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M13-37

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Bunuel
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M13-37 [#permalink] M13-37   16 Sep 2014, 00:50
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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. \(\frac{6}{9^6}\)
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C
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Bunuel
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Re M13-37 [#permalink] M13-37   16 Sep 2014, 00:50
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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
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Bunuel
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Re: M13-37 [#permalink] M13-37   24 Aug 2023, 02:03
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Mohit1994
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Re: M13-37 [#permalink] M13-37   01 Oct 2023, 00:41
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.???
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Re: M13-37 [#permalink] M13-37   01 Oct 2023, 00:49
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Mohit1994 wrote:
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?
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Re: M13-37 [#permalink] M13-37   01 Oct 2023, 01:23
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 ??
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Re: M13-37 [#permalink] M13-37   01 Oct 2023, 01:30
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Mohit1994 wrote:
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.
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M13-37 [#permalink] M13-37   14 Feb 2024, 01:27
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 wrote:
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


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Bunuel
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Re: M13-37 [#permalink] M13-37   14 Feb 2024, 01:37
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CoffeeNCream wrote:
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 wrote:
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.
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M13-37 [#permalink] M13-37   Updated on: 23 May 2024, 00:22
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.

Originally posted by sayan640 on 22 May 2024, 22:39.
Last edited by sayan640 on 23 May 2024, 00:22, edited 1 time in total.
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Re: M13-37 [#permalink] M13-37   22 May 2024, 23:56
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sayan640 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.

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.­
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Re: M13-37 [#permalink] M13-37   08 Jun 2024, 07:35
1st couple choices =  9*1
2nd couple choices=  9*1
3rd Couple Choices =  9*1

Total Choices = 9*1*3
3 couples means 6 members, 9 Choices are available to each member
Total choices available to 3 couples or 6 members= \(9^{6}\)
Required Probability = \(\frac{9*1*3 }{ 9^{6}}\) = \(\frac{3}{9^{5}}\)

(Bunuel Although this is not an available answer choice, I want to understand why we are taking product in the numerator instead of the sum?)­
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Re: M13-37 [#permalink] M13-37   08 Jun 2024, 07:59
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Anugmat740 wrote:
1st couple choices =  9*1
2nd couple choices=  9*1
3rd Couple Choices =  9*1

Total Choices = 9*1*3
3 couples means 6 members, 9 Choices are available to each member
Total choices available to 3 couples or 6 members= \(9^{6}\)
Required Probability = \(\frac{9*1*3 }{ 9^{6}}\) = \(\frac{3}{9^{5}}\)

(Bunuel Although this is not an available answer choice, I want to understand why we are taking product in the numerator instead of the sum?)­

Unfortunately, I don't understand the logic you're using, so I can't identify the problem. I suggest checking other solutions here: https://gmatclub.com/forum/around-the-world-in-80-questions-day-4-at-a-restaurant-three-415779.html. This might help.
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Re: M13-37 [#permalink]
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