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Around the World in 80 Questions (Day 4): At a restaurant, three
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20 Jul 2023, 08:00
20 Jul 2023, 08:00
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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. 6/9^3
B. 3/9^3
C. 1/9^3
D. 1/9^6
E. 6/9^6
A. 6/9^3
B. 3/9^3
C. 1/9^3
D. 1/9^6
E. 6/9^6
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Re: Around the World in 80 Questions (Day 4): At a restaurant, three
[#permalink]
20 Jul 2023, 08:17
20 Jul 2023, 08:17
5
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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
IMO C
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
Re: Around the World in 80 Questions (Day 4): At a restaurant, three
[#permalink]
24 Aug 2023, 02:04
24 Aug 2023, 02:04
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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
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
