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

vingmat001

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Updated on: 17 Dec 2018, 02:41

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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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

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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?

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KarishmaB

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02 Mar 2014, 22:22

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vingmat001

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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08 Jul 2017, 20:41

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rbramkumar

"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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[3]

03 Mar 2014, 16:30

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Wow !! Thanks for the explanation Karishma!- Just add color to the dice and it all becomes clear. :-D

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03 Mar 2014, 22:47

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vingmat001

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.

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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]
Downloaded 397 times

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08 Jul 2017, 19:01

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rbramkumar

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.

Thats why the probability is 5/36 and not 1/11 as your had asked.

Hope that helps!

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rbramkumar

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 3 can occur in 2 way - (1, 2) and (2, 1).

Hope it helps.

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09 Jul 2017, 05:19

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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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17 Oct 2017, 05:20

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VeritasPrepKarishma

vingmat001

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.

May someone please help?

Thank you!

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17 Oct 2017, 21:39

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phanthibaovien

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vingmat001

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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27 Jan 2018, 17:30

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vingmat001

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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22 Dec 2019, 10:50

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Hi All,

We're told that two number cubes with faces numbered 1 to 6 are rolled. We're asked for the probability that that SUM of the two numbers rolled is 8. This is a fairly straight-forward Probability question (and if you have ever been to a casino, then you might recognize that this situation is based on a "Craps table."

Probability = (number of ways that you "WANT")/(total number of ways that are possible).

Since there are two 6-sided dice (with the numbers 1 to 6, inclusive, written on them), the TOTAL number of possible ways to roll those dice is (6)(6) = 36

To end up with a SUM of 8, we can simply list out the options. Here, it helps to think of a 'first die' and 'second die' since the number of the first die defines what the number of the second die would need to be for us to get a sum of 8.

2-6
3-5
4-4
5-3
6-2

5 total options that give us what we 'want' out of 36 possible options is 5/36

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23 Dec 2019, 00:13

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varotkorn

VeritasKarishma

vingmat001

Dear VeritasKarishma Bunuel EMPOWERgmatRichC,
The question does not state that the 2 dices are distinct. The two may be of the same color/shape/...

How do we know that the two are distinct? And why does the order matter in this question??

Thank you

We are not given that they are identical either. When working with probability, it wouldn't matter whether they are distinct or identical since the answer would be the same in either case. But assuming them to be distinct will help us solve the question faster.

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KarishmaB

vingmat001

KarishmaB

Thank you so much for this helpful explanation. I tried to apply your "coloring concept" to an arranging groups problem:
"How many arrangements are there of the letters in the word EEL?"
--> If you colored E1 and E2, using this coloring approach, wouldn't you have six arrangements then and not the correct answer of three arrangements?

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vingmat001

If the two dice were given to be identical, (2, 4) would be the same as (4, 2). One die would show 2 and the other would show 4. In that case, you would have total 21 cases only.
Since the two Es are the same, E1LE2 is the same as E2LE1.

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Thank you so much for your prompt reply. How would you know if they are identical or not? What in the problem would be an indicator that you count the (2,4) and (4,2) as separate entities or the two Es as separate? Thank you again.

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rohit8865

rbramkumar

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....

this makes way more sense, however why don't you count 4,4 as 2 like the other sets to make it 6 = desired outcome thus 1/6?

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