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What is the probability that the sum of two dice will yield

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What is the probability that the sum of two dice will yield a 7, and then when both are thrown again, their sum will again yield a 7? assume that each die has 6 sides with faces numbered 1 to 6.

A. \(1/144\)
B. \(1/36\)
C. \(1/12\)
D. \(1/6\)
E. \(1/3\)

OA B
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by deloitte247 » Sat Jul 06, 2019 3:11 am
Total number of ways in which 2 dice can be thrown = 6 * 6 = 36
The combination that will yield a sum of 7 are;
$$1+6$$
$$2+5$$
$$3+4$$
$$4+3$$
$$5+2$$
$$6+1$$
All the 6 different combinations probability of rolling a 7 is 6/36 or 1/6.
To find the probability of this happening twice in a row =>
$$\frac{1}{6}\cdot\frac{1}{6}=\frac{1}{36}\ \ \ \ \ \ Answer=option\ B$$
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by Ian Stewart » Sat Jul 06, 2019 4:12 am
If you roll one die, you can always make a sum of 7 if you get the perfect roll on the second die. There's a 1/6 chance you roll the perfect number, so that's the probability of getting a sum of 7. The probability we do that twice in a row is (1/6)(1/6) = 1/36.

You could get down to A or B instantly by noticing that the answer needs to be the square of some very simple fraction of integers.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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by Scott@TargetTestPrep » Thu Jul 11, 2019 6:58 pm
AAPL wrote:Manhattan Prep

What is the probability that the sum of two dice will yield a 7, and then when both are thrown again, their sum will again yield a 7? assume that each die has 6 sides with faces numbered 1 to 6.

A. \(1/144\)
B. \(1/36\)
C. \(1/12\)
D. \(1/6\)
E. \(1/3\)

OA B
Since there are a total of 36 outcomes, and 6 of them yield a sum of 7 (notice that these 6 outcomes are: {1, 6}, {2, 5}, {3, 4}, {4, 3}, {5, 2} and {6, 1}), the probability that the two numbers on the dice yield a sum of 7 is 6/36 = 1/6. The probability that their sum will again yield a 7 is 1/6. Therefore, the probability that the two dice will yield a sum of 7 twice is 1/6 x 1/6 = 1/36.

Answer: B

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