $$\frac{0}{6}$$ possibility of it making a sum of 8 when combined with a number in the second dice.

2: $$\frac{1}{6}$$ chance (need 6 on the second dice) it will make a sum of 8.

3: $$\frac{1}{6}$$ chance (need 5 on the second dice) it will make a sum of 8.

4: $$\frac{1}{6}$$ chance (need 4 on the second dice) it will make a sum of 8.

5: $$\frac{1}{6}$$ chance (need 3 on the second dice) it will make a sum of 8.

6: $$\frac{1}{6}$$ chance (need 2 on the second dice) it will make a sum of 8.

Each digit in the first dice makes 6 combinations (one with each digit in second dice) and added up, they all make 36 combinations. So each digit's weight or contribution in total probability is 1/6. Adding probability sums.

$$\frac{0}{6}*\frac{1}{6} + \frac{1}{6}*\frac{1}{6} + \frac{1}{6}*\frac{1}{6} + \frac{1}{6}*\frac{1}{6} + \frac{1}{6}*\frac{1}{6} + \frac{1}{6}*\frac{1}{6}$$

$$= 0 + \frac{1}{36} + \frac{1}{36} + \frac{1}{36} + \frac{1}{36} + \frac{1}{36} = \frac{5}{36}$$

It is important to understand that of the 6 digits on a dice, only 5 make a combination that sums up to 8. The digit 1 cannot possibly add up to make 8 as the maximum digit in the second dice is 6 which would make the sum 7.
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Re: Two number cubes with faces numbered 1 to 6 are rolled. What is the   [#permalink] 27 Jan 2018, 17:30
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