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If two fair six-sided dice are thrown, what is the probabili

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megafan
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If two fair six-sided dice are thrown, what is the probabili [#permalink] New post   Updated on: 19 Jul 2013, 14:24
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If two fair six-sided dice are thrown, what is the probability that the sum of the numbers showing on the dice is a multiple of 3 ?

(A) \(\frac{1}{4}\)

(B) \(\frac{3}{11}\)

(C) \(\frac{5}{18}\)

(D) \(\frac{1}{3}\)

(E) \(\frac{4}{11}\)
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D

Originally posted by megafan on 06 Mar 2013, 19:21.
Last edited by Bunuel on 19 Jul 2013, 14:24, edited 1 time in total.
Edited the OA.
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If two fair six-sided dice are thrown, what is the probabili [#permalink] New post   Updated on: 11 Oct 2022, 02:20
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megafan wrote:
If two fair six-sided dice are thrown, what is the probability that the sum of the numbers showing on the dice is a multiple of 3 ?

(A) \(\frac{1}{4}\)

(B) \(\frac{3}{11}\)

(C) \(\frac{5}{18}\)

(D) \(\frac{1}{3}\)

(E) \(\frac{4}{11}\)


Method 1: Useful for more than 2 dice too.


No of ways of obtaining a sum of 2 = No of ways of obtaining a sum of 12
No of ways of obtaining a sum of 3 = No of ways of obtaining a sum of 11
No of ways of obtaining a sum of 4 = No of ways of obtaining a sum of 10
No of ways of obtaining a sum of 5 = No of ways of obtaining a sum of 9
No of ways of obtaining a sum of 6 = No of ways of obtaining a sum of 8

No of ways of obtaining:

A sum of 3: 2C1 = 2
A sum of 6: 5C1 = 5
A sum of 9 = A sum of 5: 4C1 = 4
A sum of 12 = A sum of 2: 1C1 = 1

Total no of ways of obtaining a sum which is a multiple of 3 = 2 + 5 + 4 + 1 = 12

Total no of ways = 6*6 = 36

Probability of the sum being a multiple of 3 = 12/36 = 1/3

Answer (D)

Method 2: You could simply count the number of ways of getting 3, 6, 9 and 12

Another Method: Since we have only 2 dice.
Check out this table of the 36 different outcomes:

....1 ......2 ......3 .....4 ......5 ......6
1 (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)
2 (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)
3 (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)
4 (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)
5 (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)
6 (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)

Each column gives you 6 different outcomes which vary by 1 each. Hence there will be exactly 2 outcomes with sum as multiple of 3 (take any 6 consecutive numbers. Exactly 2 of them will be a multiple of 3)

Hence required probability = 1/3

Originally posted by KarishmaB on 06 Mar 2013, 21:36.
Last edited by KarishmaB on 11 Oct 2022, 02:20, edited 1 time in total.
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If two fair six-sided dice are thrown, what is the probabili [#permalink] New post   Updated on: 08 Oct 2022, 12:17
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Given that two fair six-sided dice are thrown and we need to find what is the probability that the sum of the numbers showing on the dice is a multiple of 3 ?

As we are rolling two dice => Number of cases = \(6^2\) = 36

Multiple of three between 1+1 (=2) and 6+6(=12) are 3, 6, 9 , 12

Lets start writing the possible cases where sum of the two rolls = 3, 6, 9, 12. Following are the possible cases:
(1,2), (1,5)
(2,1), (2,4)
(3,3), (3,9)
(4,2), (4,5)
(5,1), (5,4)
(6,3), (6,6)

=> 12 cases

=> Probability that sum of two rolls is a multiple of 3 = \(\frac{12}{36}\) = \(\frac{1}{3}\)

So, Answer will be D
Hope it helps!

Watch the following video to learn How to Solve Dice Rolling Probability Problems


Originally posted by BrushMyQuant on 06 Mar 2013, 21:31.
Last edited by BrushMyQuant on 08 Oct 2022, 12:17, edited 1 time in total.
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Re: If two fair six-sided dice are thrown, what is the probabili [#permalink] New post   06 Mar 2013, 21:42
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nktdotgupta wrote:
Total possible values of sum are
2,3,4,5,6,7,8,9,10,11,12
-> Total values = 11
-> number of multiples of 3 are 3,6,9,12 = 4
Probability that sum is a multiple of 3 is 4/11

So, Answer will be E according to me
Am not sure where am going wrong(if i am?)?


Where you are going wrong is that the probability of getting a sum of 3 is not the same as the probability of getting 4. You are assuming that each sum is obtained with equal probability. There are 2 ways of obtaining 3 but 3 ways of obtaining 4. Hence, it is more probable that you will get a sum of 4 as compared to a sum of 3.
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Re: If two fair six-sided dice are thrown, what is the probabili [#permalink] New post   07 Mar 2013, 01:48
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megafan wrote:
If two fair six-sided dice are thrown, what is the probability that the sum of the numbers showing on the dice is a multiple of 3 ?

(A) \(\frac{1}{4}\)

(B) \(\frac{3}{11}\)

(C) \(\frac{5}{18}\)

(D) \(\frac{1}{3}\)

(E) \(\frac{4}{11}\)


Very good discussion about the same concept: mary-and-joe-are-to-throw-three-dice-each-the-score-is-the-126407.html

Hope it helps.
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Re: If two fair six-sided dice are thrown, what is the probabili [#permalink] New post   09 Mar 2013, 01:40
Hi,

I too got the answer as 1/3 however the OA , points to a different choice.. can someone pls correct/clarify???


Bunuel wrote:
megafan wrote:
If two fair six-sided dice are thrown, what is the probability that the sum of the numbers showing on the dice is a multiple of 3 ?

(A) \(\frac{1}{4}\)

(B) \(\frac{3}{11}\)

(C) \(\frac{5}{18}\)

(D) \(\frac{1}{3}\)

(E) \(\frac{4}{11}\)


Very good discussion about the same concept: mary-and-joe-are-to-throw-three-dice-each-the-score-is-the-126407.html

Hope it helps.
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Re: If two fair six-sided dice are thrown, what is the probabili [#permalink] New post   03 Apr 2015, 03:56
the easy way to do this is,
numbers multiple of 3 are : 3,6,9 and 12, those can appear on twice dice throw,
3 can come on dice in two ways as (1,2) and(2,1)
6 can be there in 5 ways:( 1,5), (5,1), (2,4), (4,2) and (3,3)
9 can be in 4 ways: (4,5), (5,4),(3,6) and (6,3)
now 12 can appear only in one way i.e, (6,6)
adding each probability it has 12 ways so dividing it by 36 total throws will give 1/3
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Re: If two fair six-sided dice are thrown, what is the probabili [#permalink] New post   05 Apr 2015, 20:37
rohitthakur wrote:
the easy way to do this is,
numbers multiple of 3 are : 3,6,9 and 12, those can appear on twice dice throw,
3 can come on dice in two ways as (1,2) and(2,1)
6 can be there in 5 ways:( 1,5), (5,1), (2,4), (4,2) and (3,3)
9 can be in 4 ways: (4,5), (5,4),(3,6) and (6,3)
now 12 can appear only in one way i.e, (6,6)
adding each probability it has 12 ways so dividing it by 36 total throws will give 1/3



I had the same approach as you, but I got stuck because I wrote the possibility of getting 6 as one too many, since I put down (3,3) twice. I see that was wrong now, but why is that wrong when (5,1) and (1,5) are two separate possibilities? Thanks if you are able to explain that to me!!
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Re: If two fair six-sided dice are thrown, what is the probabili [#permalink] New post   05 Apr 2015, 21:02
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sabineodf wrote:
rohitthakur wrote:
the easy way to do this is,
numbers multiple of 3 are : 3,6,9 and 12, those can appear on twice dice throw,
3 can come on dice in two ways as (1,2) and(2,1)
6 can be there in 5 ways:( 1,5), (5,1), (2,4), (4,2) and (3,3)
9 can be in 4 ways: (4,5), (5,4),(3,6) and (6,3)
now 12 can appear only in one way i.e, (6,6)
adding each probability it has 12 ways so dividing it by 36 total throws will give 1/3



I had the same approach as you, but I got stuck because I wrote the possibility of getting 6 as one too many, since I put down (3,3) twice. I see that was wrong now, but why is that wrong when (5,1) and (1,5) are two separate possibilities? Thanks if you are able to explain that to me!!


Imagine that one die is red and the other is yellow. A 5 on red and 1 on yellow is different from a 1 on red and 5 on yellow.
But a 3 on red and 3 on yellow is the same as 3 on red and 3 on yellow. Hence, you count it only once.
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Re: If two fair six-sided dice are thrown, what is the probabili [#permalink] New post   05 Apr 2015, 22:02
VeritasPrepKarishma wrote:
sabineodf wrote:
rohitthakur wrote:
the easy way to do this is,
numbers multiple of 3 are : 3,6,9 and 12, those can appear on twice dice throw,
3 can come on dice in two ways as (1,2) and(2,1)
6 can be there in 5 ways:( 1,5), (5,1), (2,4), (4,2) and (3,3)
9 can be in 4 ways: (4,5), (5,4),(3,6) and (6,3)
now 12 can appear only in one way i.e, (6,6)
adding each probability it has 12 ways so dividing it by 36 total throws will give 1/3



I had the same approach as you, but I got stuck because I wrote the possibility of getting 6 as one too many, since I put down (3,3) twice. I see that was wrong now, but why is that wrong when (5,1) and (1,5) are two separate possibilities? Thanks if you are able to explain that to me!!


Imagine that one die is red and the other is yellow. A 5 on red and 1 on yellow is different from a 1 on red and 5 on yellow.
But a 3 on red and 3 on yellow is the same as 3 on red and 3 on yellow. Hence, you count it only once.


Thank you, that was very helpful!
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Re: If two fair six-sided dice are thrown, what is the probabili [#permalink] New post   06 Apr 2015, 00:37
sabineodf wrote:
I had the same approach as you, but I got stuck because I wrote the possibility of getting 6 as one too many, since I put down (3,3) twice. I see that was wrong now, but why is that wrong when (5,1) and (1,5) are two separate possibilities? Thanks if you are able to explain that to me!!

When we say that the total number of combinations possible are 36, it includes combinations such as (1,5),(5,1), (2,3),(3,2),(4,1),(1,4) etc.

So, this itself shows that (5,1) and (1,5) are two separate possibilities.
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Re: If two fair six-sided dice are thrown, what is the probabili [#permalink] New post   06 Apr 2015, 00:44
gmatgrl wrote:
sabineodf wrote:
I had the same approach as you, but I got stuck because I wrote the possibility of getting 6 as one too many, since I put down (3,3) twice. I see that was wrong now, but why is that wrong when (5,1) and (1,5) are two separate possibilities? Thanks if you are able to explain that to me!!

When we say that the total number of combinations possible are 36, it includes combinations such as (1,5),(5,1), (2,3),(3,2),(4,1),(1,4) etc.

So, this itself shows that (5,1) and (1,5) are two separate possibilities.


Yes, I am aware of that. My questions was why rolling (3,3) was not considered to be two separate possibilities... Which was explained to me above :wink:
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