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