Question

An unbiased die is thrown three times what is the probability that the sum will be equal to 10

Answer 1

Answer: $\frac{1}{8}$

Step-by-step explanation:

An unbiased die is thrown 3 times, so the sum could vary from 3 to 18. There are 216 = 6×6×6 total number of ways.

The possibilities for the sum to be equal to 10 are listed in triplets

(1, 3, 6) or its permutations---total 6 in number

(1, 4, 5) or its permutations---total 6 in number

(2, 2, 6) or its permutations---total 3 in number

(2, 3, 5) or its permutations---total 6 in number

(2, 4, 4) or its permutations---total 3 in number

(3, 3, 4) or its permutations---total 3 in number.

Thus total number of favourable ways are 27.

Probability = (total number of favourable ways)/(total number of  ways)

                   = $\frac{27}{216}$

                   = $\frac{1}{8}$ .

Answer 2

Answer:

Step-by-step explanation:

An unbiased die is thrown 3 times, so the sum could vary from 3 to 18. There are 216 = 6×6×6 total number of ways.

The possibilities for the sum to be equal to 10 are listed in triplets

(1, 3, 6) or its permutations---total 6 in number

(1, 4, 5) or its permutations---total 6 in number

(2, 2, 6) or its permutations---total 3 in number

(2, 3, 5) or its permutations---total 6 in number

(2, 4, 4) or its permutations---total 3 in number

(3, 3, 4) or its permutations---total 3 in number.

Thus total number of favourable ways are 27.

Probability = (total number of favourable ways)/(total number of ways)