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

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

The probability that a couple orders the same dessert is 1/9. Therefore for 3 couples, it would be 1/9^3.
Therefore C.

Posted from my mobile device

total options are 9
total people are 6 ( 3 couples)
condition given each couple is required to order the same dessert, this dessert does not need to match the one ordered by the other couples.

we have total 9^6 ways to choose deserts
and for 3 couples it will be total 6 options
6/9^6
option E

=> P (H&W of one couple order the same dessert = (9*1)/(9*9) = 1/9

Similarly each couple will have the probability of 1/9
=> P (husband and wife of each couple will order the same dessert) = (1/9)^3

Note: Although, the dessert does not need to match the for all couples, it is not a restriction.

Total possiblities = 9 ^6

desired possibilities =
lets take one couple and assign 1 ice cream out of 9 = 9 ways
for next couple,we don't have retriction = 9 ways
similary 3rd couple = 9 ways

hence
ans C = 9*9*9/9^6

My Answer: C, 1/9^3

Total Cases=9^6, since there are 6 people each with 9 choices which can be repeated.

Cases where husband and wife order same dessert= 9^3, since one person from the couple has 9 choices and the other person then has only one choice. Since there are 3 couples, number of cases=9^3

So, Probability = 9^3/9^6 = 1/9^3

Prob=9x9x9/9^6=1/9^3

C)

suppose first husband selected first; can select in 9 ways
first wife can select only 1 way
second husband can select 9 ways
second wife can choose only 1 way
third husband can choose 9 ways
third wife can choose only 1 way
hence; 9*9*9

but all the situation is each husband can select 9, each wife can select 9
so the answer is 9*9*9/(9*9*9*9*9*9) = 1/9^3 c is the correct anwser

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

The answer is C
Each couple will order the same dessert in 1/9 ways and since there are three couples it is 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. 6/9^3
B. 3/9^3
C. 1/9^3
D. 1/9^6
E. 6/9^6


now husband orders 1/9 and wife has only one option 1
thus 1/9*1/9*1/9
thus c

The total amount of order of three couples who order from 9 different desserts is 9^6.
The total order of three couple who order the same dessert is 9*9*9 = 9^3.
9^3/9^6 = 1/9^3
Option C is the answer.

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

Right Ans : 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.
____________________________________
So, there are 3 couples.
A and B
C and D
E and F
Each person has probability of choosing a desert 1/9.
Let's say A chooses one desest with 1/9 probability ( one out of 9) and his other significant part choses the same desert with probability of 1/9.
Each couple will have the same probabilities and when we find total probability , we multiply.
1/9^6.
D is the winner.

probability of each couple ordering the same dessert. Since there are 9 different desserts on the menu, the probability of each couple choosing the same dessert is 1 out of 9 (1/9).



Probability = (1/9) * (1/9) * (1/9) = 1 / 9^3

So, the correct option is (C) 1/9^3.

1st couple :-

The first person is free to chose anything from the dessert.
Second person will have to chose the same dessert for which probability is 1/9.

Same is the case for 2nd and 3rd couple.

So, probability that the husband and wife of each couple will order the same dessert = 1/9 * 1/9 * 1/9.

Hence, correct choice is C.

Total number of ways to order desser:- There are 3 couples, so 6 people in total and everyone is free to order anything she/he wants with limitless supply of dessert with 9 different items to choose from. hence, total cases: 9^6
Now, for any couple, one person can order whatever he/she wants and other person will have to eat that perticular dessert only. Hence for any couple, we have (9*1) different options. Since the dessert does not need to match the one ordered by the other couples, we have total number of cases: (9*1*9*1*9*1)=9^3
Probability: 9^3/9^6=1/9^3. C is the right answer.