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Re: If two fair six-sided dice are thrown, what is the probabili [#permalink]
06 Mar 2013, 20:31

1

This post received KUDOS

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

My solution is wrong! Outcomes are (1,2,3,4,5,6) (1,2,3,4,5,6) Possible values of sum are 1+1=2, 1+2=3,1+3=4,1+4=5, 1+5=6,1+6=7 2+1,2+2,2+3,2+4,2+5 are already taken into consideration above so not counted again 2+6=8,2+7=9 3+1,3+2,3+3,3+4,3+5,3+6 are already taken into consideration above so not counted again 4+1,4+2,4+3,4+4,4+5 are already taken into consideration above so not counted again 4+6 = 10 5+1,5+2,5+3,5+4,5+5 are already taken into consideration above so not counted again 5+6 = 11 6+1,6+2,6+3,6+4,6+5 are already taken into consideration above so not counted again 6+6 = 12

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?)? _________________

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:

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 _________________

Re: If two fair six-sided dice are thrown, what is the probabili [#permalink]
06 Mar 2013, 20:42

1

This post received KUDOS

Expert's post

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. _________________

Re: If two fair six-sided dice are thrown, what is the probabili [#permalink]
04 Jul 2014, 11:53

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Re: If two fair six-sided dice are thrown, what is the probabili [#permalink]
03 Apr 2015, 02:56

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

Re: If two fair six-sided dice are thrown, what is the probabili [#permalink]
05 Apr 2015, 19:37

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!!

Re: If two fair six-sided dice are thrown, what is the probabili [#permalink]
05 Apr 2015, 20:02

Expert's post

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. _________________

Re: If two fair six-sided dice are thrown, what is the probabili [#permalink]
05 Apr 2015, 21:02

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.

Re: If two fair six-sided dice are thrown, what is the probabili [#permalink]
05 Apr 2015, 23:37

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.

Re: If two fair six-sided dice are thrown, what is the probabili [#permalink]
05 Apr 2015, 23:44

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

gmatclubot

Re: If two fair six-sided dice are thrown, what is the probabili
[#permalink]
05 Apr 2015, 23:44

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