Problem Solving

A 6 sided dice is thrown 3 times. What is the probabilty that the sum of the throws is less than 6?

A) 1/18
B)5/108
c) 5/27
d)1/27

Is the answer {B}?

Using Separator method

Three cases:

Case 1: Total Sum to distribute = 5 (1,1,1,1,1)

==> 4!/2!*2! = 6

Case 2: Total Sum to distribute = 4 (1,1,1,1)
==> 3!/2! = 3

Case 3: Sum is 3 ==> 1

So, Probability = (6+3+1)/6*6*6 = 10/216 = 5/108

AIM TO CRACK GMAT wrote:A 6 sided dice is thrown 3 times. What is the probabilty that the sum of the throws is less than 6?

A) 1/18
B)5/108
c) 5/27
d)1/27

Although it's palpably NOT an original GMAT problem, yet it deserves a shot by GMATians.

When 3 dice are thrown, total outcomes are (6)(6)(6) = 216.

For the sum to be less than 6, we've only three possible sums to accept and those are 3, 4, or 5.

A sum of 3 can be had in just one way, and i.e. 1 1 1.

A sum of 4 can be had in just three ways, and those are 1 1 2, 1 2 1, and 2 1 1.

A sum of 5 can be had in just six ways, and those are 1 1 3, 1 3 1, 3 1 1, 1 2 2, 2 1 2, and 2 2 1.

Thus the favorable number of outcomes = 1 + 3 + 6 = 10.

Hence, the required probability = [spoiler]10/216 = 5/108.

Pick B[/spoiler]