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

Question

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}

KarishmaB · Accepted Answer

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

BrushMyQuant · Answer

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

https://www.youtube.com/watch?v=5PVZYQXet18

KarishmaB · Answer

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.

Bunuel · Answer

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.

SpotlessMind · Answer

Hi,

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

rohitthakur · Answer

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

sabineodf · Answer

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!!

KarishmaB · Answer

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.

sabineodf · Answer

Thank you, that was very helpful!

gmatgrl · Answer

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.

sabineodf · Answer

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:

SajalBatwara · Answer

KarishmaB  can you please explain method 1 in detail.

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

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