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# Alejandro has a six-sided die with faces numbered 1 through

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Alejandro has a six-sided die with faces numbered 1 through [#permalink]  03 Jun 2017, 02:29
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72% (01:00) correct 27% (01:10) wrong based on 74 sessions

Alejandro has a six-sided die with faces numbered 1 through 6. He rolls the die twice.

 Quantity A Quantity B The probability that both rolls are even The probability that neither roll is a multiple of 3

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
[Reveal] Spoiler: OA

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Re: Alejandro has a six-sided die with faces numbered 1 through [#permalink]  21 Sep 2017, 04:39
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The probability that the two rolls are even is 3/6*3/6 since there are three numbers out of six that are even (2, 4, 6). The probability that neither roll is a multiple of 3 is 4/6*4/6 because there are only two numbers who are multiple of 3, 3 and 6. Thus comparing 1/4 to 4/9 leads to the choice B!
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Re: Alejandro has a six-sided die with faces numbered 1 through [#permalink]  20 Dec 2018, 16:10
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A: All possibilities: 6*6 = 36
The ones which both parameters are even:
([2 | 4 | 6] , 2) -> 3
([2 | 4 | 6] , 4) -> 3
([2 | 4 | 6] ,6) -> 3
Probability = 9/36

B: All possibilities: 6*6 = 36
The ones which neither roll is a multiple of 3:
We need to eliminate 3 and 6 from our numbers, so we will have: 4*4 = 16
Probability=16/36

B is bigger than A.
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Re: Alejandro has a six-sided die with faces numbered 1 through [#permalink]  20 Dec 2018, 18:37
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Expert's post
Carcass wrote:

Alejandro has a six-sided die with faces numbered 1 through 6. He rolls the die twice.

 Quantity A Quantity B The probability that both rolls are even The probability that neither roll is a multiple of 3

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.

Without getting into probability, the answer will be the choice that has more ways..
The probability that both rolls are even - so there are 3 ways in which it can be even .. the ways = 3*3=9
The probability that neither roll is a multiple of 3 - so the wording has changed from both to neither. thus there are 4 ways of not having a multiple of 3, => 4*4=16

Thus 16>9 or B>A
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Some useful Theory.
1. Arithmetic and Geometric progressions : https://greprepclub.com/forum/progressions-arithmetic-geometric-and-harmonic-11574.html#p27048
2. Effect of Arithmetic Operations on fraction : https://greprepclub.com/forum/effects-of-arithmetic-operations-on-fractions-11573.html?sid=d570445335a783891cd4d48a17db9825
3. Remainders : https://greprepclub.com/forum/remainders-what-you-should-know-11524.html
4. Number properties : https://greprepclub.com/forum/number-property-all-you-require-11518.html
5. Absolute Modulus and Inequalities : https://greprepclub.com/forum/absolute-modulus-a-better-understanding-11281.html

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Re: Alejandro has a six-sided die with faces numbered 1 through [#permalink]  21 Dec 2018, 10:20
The probability that the two rolls are even is $$\frac{3}{6} * \frac{3}{6} which is \frac{1}{4}$$
possible even nos are :2, 4 and 6.
The probability that neither roll is a multiple of 3 becomes
$$\frac{4}{6} * \frac{4}{6}$$ which is $$\frac{4}{9}$$
Re: Alejandro has a six-sided die with faces numbered 1 through   [#permalink] 21 Dec 2018, 10:20
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