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777dan777 [17]
3 years ago
12

A class of 24 students wants to choose 3 students at random to bring food for a class party. Any set of 3 students should have a

n equal chance of being chosen. Which of the following strategies will result in a fair decision?
A. Assign a number to each student. Write the numbers on slips of paper and put them all in a hat. Randomly choose three slips of paper. The students with those three number can bring the food.
B. Arrange the students in a line. Start at one end and have each student flip a coin. The first three students to flip heads can bring the food.
C. Ask the students to volunteer. The first three students to raise their hands can bring the food.
D. None of the above
Mathematics
1 answer:
belka [17]3 years ago
8 0
I would think a would be the fairest 
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For each children, there are only two possible outcomes. Either it is a girl, or they are not. The probability of a child being a girl is independent of any other child. This means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

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0.5 probability of a children being a girl:

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Four children means that n = 4. So

P(X < 2) = P(X = 0) + P(X = 1)

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P(X = 0) = C_{4,0}.(0.5)^{0}.(0.5)^{4} = 0.0625

P(X = 1) = C_{4,1}.(0.5)^{1}.(0.5)^{3} = 0.25

P(X < 2) = P(X = 0) + P(X = 1) = 0.0625 + 0.25 = 0.3125

0.3125 = 31.25% probability that it is not ideal.

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