Question

Three balanced dice are tossed. Find the probability of obtaining a nine, given:

a. The sum is odd.

b. The sum is less than or equal to nine.

c. None of the dice are odd.

d. At least one of the dice is odd.

e. At least two of the dice are odd.

f. All dice are odd.

g. All dice are different.

h. Two of the dice are the same.

i. All dice are the same.

Answer 1

When three balanced die rolled possible possible number of outcomes are 6*6*6 = 216. Following table shows the possible sums and number of ways we can obtained this sum:

Number of ways Sum 10 15 21 25 10 25 21 15 10 12 13 14 15 16 17 18 216 Total

For example; The sum 4 can be obtained in following outcomes:

(1,1,2), (1,2,1) and (2,1,1)

(a)

Following table shows the odd sums:

Out of 108 outcomes, showing odd sum, 25 outcomes show sum 9 so required probability is

P(sum 9| sum is odd) = 25 /108 = 0.2315

(b)

Following table shows the sums less than equal to 9:

Out of 81 outcomes, showing odd less than equal to 9, 25 outcomes show sum 9 so required probability is

P(sum 9| sum is less than equal to 9) = 25 / 81 = 0.3086

(c)

If none of the dice shows odd number then it is not possible to get sum 9 so

P(sum 9| none dice shows odd) = 0

(d)

Here first we need to find the number of outcomes in which all die show even numbers, like outcomes (2,2,2), (2,4,6) etc.

Number of such outcomes are:

(2,2,2) in 1 way

(2,2,4) in 3 ways

(6,2,2) in 3 ways

(4,4,2) in 3 ways

(4,4,4) in 1 way

(6,4,2) in 6 ways

(6,6,2) in 3 ways

(4,4,6) in 3 ways

(6,6,4) in 3 ways

(6,6,6) in 1 ways

Total number of ways: 27

So number of outcomes, in which at least one dice shows an odd is 216 -27 = 189

Out of 189 outcomes, 25 outcomes show sum 9 so required probability is

P(sum 9| at least one die show odd) = 25 / 189 = 0.1323