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# John rolls a fair six-sided die with faces numbered 1 throug

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John rolls a fair six-sided die with faces numbered 1 throug [#permalink]  20 Dec 2018, 18:42
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Question Stats:

39% (00:43) correct 60% (00:31) wrong based on 63 sessions

John rolls a fair six-sided die with faces numbered 1 through 6 twice.

 Quantity A Quantity B The probability that both rolls are even The probability that both rolls are not odd

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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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: John rolls a fair six-sided die with faces numbered 1 throug [#permalink]  21 Dec 2018, 09:34
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the question says
'The probability that both rolls are even'
meaning either 2 or 4 or 6 on both the die
ie 3/6 * 3/6 which is 1/4(A)
whereas
'The probability that both rolls are not odd'
means that the dies cannot have odd on the both of them together
so 1 - probability that both rolls are odd

ie
1 - (3/6 * 3/6) which is 1 - 1/4
which is 3/4(B) thus B is bigger than option A
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Re: John rolls a fair six-sided die with faces numbered 1 throug [#permalink]  23 Dec 2018, 18:27
1
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The first thing that should be considered here is that these two statements are not complementary:
Probability of a side to be even: P(E)
Probability of a side to be odd: P(O)
A: P(E and E)
B: P(E and O) + P(O and E) + P(E and E) = 1 - P(O and O)
Total possible occurrences: 6*6 = 36

As P(E and O) and P(O and E) can’t be negative, seemingly B is larger than A. But let’s try:
A: P(E and E) = 9/36 = 1/4
(2 | 4 | 6, 2| 4| 6) -> (2,2), (2,4),(2,6), (4,2), …, (6,6) [9 ones]

B: P(E and O) + P(O and E) + P(E and E) = 3 * 9/36 = 3/4

P(E and O) = 9/36
(2|4|6, 1,3,5) -> 3*3 = 9
P(O and E) = 9/36
(1,3,5, 2|4|6) -> 3*3 = 9
P(E and E)= 9/36

So B is bigger than A.
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Re: John rolls a fair six-sided die with faces numbered 1 throug [#permalink]  01 Jun 2019, 04:41
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chetan2u wrote:

John rolls a fair six-sided die with faces numbered 1 through 6 twice.

 Quantity A Quantity B The probability that both rolls are even The probability that both rolls are not odd

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.

--------CONCEPT---------

In probability, especially dices, there are always very limited outcomes with a particular logic
So, only 4 possibilities in throwing 2 dices exist:

1. Both even
2. Both odd
3. First even & Then odd
4. First odd then even

So, both even probability = 1/4
And both not odd probability = 4/4 - 1/4 = 3/4

Hence, B>A......option B
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Re: John rolls a fair six-sided die with faces numbered 1 throug [#permalink]  01 Jun 2019, 05:29
GREsucks wrote:
chetan2u wrote:

John rolls a fair six-sided die with faces numbered 1 through 6 twice.

 Quantity A Quantity B The probability that both rolls are even The probability that both rolls are not odd

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.

--------CONCEPT---------

In probability, especially dices, there are always very limited outcomes with a particular logic
So, only 4 possibilities in throwing 2 dices exist:

1. Both even
2. Both odd
3. First even & Then odd
4. First odd then even

So, both even probability = 1/4
And both not odd probability = 4/4 - 1/4 = 3/4

Hence, B>A......option B

That's great concept.

But I would like to mention one point

Since the number of odds = no. of even, in the fair six sided dice

SO once we get the probability that both are even = $$\frac{1}{4}$$= probability of both are odd

Hence the probability of both not odd (As per complement rule) = $$1 - \frac{1}{4} = \frac{3}{4}$$
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Re: John rolls a fair six-sided die with faces numbered 1 throug [#permalink]  02 Jun 2019, 02:53
nice question
Re: John rolls a fair six-sided die with faces numbered 1 throug   [#permalink] 02 Jun 2019, 02:53
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