Answer: B
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.

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) = \(1 - \frac{1}{4} = \frac{3}{4}\)

nice question

this question is GRE-level. Nice

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.[/quote]

--------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[/quote]


very helpful logic and concept.
Thanks for sharing.

This question is worded poorly from a grammar perspective, such that one could plainly read it as meaning the probability that neither role contains an odd, as opposed to one odd and one even being permitted. In plain English, if someone says to me, "both of my socks are not green," I would not read that as one could be green and one could be red. I would read that as them telling me neither is green. The GRE would not write a question which was ambiguous in this manner. I think what they mean for B, if they want to allow for the possibility of one odd and one even, is "the probability that at least one of the rolls is not odd," which is not ambiguous.

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

As we are rolling two dice => Number of cases = \(6^2\) = 36

Quantity A: The probability that both rolls are even

Probability of getting an even number in any roll = \(\frac{3}{6}\) (As there are three even numbers 2, 4, 6 out of the 6 numbers) = \(\frac{1}{2}\)

=> Probability that both rolls are even = \(\frac{1}{2}\) * \(\frac{1}{2}\) = \(\frac{1}{4}\) = 0.25

Quantity B: The probability that both rolls are not odd

Probability that both rolls are not odd = 1 - P(Both odd)

Probability of getting an odd number in any roll = \(\frac{3}{6}\) (As there are three even numbers 2, 4, 6 out of the 6 numbers) = \(\frac{1}{2}\)

=> Probability that both rolls are odd = \(\frac{1}{2}\) * \(\frac{1}{2}\) = \(\frac{1}{4}\)

=> Probability that both rolls are not odd = 1 - P(Both odd) = 1 - \(\frac{1}{4}\) = \(\frac{3}{4}\) = 0.75

Clearly, Quantity B(0.75) > Quantity A(0.25)

So, Answer will be B
Hope it helps!

Watch the following video to learn How to Solve Dice Rolling Probability Problems

C.
"Even" and "not odd" are synonymous and express the exact same meaning. This is because a number on a die can only be either even or odd, with no other possibilities. Thus, the two statements are syntactically identical

Notice that the word "both" under quantity B is not negated.

Part B in this question says "The probability that both rolls are not odd".
Would the answer change if Part B is worded as "The probability that not both rolls are odd".

I would argue that these two statements don't mean the same, and the solution provided above might apply more to the second statement "The probability that not both rolls are odd" . Given the insufficiency of data, I would be open to the argument that D could be a potential answer as well.

I think is C

Step 1: Interpret the problem

We have a fair six-sided die: faces $\(1,2,3,4,5,6\)$.
John rolls it twice.
Even numbers on die $\(=\{2,4,6\} \rightarrow 3\)$ even faces.
Odd numbers on die $\(=\{1,3,5\} \rightarrow 3\)$ odd faces.

Step 2: Quantity A - Probability that both rolls are even

$$
\(\begin{gathered}
P(\text { first even })=\frac{3}{6}=\frac{1}{2} \\
P(\text { second even })=\frac{3}{6}=\frac{1}{2} \\
P(\text { both even\) })=\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}
\end{gathered}
$$


So Quantity $\(\mathbf{A}=\frac{1}{4}\)$.


Step 3: Quantity B - Probability that both rolls are not odd

Not odd means even (since die numbers are either even or odd).
"Both rolls are not odd" = "Both rolls are even".

That's exactly the same as Quantity A!
So Quantity B $\(=\frac{1}{4}\)$.

Step 4: Compare
Quantity $\(\mathrm{A}=\frac{1}{4}\)$
Quantity B $\(=\frac{1}{4}\)$