Question

3 dice are rolled. What is the probability that you will get the sum of the no’s as 10

Answer 1

n(s)=6^3=216
no.of ways to get 10as sum
(2,2,6)->3!/2!=3
(2,4,4)->3!/2!=3
(1,3,6)->3!=6
(1,4,5)->3!=6
(2,3,5)->3!=6
(3,3,4)->3!/2!=3
Total n(e)=27
probability=n(e)/n(s)=27/216=1/8

Answer 2

Answer:

Given,

3 dices are rolled

to find: probability that the sum will be 10

Explanation:

Total outcomes =

${6}^{3} = 216$

because in each dice there are 6 numbers

sum is 10 in the following cases:

(1,3,6) = these are three numbers but they can be get in any sequence so the outcome will be 3! = 6

(2,4,4) = these are three numbers but they can be get in any sequence so the outcome will be 3! = 6 but there is repetition one number so it will be 3!/2! = 3

(2,3,5) = these are three numbers but they can be get in any sequence so the outcome will be 3! = 6

(1,4,5) = these are three numbers but they can be get in any sequence so the outcome will be 3! = 6

(2,2,6) = these are three numbers but they can be get in any sequence so the outcome will be 3! = 6 but there is repetition one number so it will be 3!/2! = 3

(3,4,3) = these are three numbers but they can be get in any sequence so the outcome will be 3! = 6 but there is repetition one number so it will be 3!/2! = 3

total number of possibilities = 27

probability = total number of possibilities / total number of outcomes

Probability = 27/216

Probability = 1/8