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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Karishma

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