Question

An experiment consists of rolling three fair dice --- a red die, a blue die, and a white die --- and recording the number rolled on each die. Assume that the dice are fair, so that all outcomes are equally likely. (1) What probability should be assigned to each outcome? equation editorEquation Editor (2) What is the probability that the sum of the numbers rolled is 5? equation editorEquation Editor (3) What is the probability that the sum of the numbers rolled is at most 6? equation editorEquation Editor

Answer 1

Given that three fair dice --- a red die, a blue die, and a white die are rolled

Number of outcomes = 6 * 6 * 6 = 216

Question (1)

Since all the three dice are fair, all the 216 outcomes will be equally likey

So Probability of each outcome = 1/216 = 0.0046296

= 0.0046 rounded to 4 decimal places

Question (2)

Probability that the sum of numbers rolled on three dices is 5

The possible outcomes where the sum of numbers rolled on three dices is 5 are given below

(1,1,3) (1,3,1) (3,1,1) (1,2,2) (2,1,2) (2,2,1)

So there are 6 outcomes possible where the sum of numbers rolled on three dices is 5

Probability that the sum of numbers rolled on three dices is 5 = 6/216 = 1/36

= 0.027778

Question (3)

Probability that the sum of numbers rolled on three dices is at most 6

So we need to look at outcomes where the sum of numbers rolled on three dices is 3,4,5,6

The possible outcomes where the sum of numbers rolled on three dices is at most 6 are given below

(1,1,1) (1,1,2) (1,2,1) (2,1,1)  (1,1,3) (1,3,1) (3,1,1) (1,2,2) (2,1,2) (2,2,1) (1,2,3) (3,2,1) (1,3,2) (2,3,1) (2,1,3) (3,1,2) (1,4,1) (4,1,1) (1,1,4) (2,2,2)

So there are 20 outcomes possible where the sum of numbers rolled on three dices is at most 6

Probability that the sum of numbers rolled on three dices is at most 6 = 20/216 = 5/54

= 0.092593

= 0.0926 rounded to 4 decimal places