Last visit was: 23 Jun 2024, 20:33 It is currently 23 Jun 2024, 20:33
Toolkit
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

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.

# Around the World in 80 Questions (Day 4): At a restaurant, three

SORT BY:
Tags:
Show Tags
Hide Tags
Math Expert
Joined: 02 Sep 2009
Posts: 93964
Own Kudos [?]: 634381 [18]
Given Kudos: 82420
Math Expert
Joined: 02 Sep 2009
Posts: 93964
Own Kudos [?]: 634381 [0]
Given Kudos: 82420
General Discussion
Senior Manager
Joined: 10 Mar 2015
Posts: 252
Own Kudos [?]: 235 [2]
Given Kudos: 175
Location: India
Concentration: Strategy, Marketing
GPA: 3.5
GMAT Club Legend
Joined: 03 Jun 2019
Posts: 5226
Own Kudos [?]: 4087 [4]
Given Kudos: 160
Location: India
GMAT 1: 690 Q50 V34
WE:Engineering (Transportation)
Re: Around the World in 80 Questions (Day 4): At a restaurant, three [#permalink]
4
Kudos
Given: At a restaurant, three couples are preparing to order dessert.
Asked: 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.

Total number of ways to order desserts = 9^6
Since all 6 persons can order any of the 9 desserts

Number of ways such that the husband and wife of each couple will order the same dessert = 9*1*9*1*9*1 = 9^3
Since one out of husband or wife of each couples orders a dessert and other orders the same.

The probability that the husband and wife of each couple will order the same dessert = 9^3/9^6 = 1/9^3

IMO C
Manager
Joined: 07 May 2023
Posts: 198
Own Kudos [?]: 242 [1]
Given Kudos: 47
Location: India
Re: Around the World in 80 Questions (Day 4): At a restaurant, three [#permalink]
1
Kudos
Bunuel wrote:
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. 6/9^3
B. 3/9^3
C. 1/9^3
D. 1/9^6
E. 6/9^6

 This question was provided by GMAT Club for the Around the World in 80 Questions Win over $20,000 in prizes: Courses, Tests & more Husband1 - Wife1 Husband2 - Wife2 Husband3 - Wife3 Husband1 can select 1 dessert in 1/9 ways. Wife1 can select only in 1 way. Husband2 can select 1 dessert in 1/9 ways. Wife2 can select only in 1 way. Husband3 can select 1 dessert in 1/9 ways. Wife3 can select only in 1 way. Total = 1/9 * 1/9 * 1/9 = 1/9^3 IMO C General GMAT Forum Moderator Joined: 03 Aug 2017 Posts: 280 Own Kudos [?]: 238 [0] Given Kudos: 198 Re: Around the World in 80 Questions (Day 4): At a restaurant, three [#permalink] Bunuel Can you or someone please help what I am missing in this approach? (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.) 1st couple can choose any desert out of 9 deserts available. 2nd couple can choose among the remaining 8 deserts available . ( 1 desert has already been picked up by the 1st couple) 3rd couple can choose among the remaining 7 desert available .( 2 deserts have already been picked by 1st and 2nd couple). Total ways in which the husband and wife of each couple will order the same dessert= 9*8*7 (no repetition as the deserts does not need to match the one ordered by the other couples). Every person in the group out of 6 people(3 couples) can order 9 deserts. Total no. of ways to order desert = 9^6 So required Probability = ($$\frac{9*8*7}{9^6}$$). Math Expert Joined: 02 Sep 2009 Posts: 93964 Own Kudos [?]: 634381 [1] Given Kudos: 82420 Re: Around the World in 80 Questions (Day 4): At a restaurant, three [#permalink] 1 Kudos Expert Reply Catman wrote: Bunuel Can you or someone please help what I am missing in this approach? (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.) 1st couple can choose any desert out of 9 deserts available. 2nd couple can choose among the remaining 8 deserts available . ( 1 desert has already been picked up by the 1st couple) 3rd couple can choose among the remaining 7 desert available .( 2 deserts have already been picked by 1st and 2nd couple). Total ways in which the husband and wife of each couple will order the same dessert= 9*8*7 (no repetition as the deserts does not need to match the one ordered by the other couples). Every person in the group out of 6 people(3 couples) can order 9 deserts. Total no. of ways to order desert = 9^6 So required Probability = ($$\frac{9*8*7}{9^6}$$). 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. The above means that all three couples, or two of the three, can select the same desert. Tutor Joined: 26 Jun 2014 Status:Mentor & Coach | GMAT Q51 | CAT 99.98 Posts: 450 Own Kudos [?]: 792 [1] Given Kudos: 8 Re: Around the World in 80 Questions (Day 4): At a restaurant, three [#permalink] 1 Kudos Expert Reply Bunuel wrote: 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. 6/9^3 B. 3/9^3 C. 1/9^3 D. 1/9^6 E. 6/9^6  This question was provided by GMAT Club for the Around the World in 80 Questions Win over$20,000 in prizes: Courses, Tests & more

For the first couple: The wife can choose any desert - probability of that choice = 1
Probability that the husband chooses the same desert = 1/9
Thus, probability that they choose the same desert = 1 * 1/9 = 1/9

In the same way, for each couple, the probability is 1/9

Thus, for all couples, the probability is = 1/9 * 1/9 * 1/9 = (1/9)^3