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Two fair dice are thrown.
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31 Aug 2022, 04:06
31 Aug 2022, 04:06
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66% (02:08) correct
33% (02:15) wrong based on 6 sessions
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Two fair dice are thrown. What is the probability that the outcome will either total 7 or include a 3?
(A) 7/12
(B) 5/12
(C) 8/9
(D) 2/3
(E) 1/2
(A) 7/12
(B) 5/12
(C) 8/9
(D) 2/3
(E) 1/2
Re: Two fair dice are thrown.
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10 Sep 2022, 22:18
10 Sep 2022, 22:18
1
here total number of outcomes = (1,1), (1,2), (1,3), ......., (6,4), (6,5), (6,6) = 36
desired outcomes that result in a sum to be 7 = (1,6), (6,1), (2,5), (5,2), (3,4), (4,3) = 6
P (sum 7) = 6/36 = 1/6
outcomes with a 3 = (1,3), (2,3), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,3), (5,3), (6,3) = 11
(3,4) and (4,3) are already included so we do not count them to calculate the probability of getting a 3.
therefore, desired outcomes = 11-2 = 9
P (a 3) = 9/36 = 1/4
now, P (sum 7 or a 3) = 1/6 + 1/4 = 10/24 = 5/12
desired outcomes that result in a sum to be 7 = (1,6), (6,1), (2,5), (5,2), (3,4), (4,3) = 6
P (sum 7) = 6/36 = 1/6
outcomes with a 3 = (1,3), (2,3), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,3), (5,3), (6,3) = 11
(3,4) and (4,3) are already included so we do not count them to calculate the probability of getting a 3.
therefore, desired outcomes = 11-2 = 9
P (a 3) = 9/36 = 1/4
now, P (sum 7 or a 3) = 1/6 + 1/4 = 10/24 = 5/12
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Re: Two fair dice are thrown.
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15 Oct 2022, 10:23
15 Oct 2022, 10:23
Given that Two fair dice are thrown and We need to find What is the probability that the outcome will either total 7 or include a 3?
As we are rolling two dice => Number of cases = \(6^2\) = 36
Now for the sum to be 7 or include a 3, we will have two cases
1. Sum is 7 and neither of the two outcomes are 3
2. Either of the two outcomes is 3
Sum is 7 and neither of the two outcomes are 3
Following outcomes are possible
(1,6), (2,5), (5,2), (6,1) = 4 cases
Either of the two outcomes is 3 and sum is not 7
First outcome can be 3 and second outcome can be any number from 1 to 6 => 6 cases
Second outcome can be 3 and first outcome can be any number apart from 3 (as we have already considered that in first case) => 5 cases (First outcome can be 1, 2, 4, 5, 6)
=> 6 + 5 = 11 cases
=> Total cases = 4 + 11 = 15 cases
=> Probability that the outcome will either total 7 or include a 3 = \(\frac{15}{36}\) = \(\frac{5}{12}\)
So, Answer will be B
Hope it helps!
Watch the following video to learn How to Solve Dice Rolling Probability Problems
As we are rolling two dice => Number of cases = \(6^2\) = 36
Now for the sum to be 7 or include a 3, we will have two cases
1. Sum is 7 and neither of the two outcomes are 3
2. Either of the two outcomes is 3
Sum is 7 and neither of the two outcomes are 3
Following outcomes are possible
(1,6), (2,5), (5,2), (6,1) = 4 cases
Either of the two outcomes is 3 and sum is not 7
First outcome can be 3 and second outcome can be any number from 1 to 6 => 6 cases
Second outcome can be 3 and first outcome can be any number apart from 3 (as we have already considered that in first case) => 5 cases (First outcome can be 1, 2, 4, 5, 6)
=> 6 + 5 = 11 cases
=> Total cases = 4 + 11 = 15 cases
=> Probability that the outcome will either total 7 or include a 3 = \(\frac{15}{36}\) = \(\frac{5}{12}\)
So, Answer will be B
Hope it helps!
Watch the following video to learn How to Solve Dice Rolling Probability Problems
