Last visit was: 19 May 2025, 14:59 It is currently 19 May 2025, 14:59
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.
Close
Request Expert Reply
Confirm Cancel
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 May 2025
Posts: 101,532
Own Kudos:
Given Kudos: 93,558
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 101,532
Kudos: 725,563
 [32]
4
Kudos
Add Kudos
28
Bookmarks
Bookmark this Post
Most Helpful Reply
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 May 2025
Posts: 101,532
Own Kudos:
Given Kudos: 93,558
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 101,532
Kudos: 725,563
 [14]
6
Kudos
Add Kudos
7
Bookmarks
Bookmark this Post
General Discussion
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 May 2025
Posts: 101,532
Own Kudos:
Given Kudos: 93,558
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 101,532
Kudos: 725,563
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
Mohit1994
Joined: 12 Dec 2022
Last visit: 14 Dec 2024
Posts: 20
Own Kudos:
3
 [1]
Given Kudos: 82
Location: India
Posts: 20
Kudos: 3
 [1]
Kudos
Add Kudos
Bookmarks
Bookmark this Post
i think ans should be -

Probability- 1/9 *1/8*1/7

as 1st couple is already selected 1 desert and now 2nd couple need to select 1 desert from 8 and for 3rd couple need to select 1 desert from 7.???
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 May 2025
Posts: 101,532
Own Kudos:
725,563
 [2]
Given Kudos: 93,558
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 101,532
Kudos: 725,563
 [2]
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Mohit1994
i think ans should be -

Probability- 1/9 *1/8*1/7

as 1st couple is already selected 1 desert and now 2nd couple need to select 1 desert from 8 and for 3rd couple need to select 1 desert from 7.???

Why? Why cannot all three couples order the same dessert? Are desserts running out?
User avatar
Mohit1994
Joined: 12 Dec 2022
Last visit: 14 Dec 2024
Posts: 20
Own Kudos:
Given Kudos: 82
Location: India
Posts: 20
Kudos: 3
Kudos
Add Kudos
Bookmarks
Bookmark this Post
As mentioned in the question that -- this dessert does not need to match the one ordered by the other couples.
so each desert selected by every couple shld be different right ??
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 May 2025
Posts: 101,532
Own Kudos:
725,563
 [2]
Given Kudos: 93,558
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 101,532
Kudos: 725,563
 [2]
1
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Mohit1994
As mentioned in the question that -- this dessert does not need to match the one ordered by the other couples.
so each desert selected by every couple shld be different right ??

No, the statement in the question "Note that while each couple is required to order the same dessert, this dessert does not need to match the one ordered by the other couples" implies that within a single couple, both members must order identical desserts. Separate couples can indeed choose the same dessert, but they are not necessarily required to do so. This means that while individuals within a couple must have matching desserts, there is no requirement either preventing or mandating different couples from selecting the same dessert.
User avatar
CoffeeNCream
Joined: 10 Nov 2022
Last visit: 13 Sep 2024
Posts: 9
Own Kudos:
Given Kudos: 11
Location: Australia
Posts: 9
Kudos: 2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hi Bunuel, beautiful question.

Question - In approaching this, I had 6 slots, 2 for each couple of husband and wife

First slot, husband has 1/9 probability of choosing a specific dessert, wife also has 1/9 of choosing that same dessert.

Why is the dessert the husband chooses 1 instead of 1/9?

Thank you for your time. I’ve been doing GMATClub quant tests and if anyone is reading this, don’t hesitate, buy it and study it. It is an amazing resource and the solutions and then discussions are amazing. Thank you for all you do Bunuel and team GMATClub!

Bunuel
Official Solution:


At a restaurant, three couples are preparing to order dessert. If there are 9 different desserts on the menu, what is the probability that the husband and wife of each couple will order the same dessert? Note that while each couple is required to order the same dessert, this dessert does not need to match the one ordered by the other couples.


A. \(\frac{6}{9^3}\)
B. \(\frac{3}{9^3}\)
C. \(\frac{1}{9^3}\)
D. \(\frac{1}{9^6}\)
E. \(\frac{6}{9^6}\)


For each couple, there's a 1/9 chance the wife will order the same dessert as her husband. With three couples, the combined probability is:

\(\frac{1}{9} *\frac{1}{9} *\frac{1}{9} = \frac{1}{9^3} \).


Answer: C

Posted from my mobile device
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 May 2025
Posts: 101,532
Own Kudos:
725,563
 [4]
Given Kudos: 93,558
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 101,532
Kudos: 725,563
 [4]
3
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
CoffeeNCream
Hi Bunuel, beautiful question.

Question - In approaching this, I had 6 slots, 2 for each couple of husband and wife

First slot, husband has 1/9 probability of choosing a specific dessert, wife also has 1/9 of choosing that same dessert.

Why is the dessert the husband chooses 1 instead of 1/9?

Thank you for your time. I’ve been doing GMATClub quant tests and if anyone is reading this, don’t hesitate, buy it and study it. It is an amazing resource and the solutions and then discussions are amazing. Thank you for all you do Bunuel and team GMATClub!

Bunuel
Official Solution:


At a restaurant, three couples are preparing to order dessert. If there are 9 different desserts on the menu, what is the probability that the husband and wife of each couple will order the same dessert? Note that while each couple is required to order the same dessert, this dessert does not need to match the one ordered by the other couples.


A. \(\frac{6}{9^3}\)
B. \(\frac{3}{9^3}\)
C. \(\frac{1}{9^3}\)
D. \(\frac{1}{9^6}\)
E. \(\frac{6}{9^6}\)


For each couple, there's a 1/9 chance the wife will order the same dessert as her husband. With three couples, the combined probability is:

\(\frac{1}{9} *\frac{1}{9} *\frac{1}{9} = \frac{1}{9^3} \).


Answer: C


 
­
A husband can choose ANY dessert from the 9, so the probability of that is 1. The probability that his wife will choose the same dessert is 1/9.

Or think about this in this way: the husband has a 1/9 probability of choosing some specific desserts, say pavlova, his wife has a 1/9 chance to also choose pavlova. However, we have 9 different desserts and for each, we'd have the same probability of 1/9*1/9, making the overall probability that the husband and wife will choose the same dessert is 9*1/9*1/9 = 1/9.­

Hope it's clear.
User avatar
sayan640
Joined: 29 Oct 2015
Last visit: 19 May 2025
Posts: 1,206
Own Kudos:
Given Kudos: 780
GMAT 1: 570 Q42 V28
Products:
GMAT 1: 570 Q42 V28
Posts: 1,206
Kudos: 684
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel , how do I know after somebody picks one deserts , the total count of deserts will not diminsh by one ? The total no. of deserts is same i.e 9. for three couples here.
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 May 2025
Posts: 101,532
Own Kudos:
725,563
 [1]
Given Kudos: 93,558
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 101,532
Kudos: 725,563
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
 
sayan640
At a restaurant, three couples are preparing to order dessert. If there are 9 different desserts on the menu, what is the probability that the husband and wife of each couple will order the same dessert? Note that while each couple is required to order the same dessert, this dessert does not need to match the one ordered by the other couples.

Bunuel , how do I know after somebody picks one deserts , the total count of deserts will not diminsh by one ? The total no. of deserts is same i.e 9. for three couples here. Please help. Bunuel
If you've been in a restaurant, you should know that the menu items usually do not run out. Also, the question asks about the probability that the husband and wife of each couple will order the same dessert, so it's possible to order the same dessert more than once. Moreover, at the end, the question says, "Note that while each couple is required to order the same dessert, this dessert does not need to match the one ordered by the other couples," which also indicates that each couple can independently choose from the 9 available desserts without affecting the total number.

P.S. Please do NOT tag a person more than once in one post.­
User avatar
lnyngayan
Joined: 09 Mar 2023
Last visit: 13 Oct 2024
Posts: 29
Own Kudos:
Given Kudos: 240
Posts: 29
Kudos: 1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
Official Solution:


At a restaurant, three couples are preparing to order dessert. If there are 9 different desserts on the menu, what is the probability that the husband and wife of each couple will order the same dessert? Note that while each couple is required to order the same dessert, this dessert does not need to match the one ordered by the other couples.


A. \(\frac{6}{9^3}\)
B. \(\frac{3}{9^3}\)
C. \(\frac{1}{9^3}\)
D. \(\frac{1}{9^6}\)
E. \(\frac{6}{9^6}\)


For each couple, there's a 1/9 chance the wife will order the same dessert as her husband. With three couples, the combined probability is:

\(\frac{1}{9} *\frac{1}{9} *\frac{1}{9} = \frac{1}{9^3} \).


Answer: C

Why the wife and the husband of the same couple choosing the same dessert is 1/9 instead of (1/9)^2?
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 May 2025
Posts: 101,532
Own Kudos:
Given Kudos: 93,558
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 101,532
Kudos: 725,563
Kudos
Add Kudos
Bookmarks
Bookmark this Post
lnyngayan
Bunuel
Official Solution:


At a restaurant, three couples are preparing to order dessert. If there are 9 different desserts on the menu, what is the probability that the husband and wife of each couple will order the same dessert? Note that while each couple is required to order the same dessert, this dessert does not need to match the one ordered by the other couples.


A. \(\frac{6}{9^3}\)
B. \(\frac{3}{9^3}\)
C. \(\frac{1}{9^3}\)
D. \(\frac{1}{9^6}\)
E. \(\frac{6}{9^6}\)


For each couple, there's a 1/9 chance the wife will order the same dessert as her husband. With three couples, the combined probability is:

\(\frac{1}{9} *\frac{1}{9} *\frac{1}{9} = \frac{1}{9^3} \).


Answer: C

Why the wife and the husband of the same couple choosing the same dessert is 1/9 instead of (1/9)^2?

I tried addressing this doubt here. If it's still not clear, you can check alternative solutions in this thread.
User avatar
Sidharth20
Joined: 21 Mar 2024
Last visit: 13 May 2025
Posts: 17
Own Kudos:
Posts: 17
Kudos: 3
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hi Bunuel,

Can you tell me why didn't we consider the order over here? i.e. - For couple 1 - Husband selects the first desert and wife selects the same OR Wife selects the first desert and then husband selects the same: 1*1/9 + 1*1/9

So for 3 couples - (1*1/9 + 1*1/9)^3
Bunuel
lnyngayan
Bunuel
Official Solution:


At a restaurant, three couples are preparing to order dessert. If there are 9 different desserts on the menu, what is the probability that the husband and wife of each couple will order the same dessert? Note that while each couple is required to order the same dessert, this dessert does not need to match the one ordered by the other couples.


A. \(\frac{6}{9^3}\)
B. \(\frac{3}{9^3}\)
C. \(\frac{1}{9^3}\)
D. \(\frac{1}{9^6}\)
E. \(\frac{6}{9^6}\)


For each couple, there's a 1/9 chance the wife will order the same dessert as her husband. With three couples, the combined probability is:

\(\frac{1}{9} *\frac{1}{9} *\frac{1}{9} = \frac{1}{9^3} \).


Answer: C

Why the wife and the husband of the same couple choosing the same dessert is 1/9 instead of (1/9)^2?

I tried addressing this doubt here. If it's still not clear, you can check alternative solutions in this thread.
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 May 2025
Posts: 101,532
Own Kudos:
Given Kudos: 93,558
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 101,532
Kudos: 725,563
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Sidharth20
Hi Bunuel,

Can you tell me why didn't we consider the order over here? i.e. - For couple 1 - Husband selects the first desert and wife selects the same OR Wife selects the first desert and then husband selects the same: 1*1/9 + 1*1/9

So for 3 couples - (1*1/9 + 1*1/9)^3

This is because {wife = x, husband = x} is the same case as {husband = x, wife = x}.
User avatar
naoliden
Joined: 30 Jun 2024
Last visit: 17 Apr 2025
Posts: 5
Own Kudos:
Given Kudos: 16
Posts: 5
Kudos: 1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hi,
I understand the (1/3)^3 but can you explain why don ́t you need to multiply by 3! too?

I though it would be:

(1/3)^3 * (3 couples) *(people in a couple) = (1/3)^3 * 3!
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 May 2025
Posts: 101,532
Own Kudos:
Given Kudos: 93,558
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 101,532
Kudos: 725,563
Kudos
Add Kudos
Bookmarks
Bookmark this Post
naoliden
Bunuel
Official Solution:


At a restaurant, three couples are preparing to order dessert. If there are 9 different desserts on the menu, what is the probability that the husband and wife of each couple will order the same dessert? Note that while each couple is required to order the same dessert, this dessert does not need to match the one ordered by the other couples.


A. \(\frac{6}{9^3}\)
B. \(\frac{3}{9^3}\)
C. \(\frac{1}{9^3}\)
D. \(\frac{1}{9^6}\)
E. \(\frac{6}{9^6}\)


For each couple, there's a 1/9 chance the wife will order the same dessert as her husband. With three couples, the combined probability is:

\(\frac{1}{9} *\frac{1}{9} *\frac{1}{9} = \frac{1}{9^3} \).


Answer: C
Hi,
I understand the (1/3)^3 but can you explain why don ́t you need to multiply by 3! too?

I though it would be:

(1/3)^3 * (3 couples) *(people in a couple) = (1/3)^3 * 3!

The order of the couples doesn’t matter in this scenario. Each couple independently chooses a dessert, so there’s no need to multiply by 3!.
User avatar
hr1212
Joined: 18 Apr 2019
Last visit: 19 May 2025
Posts: 288
Own Kudos:
295
 [1]
Given Kudos: 317
GMAT Focus 1: 775 Q90 V85 DI90
Products:
GMAT Focus 1: 775 Q90 V85 DI90
Posts: 288
Kudos: 295
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
At a restaurant, three couples are preparing to order dessert. If there are 9 different desserts on the menu, what is the probability that the husband and wife of each couple will order the same dessert? Note that while each couple is required to order the same dessert, this dessert does not need to match the one ordered by the other couples.


A. \(\frac{6}{9^3}\)
B. \(\frac{3}{9^3}\)
C. \(\frac{1}{9^3}\)
D. \(\frac{1}{9^6}\)
E. \(\frac{6}{9^6}\)
One way to think could be,

No. of total possibilities = 9*9*9*9*9*9 (where each person can order any one of 9 desserts so has 9 choices) = \(9^6\)

Possibilities where couples order same dessert = 9*1*9*1*9*1 (wife has 9 choices while her husband has to order the same dessert for each couple) = \(9^3\)

Probability = \(\frac{9^3}{9^6}\) = \(\frac{1}{9^3}\)

IMO: C
User avatar
hari0616
Joined: 13 Nov 2024
Last visit: 03 May 2025
Posts: 18
Own Kudos:
7
 [2]
Given Kudos: 37
Posts: 18
Kudos: 7
 [2]
1
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
An alternate way to think of this problem: Since the question says no 2 couples need to order the same dessert, let's start with one couple.
The P(husband and wife ordering same dessert) = 1 - P(both order different desserts)

P (both ordering different desserts) = \(\frac{9 * 8}{9^2}\) = \(\frac{1}{9}\).

So, for 3 couples, it will be \(\frac{1}{9}\) * \(\frac{1}{9}\) * \(\frac{1}{9}\) = \(\frac{1}{9^3}\).

Therefore, C.


Bunuel
At a restaurant, three couples are preparing to order dessert. If there are 9 different desserts on the menu, what is the probability that the husband and wife of each couple will order the same dessert? Note that while each couple is required to order the same dessert, this dessert does not need to match the one ordered by the other couples.


A. \(\frac{6}{9^3}\)
B. \(\frac{3}{9^3}\)
C. \(\frac{1}{9^3}\)
D. \(\frac{1}{9^6}\)
E. \(\frac{6}{9^6}\)
User avatar
Mayanksureka
Joined: 29 Jul 2024
Last visit: 19 May 2025
Posts: 4
Own Kudos:
Given Kudos: 13
GMAT Focus 1: 665 Q84 V81 DI81
GMAT Focus 2: 715 Q90 V84 DI83
Products:
GMAT Focus 2: 715 Q90 V84 DI83
Posts: 4
Kudos: 2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I like the solution - it’s helpful.
 1   2   
Moderators:
Math Expert
101532 posts
Founder
40703 posts