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

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If two fair six-sided dice are thrown, what is the probabili [#permalink] New post 06 Mar 2013, 18:21
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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}
[Reveal] Spoiler: OA

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Last edited by Bunuel on 19 Jul 2013, 13:24, edited 1 time in total.
Edited the OA.
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Re: If two fair six-sided dice are thrown, what is the probabili [#permalink] New post 06 Mar 2013, 20:31
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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}


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?)?
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Re: If two fair six-sided dice are thrown, what is the probabili [#permalink] New post 06 Mar 2013, 20:36
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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.

Check out these posts first:
http://www.veritasprep.com/blog/2012/10 ... l-picture/
http://www.veritasprep.com/blog/2012/10 ... e-part-ii/

Now, the explanation given below will make sense.

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
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Re: If two fair six-sided dice are thrown, what is the probabili [#permalink] New post 06 Mar 2013, 20:42
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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, 00:48
Expert's post
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, 00:40
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.
Re: If two fair six-sided dice are thrown, what is the probabili   [#permalink] 09 Mar 2013, 00:40
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