Question

A balance die thrown twice probability of sum of number is less than 8

Answer 1

The chances are:-

1 & 1, 1&2, 1&3, 1&4, 1&5, 1&6, 2&2, 2&3, 2&4, 2&5, 3&3, 3&4

No of chances = 12

Total = 36

P(E) = 12/36 = 1/3

Answer 2

If a die is rolled twice what is the probability the sum is greater than 8?


A easy way to solve this is to count all the possible cases with the die being thrown twice and getting a sum greater than 8. The combinations are:

3 & 6, 4&5, 4&6, 5&4, 5&5, 5&6, 6&3, 6&4, 6&5, 6&6. That’s ten cases.

The probability of getting each number on each throw is 1/6.

So, the total probability of getting a sum greater than 8 is:

(1/6 x 1/6) + (1/6 x 1/6)…. 10 times. This equals to: 10 x (1/36) :

P = 10/36 = 5/18