die with 6 faces

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die with 6 faces

by ruplun » Fri Jul 29, 2011 5:35 am
A single die with 6 faces numbered 1 through 6 is thrown twice.If the numeral that faces upward as the result of each throw is recorded , what is the probability that the sum of 2 numbers is less than 10?

a. 5/6
b. 2/3
c. 1/2
d. 1/3
e. 1/6
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by Brent@GMATPrepNow » Fri Jul 29, 2011 6:06 am
ruplun wrote:A single die with 6 faces numbered 1 through 6 is thrown twice.If the numeral that faces upward as the result of each throw is recorded , what is the probability that the sum of 2 numbers is less than 10?

a. 5/6
b. 2/3
c. 1/2
d. 1/3
e. 1/6
We can use counting methods to solve this one.

P(sum < 10) = [total number of outcomes where the sum < 10] / [total number of outcomes]

Total number of outcomes
We can list them (wouldn't take long), or we can note that there are 6 possible outcomes for the first die and there are 6 possible outcomes for the second die.
So, the total number of outcomes for both die is 6 x 6 = 36


Total number of outcomes where the sum < 10
The easier approach here is to consider the number of outcomes where the sum is > 10
There are 6 such outcomes.
They are: (4,6), (6,4), (6,5), (5,6), (6,6), (5,5)
Since there are 36 outcomes altogether, we know that the other 30 outcomes must be such that the sum is less than 10.


So, P(sum < 10) = [total number of outcomes where the sum < 10] / [total number of outcomes]
= 30 / 36
= 5/6
= A

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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