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Sagot :
Using the binomial distribution, it is found that there is a 0.0344 = 3.44% probability that at least two students choose the same president.
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For each president, there are only two possible outcomes, either they were chosen, or they were not. The probability of a president being chosen is independent of any other president, which means that the binomial probability distribution is used to solve this question.
Binomial probability distribution
It is the probability of exactly x successes on n repeated trials.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by:
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
p is the probability of a success on a single trial.
In this problem:
- 5 students, thus [tex]n = 5[/tex].
- 16 presidents equally as likely to be chosen, thus [tex]p = \frac{1}{16} = 0.0625[/tex]
The probability that at least two students choose the same president is the probability that a president is chosen more than once, that is, P(X > 1), which is:
[tex]P(X > 1) = 1 - P(X \leq 1)[/tex]
In which:
[tex]P(X \leq 1) = P(X = 0) + P(X = 1)[/tex]
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{5,0}.(0.0625)^{0}.(0.9375)^{5} = 0.7242[/tex]
[tex]P(X = 1) = C_{5,1}.(0.0625)^{1}.(0.9375)^{4} = 0.2414[/tex]
Then
[tex]P(X \leq 1) = P(X = 0) + P(X = 1) = 0.7242 + 0.2414 = 0.9656[/tex]
[tex]P(X > 1) = 1 - P(X \leq 1) = 1 - 0.9656 = 0.0344[/tex]
0.0344 = 3.44% probability that at least two students choose the same president.
A similar problem is given at https://brainly.com/question/24863377
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