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

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


 


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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

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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 = (\(\frac{9*8*7}{9^6}\)).
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Re: Around the World in 80 Questions (Day 4): At a restaurant, three [#permalink]
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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.
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Re: Around the World in 80 Questions (Day 4): At a restaurant, three [#permalink]
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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
 


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for the Around the World in 80 Questions

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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
Answer C
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Re: Around the World in 80 Questions (Day 4): At a restaurant, three [#permalink]
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