Last visit was: 19 Nov 2025, 09:24 It is currently 19 Nov 2025, 09:24
Home
GMAT
Messages
Tests
Schools
Events
Rewards
Chat
My Profile
Close
GMAT 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 GMAT 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.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
555-605 Level|   Probability|      
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,389
Own Kudos:
778,286
 [32]
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,389
Kudos: 778,286
 [32]
A fair die, with sides numbered from 1 to 6, is rolled repeatedly 24 Dec 2018, 01:00
4
Kudos
Add Kudos
28
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

35% (medium)

Question Stats:

79% (02:11) correct 21% (02:38) wrong based on 944 sessions

History

Date
Time
Result
Not Attempted Yet
A fair die, with sides numbered from 1 to 6, is rolled repeatedly until a number greater than 4 first appears. What is the probability that this occurs on the third or fourth roll?

A. 8/81

B. 4/27

C. 16/27

D. 2/9

E. 20/81

Experience GMAT Club Test Questions
Yes, you've landed on a GMAT Club Tests question
Craving more? Unlock our full suite of GMAT Club Tests here
Want to experience more? Get a taste of our tests with our free trial today
Rise to the challenge with GMAT Club Tests. Happy practicing!
GMAT Club Tests Check Our Products Try Free Tests
ShowHide Answer
Official Answer
E
Most Helpful Reply
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,389
Own Kudos:
778,286
 [6]
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,389
Kudos: 778,286
 [6]
A fair die, with sides numbered from 1 to 6, is rolled repeatedly 04 Jul 2019, 08:48
2
Kudos
Add Kudos
4
Bookmarks
Bookmark this Post
A fair die, with sides numbered from 1 to 6, is rolled repeatedly until a number greater than 4 first appears. What is the probability that this occurs on the third or fourth roll?

A. \(\frac{8}{81}\)
B. \(\frac{4}{27}\)
C. \(\frac{16}{27}\)
D. \(\frac{2}{9}\)
E. \(\frac{20}{81}\)


A die has two numbers greater than 4: 5 and 6. The probability of rolling a number greater than 4 (either 5 or 6) is \(\frac{2}{6} = \frac{1}{3}\). Consequently, the probability of rolling a 4 or less is \(1- \frac{1}{3}= \frac{2}{3}\).

We want to determine the probability that a number greater than 4 will first appear on the third or fourth roll.

The probability of a number greater than 4 first appearing on the third roll \(= P(4 \ or \ less; \ 4 \ or \ less; \ more \ than \ 4) = \frac{2}{3}* \frac{2}{3}* \frac{1}{3}=\frac{4}{27}\)

The probability of a number greater than 4 first appearing on the fourth roll \(= P(4 \ or \ less; \ 4 \ or \ less; \ 4 \ or \ less; \ more \ than \ 4) = \frac{2}{3}* \frac{2}{3}*\frac{2}{3}* \frac{1}{3}=\frac{8}{81}\)

Therefore, the combined probability that a number greater than 4 will first appear on either the third or fourth roll is \(\frac{4}{27} + \frac{8}{81} = \frac{20}{81}\).


Answer: E
General Discussion
User avatar
Archit3110
User avatar
Major Poster
Joined: 18 Aug 2017
Last visit: 19 Nov 2025
Posts: 8,422
Own Kudos:
4,981
 [2]
Given Kudos: 243
Status:You learn more from failure than from success.
Location: India
Concentration: Sustainability, Marketing
GMAT Focus 1: 545 Q79 V79 DI73
GMAT Focus 2: 645 Q83 V82 DI81
GPA: 4
WE:Marketing (Energy)
GMAT Focus 2: 645 Q83 V82 DI81
Posts: 8,422
Kudos: 4,981
 [2]
A fair die, with sides numbered from 1 to 6, is rolled repeatedly 24 Dec 2018, 02:21
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
A fair die with sides numbered 1, 2, 3, 4, 5, and 6 is to be rolled until a number greater than 4 first appears. What is the probability that a number greater than 4 will first appear on the third or fourth roll?

A. 8/81

B. 4/27

C. 16/27

D. 2/9

E. 20/81
Show more


for third roll no >4
4/6*4/6*2/6 = 4/27

for 4th roll no>4
4/6*4/6*4/6*2/6 = 8/81

4/27+8/81 = 20/81 IMOE
avatar
MPhizzle
avatar
Joined: 16 Jul 2017
Last visit: 05 May 2019
Posts: 10
Own Kudos:
2
Given Kudos: 139
Posts: 10
Kudos: 2
A fair die, with sides numbered from 1 to 6, is rolled repeatedly 28 Dec 2018, 11:32
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Archit3110
Bunuel
A fair die with sides numbered 1, 2, 3, 4, 5, and 6 is to be rolled until a number greater than 4 first appears. What is the probability that a number greater than 4 will first appear on the third or fourth roll?

A. 8/81

B. 4/27

C. 16/27

D. 2/9

E. 20/81


for third roll no >4
4/6*4/6*2/6 = 4/27

for 4th roll no>4
4/6*4/6*4/6*2/6 = 8/81

4/27+8/81 = 20/81 IMOE
Show more

Can you explain why it is 2/6? I thought it would be 4/6 for each of the three rolls.
thanks.
User avatar
Archit3110
User avatar
Major Poster
Joined: 18 Aug 2017
Last visit: 19 Nov 2025
Posts: 8,422
Own Kudos:
4,981
Given Kudos: 243
Status:You learn more from failure than from success.
Location: India
Concentration: Sustainability, Marketing
GMAT Focus 1: 545 Q79 V79 DI73
GMAT Focus 2: 645 Q83 V82 DI81
GPA: 4
WE:Marketing (Energy)
GMAT Focus 2: 645 Q83 V82 DI81
Posts: 8,422
Kudos: 4,981
A fair die, with sides numbered from 1 to 6, is rolled repeatedly 06 Feb 2019, 08:57
Kudos
Add Kudos
Bookmarks
Bookmark this Post
MPhizzle
Archit3110
Bunuel
A fair die with sides numbered 1, 2, 3, 4, 5, and 6 is to be rolled until a number greater than 4 first appears. What is the probability that a number greater than 4 will first appear on the third or fourth roll?

A. 8/81

B. 4/27

C. 16/27

D. 2/9

E. 20/81


for third roll no >4
4/6*4/6*2/6 = 4/27

for 4th roll no>4
4/6*4/6*4/6*2/6 = 8/81

4/27+8/81 = 20/81 IMOE

Can you explain why it is 2/6? I thought it would be 4/6 for each of the three rolls.
thanks.
Show more

MPhizzle

question has asked that no >4 appears after 3rd or 4th roll
so here 4/6 ; means : 1,2,3,4 digits and 2/6 refers to 5/6
so no >4 appearing on 3rd roll
4/6*4/6*2/6

and no>4 appearing after 4th roll
4/6*4/6*4/6*2/6

hope this helps..
avatar
Shishou
avatar
Joined: 10 Jun 2019
Last visit: 08 Apr 2021
Posts: 103
Own Kudos:
91
Given Kudos: 112
Products:
My Reviews
Posts: 103
Kudos: 91
A fair die, with sides numbered from 1 to 6, is rolled repeatedly 05 Jul 2019, 02:00
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
Bunuel
A fair die with sides numbered 1, 2, 3, 4, 5, and 6 is to be rolled until a number greater than 4 first appears. What is the probability that a number greater than 4 will first appear on the third or fourth roll?

A. 8/81

B. 4/27

C. 16/27

D. 2/9

E. 20/81

Official Solution:


A fair die with sides numbered 1, 2, 3, 4, 5, and 6 is to be rolled until a number greater than 4 first appears. What is the probability that a number greater than 4 will first appear on the third or fourth roll?


A. \(\frac{8}{81}\)
B. \(\frac{4}{27}\)
C. \(\frac{16}{27}\)
D. \(\frac{2}{9}\)
E. \(\frac{20}{81}\)


The numbers greater than 4 on a die are 5 and 6. The probability of getting a number greater than 4 (so 5 or 6) is \(\frac{2}{6} = \frac{1}{3}\) and the probability of getting 4 or less is therefore, \(1- \frac{1}{3}= \frac{2}{3}\)

We need to find the probability that a number greater than 4 will first appear on the third or fourth roll.

The probability of a number greater than 4 first appearing on third roll \(= P(4 \ or \ less; \ 4 \ or \ less; \ more \ than \ 4) = \frac{2}{3}* \frac{2}{3}* \frac{1}{3}=\frac{4}{27}\)

The probability of a number greater than 4 first appearing on fourth roll \(= P(4 \ or \ less; \ 4 \ or \ less; \ 4 \ or \ less; \ more \ than \ 4) = \frac{2}{3}* \frac{2}{3}*\frac{2}{3}* \frac{1}{3}=\frac{8}{81}\)

Thus, the probability that a number greater than 4 will first appear on the third or fourth roll \(= \frac{4}{27} + \frac{8}{81}=\frac{20}{81}\)


Answer: E
Show more
?

Hi Bunuel why did we not multiply the 2 probabilities by 3C2 and the second by 4C3?I am a bit confused
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,389
Own Kudos:
778,286
 [1]
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,389
Kudos: 778,286
 [1]
A fair die, with sides numbered from 1 to 6, is rolled repeatedly 05 Jul 2019, 02:07
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Shishou
Bunuel
Bunuel
A fair die with sides numbered 1, 2, 3, 4, 5, and 6 is to be rolled until a number greater than 4 first appears. What is the probability that a number greater than 4 will first appear on the third or fourth roll?

A. 8/81

B. 4/27

C. 16/27

D. 2/9

E. 20/81

Official Solution:


A fair die with sides numbered 1, 2, 3, 4, 5, and 6 is to be rolled until a number greater than 4 first appears. What is the probability that a number greater than 4 will first appear on the third or fourth roll?


A. \(\frac{8}{81}\)
B. \(\frac{4}{27}\)
C. \(\frac{16}{27}\)
D. \(\frac{2}{9}\)
E. \(\frac{20}{81}\)


The numbers greater than 4 on a die are 5 and 6. The probability of getting a number greater than 4 (so 5 or 6) is \(\frac{2}{6} = \frac{1}{3}\) and the probability of getting 4 or less is therefore, \(1- \frac{1}{3}= \frac{2}{3}\)

We need to find the probability that a number greater than 4 will first appear on the third or fourth roll.

The probability of a number greater than 4 first appearing on third roll \(= P(4 \ or \ less; \ 4 \ or \ less; \ more \ than \ 4) = \frac{2}{3}* \frac{2}{3}* \frac{1}{3}=\frac{4}{27}\)

The probability of a number greater than 4 first appearing on fourth roll \(= P(4 \ or \ less; \ 4 \ or \ less; \ 4 \ or \ less; \ more \ than \ 4) = \frac{2}{3}* \frac{2}{3}*\frac{2}{3}* \frac{1}{3}=\frac{8}{81}\)

Thus, the probability that a number greater than 4 will first appear on the third or fourth roll \(= \frac{4}{27} + \frac{8}{81}=\frac{20}{81}\)


Answer: E
?

Hi Bunuel why did we not multiply the 2 probabilities by 3C2 and the second by 4C3?I am a bit confused
Show more

Because in each case we are interested in that specific sequence of events. For example, we need \((4 \ or \ less; \ 4 \ or \ less; \ more \ than \ 4)\) and not say \((more \ than \ 4; \ 4 \ or \ less; \ 4 \ or \ less)\) because in that case condition that a number greater than 4 will first appear on the third roll would not be met.
User avatar
BrushMyQuant
User avatar
Joined: 05 Apr 2011
Last visit: 21 Oct 2025
Posts: 2,284
Own Kudos:
2,552
 [1]
Given Kudos: 100
Status:Tutor - BrushMyQuant
Location: India
Concentration: Finance, Marketing
Schools: XLRI (A)
GMAT 1: 700 Q51 V31
GPA: 3
WE:Information Technology (Computer Software)
Expert
Expert reply
Schools: XLRI (A)
GMAT 1: 700 Q51 V31
Posts: 2,284
Kudos: 2,552
 [1]
A fair die, with sides numbered from 1 to 6, is rolled repeatedly 16 Oct 2022, 01:48
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Given that A fair die with sides numbered 1, 2, 3, 4, 5, and 6 is to be rolled until a number greater than 4 first appears and We need to find What is the probability that a number greater than 4 will first appear on the third or fourth roll?

At each stage The Probability of Getting a number Greater than 4 = \(\frac{2}{6}\) (As there are two numbers greater than 4 i.e. 5 and 6 out of 1 to 6) = \(\frac{1}{3}\)

Similarly, Probability of Getting a number less than or equal to 4 = \(\frac{4}{6}\) = \(\frac{2}{3}\)

Probability that a number greater than 4 will first appear on the third or fourth roll = P(number greater than 4 will first appear on the third roll) + P(number greater than 4 will first appear on the fourth roll)

P(number greater than 4 will first appear on the third roll) = P(First Roll will give ≤ 4) * P(Second Roll will give ≤ 4) * P(Third Roll will give > 4) = \(\frac{2}{3}\) * \(\frac{2}{3}\) * \(\frac{1}{3}\) = \(\frac{4}{27}\)

P(number greater than 4 will first appear on the fourth roll) = P(First Roll will give ≤ 4) * P(Second Roll will give ≤ 4) * P(Third Roll will give ≤ 4) * P(Fourth Roll will give > 4) = \(\frac{2}{3}\) * \(\frac{2}{3}\) * \(\frac{2}{3}\) *\(\frac{1}{3}\) = \(\frac{8}{81}\)

Probability that a number greater than 4 will first appear on the third or fourth roll = P(number greater than 4 will first appear on the third roll) + P(number greater than 4 will first appear on the fourth roll) = \(\frac{4}{27}\) + \(\frac{8}{81}\) = \(\frac{(12 + 8)}{81}\) = \(\frac{20}{81}\)

So, Answer will be E
Hope it helps!

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

User avatar
pierjoejoe
User avatar
Joined: 30 Jul 2024
Last visit: 29 Jul 2025
Posts: 127
Own Kudos:
55
Given Kudos: 425
Location: Italy
Concentration: Accounting, Finance
GMAT Focus 1: 645 Q84 V84 DI78
GPA: 4
WE:Research (Technology)
GMAT Focus 1: 645 Q84 V84 DI78
Posts: 127
Kudos: 55
A fair die, with sides numbered from 1 to 6, is rolled repeatedly 09 Sep 2024, 02:46
Kudos
Add Kudos
Bookmarks
Bookmark this Post
in order to role a number greater than 4 at third or fourth role we need

third role P(x<=4)^2 * (1-P(x<=4))
fourth role P(x<=4)^3 * (1-P(x<=4))

and sum them together
P(x<=4) = 1/6 + 1/6 + 1/6 + 1/6 = 2/3

so the result is
P(third) = (2/3)^2 * 1/3 = 4/27
P(fourth) = (2/3)^3 * 1/3 = 8/81

P(third or fourth) = P(third) + P(fourth) = 4/27 + 8/81 = 20/81
User avatar
diva_jaxn
User avatar
Joined: 01 Nov 2023
Last visit: 19 Nov 2025
Posts: 24
Own Kudos:
4
Given Kudos: 83
Location: India
Schools: ISB '26 Booth (Chicago) CBS (S) HEC
GPA: 9.35
Products:
Forum Quiz
Schools: ISB '26 Booth (Chicago) CBS (S) HEC
Posts: 24
Kudos: 4
A fair die, with sides numbered from 1 to 6, is rolled repeatedly 19 Sep 2025, 05:32
Kudos
Add Kudos
Bookmarks
Bookmark this Post
here why do we multiply and not add
Bunuel
A fair die, with sides numbered from 1 to 6, is rolled repeatedly until a number greater than 4 first appears. What is the probability that this occurs on the third or fourth roll?

A. \(\frac{8}{81}\)
B. \(\frac{4}{27}\)
C. \(\frac{16}{27}\)
D. \(\frac{2}{9}\)
E. \(\frac{20}{81}\)


A die has two numbers greater than 4: 5 and 6. The probability of rolling a number greater than 4 (either 5 or 6) is \(\frac{2}{6} = \frac{1}{3}\). Consequently, the probability of rolling a 4 or less is \(1- \frac{1}{3}= \frac{2}{3}\).

We want to determine the probability that a number greater than 4 will first appear on the third or fourth roll.

The probability of a number greater than 4 first appearing on the third roll \(= P(4 \ or \ less; \ 4 \ or \ less; \ more \ than \ 4) = \frac{2}{3}* \frac{2}{3}* \frac{1}{3}=\frac{4}{27}\)

The probability of a number greater than 4 first appearing on the fourth roll \(= P(4 \ or \ less; \ 4 \ or \ less; \ 4 \ or \ less; \ more \ than \ 4) = \frac{2}{3}* \frac{2}{3}*\frac{2}{3}* \frac{1}{3}=\frac{8}{81}\)

Therefore, the combined probability that a number greater than 4 will first appear on either the third or fourth roll is \(\frac{4}{27} + \frac{8}{81} = \frac{20}{81}\).


Answer: E
Show more
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,389
Own Kudos:
778,286
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,389
Kudos: 778,286
A fair die, with sides numbered from 1 to 6, is rolled repeatedly 19 Sep 2025, 05:55
Kudos
Add Kudos
Bookmarks
Bookmark this Post
diva_jaxn
here why do we multiply and not add

Show more

You multiply because the events must happen in sequence: for the third roll to be the first success, roll 1 must fail, roll 2 must fail, and roll 3 must succeed. Independent events in sequence are multiplied, not added.

You add only at the very end, when combining the probability of two different cases (third roll success or fourth roll success).
Moderators:
Bunuel
Math Expert
105389 posts
Krunaal
Tuck School Moderator
805 posts