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22 Apr 2025, 04:22
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Total outcomes:
9^6. Each couple has nine options to choose from.
Hence total possibilities 9^6.
Favourable outcomes
AB/CD/EF (3 couples)
Lets say A chose 1 dessert out of 9. Now B can have only that specific one as dessert. Sorry B
Hence for A there are 9 favourable ways but for B only 1 way. Same goes for other couples
(9*1)/(9*1)/(9*1)
Probability= (9*9*9)/9^6
9^3/9^6
Therefore : 1/9^3. Option C.
9^6. Each couple has nine options to choose from.
Hence total possibilities 9^6.
Favourable outcomes
AB/CD/EF (3 couples)
Lets say A chose 1 dessert out of 9. Now B can have only that specific one as dessert. Sorry B
Hence for A there are 9 favourable ways but for B only 1 way. Same goes for other couples
(9*1)/(9*1)/(9*1)
Probability= (9*9*9)/9^6
9^3/9^6
Therefore : 1/9^3. Option C.
04 May 2025, 05:21
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I solved it like this
total = (9)^6
and now wanted =
Couple 1 = 9c1 *1 = 9
Couple 2 = 9c1 *1 = 9
Couple 3 = 9c1 *1 = 9
so 9^3 is wanted
hence 9^3/9^6 = 1/9^3
I solved it with cool mind after getting it wrong on test & got it logically correct.
total = (9)^6
and now wanted =
Couple 1 = 9c1 *1 = 9
Couple 2 = 9c1 *1 = 9
Couple 3 = 9c1 *1 = 9
so 9^3 is wanted
hence 9^3/9^6 = 1/9^3
I solved it with cool mind after getting it wrong on test & got it logically correct.
Bunuel
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. \(\frac{6}{9^6}\)
A. \(\frac{6}{9^3}\)
B. \(\frac{3}{9^3}\)
C. \(\frac{1}{9^3}\)
D. \(\frac{1}{9^6}\)
E. \(\frac{6}{9^6}\)
