Last visit was: 17 Dec 2024, 21:57 It is currently 17 Dec 2024, 21:57

Close

GRE Prep Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GRE score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel
avatar
Supreme Moderator
Joined: 01 Nov 2017
Posts: 371
Own Kudos [?]: 471 [17]
Given Kudos: 0
Send PM
Most Helpful Community Reply
avatar
Intern
Intern
Joined: 24 May 2019
Posts: 28
Own Kudos [?]: 39 [7]
Given Kudos: 0
Send PM
General Discussion
avatar
Intern
Intern
Joined: 21 Dec 2018
Posts: 20
Own Kudos [?]: 8 [2]
Given Kudos: 0
Send PM
avatar
Manager
Manager
Joined: 22 Feb 2018
Posts: 163
Own Kudos [?]: 215 [2]
Given Kudos: 0
Send PM
Re: John rolls a fair six-sided die with faces numbered 1 throug [#permalink]
2
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.
avatar
Retired Moderator
Joined: 20 Apr 2016
Posts: 1307
Own Kudos [?]: 2280 [1]
Given Kudos: 251
WE:Engineering (Energy and Utilities)
Send PM
Re: John rolls a fair six-sided die with faces numbered 1 throug [#permalink]
1
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) = \(1 - \frac{1}{4} = \frac{3}{4}\)
avatar
Intern
Intern
Joined: 26 May 2018
Posts: 37
Own Kudos [?]: 9 [0]
Given Kudos: 0
Send PM
Re: John rolls a fair six-sided die with faces numbered 1 throug [#permalink]
nice question
Manager
Manager
Joined: 02 Sep 2019
Posts: 181
Own Kudos [?]: 144 [0]
Given Kudos: 94
Concentration: Finance
GRE 1: Q151 V148
GPA: 3.14
Send PM
Re: John rolls a fair six-sided die with faces numbered 1 throug [#permalink]
this question is GRE-level. Nice
Intern
Intern
Joined: 19 Mar 2022
Posts: 43
Own Kudos [?]: 30 [1]
Given Kudos: 114
GRE 1: Q163 V156
Send PM
Re: John rolls a fair six-sided die with faces numbered 1 throug [#permalink]
1
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.
Intern
Intern
Joined: 08 Aug 2022
Posts: 49
Own Kudos [?]: 38 [2]
Given Kudos: 98
Send PM
John rolls a fair six-sided die with faces numbered 1 throug [#permalink]
2
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.
Moderator
Moderator
Joined: 02 Jan 2020
Status:GRE Quant Tutor
Posts: 1115
Own Kudos [?]: 974 [0]
Given Kudos: 9
Location: India
Concentration: General Management
Schools: XLRI Jamshedpur, India - Class of 2014
GMAT 1: 700 Q51 V31
GPA: 2.8
WE:Engineering (Computer Software)
Send PM
John rolls a fair six-sided die with faces numbered 1 throug [#permalink]
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

Intern
Intern
Joined: 29 Jun 2022
Posts: 4
Own Kudos [?]: 1 [1]
Given Kudos: 1
Send PM
Re: John rolls a fair six-sided die with faces numbered 1 throug [#permalink]
1
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.
Prep Club for GRE Bot
Re: John rolls a fair six-sided die with faces numbered 1 throug [#permalink]
Moderators:
GRE Instructor
88 posts
GRE Forum Moderator
37 posts
Moderator
1115 posts
GRE Instructor
234 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne