An old-fashioned Chinese restaurant offers a family dinner where you get to choose one dish from “column A” (which has 8 dishes) , one dish from “column B” (which has 10 dishes) and one dish from “column C” (which has 5 dishes). How many different family dinners can be chosen?
2 answers:
Final answer: 400
Detailed explanation:
Information provided: Dish count in column A = 8
Dish count in column B = 10
Dish count in column C = 5
As each dinner requires selecting one dish from each column, then
The overall number of possible dish combinations for dinner is:-
Thus, total family dinners = 400
The result is calculated as 10 multiplied by 8 multiplied by 5, which equals 400.
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In the absence of a specific question posed, below are the potential inquiries along with their respective answers:
P(fewer than 4 tosses)
= P(one toss) + P(two tosses) + P(three tosses)
= (3/4) + (3/4)(1/4) + (3/4)(1/4)^2
= 0.984375
Expected value
= 1 / p
= 1 / (3/4)
= 4 / 3
Variance
= (1 - p) / p^2
= (1 - (3/4)) / (3/4)^2
= (1/4) / (9/16)
= 4 / 9
Standard deviation
= sqrt(Variance)
= sqrt(4 / 9)
= 2 / 3