Answer :
Using the binomial distribution, it is found that there is a:
a) 0.0625 = 6.25% probability that the first four players that arrive are all assigned to the north court.
b) 12.50% probability that the first four players that arrive are all assigned to the same court.
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For each men, there are only two possible outcomes. Either they play on the north court, or they do not. The probability of a men playing on a court is independent of any other other men, which means that the binomial distribution is used to solve this question.
Binomial probability distribution
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
The parameters are:
- x is the number of successes.
- n is the number of trials.
- p is the probability of a success on a single trial.
In this problem:
- 4 players, thus [tex]n = 4[/tex]
- Equally as likely to choose any court, thus [tex]p = 0.5[/tex].
Item a:
The probability that all are assigned to the north court is P(X = 4), thus:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 4) = C_{4,4}.(0.5)^{4}.(0.5)^{0} = 0.0625[/tex]
0.0625 = 6.25% probability that the first four players that arrive are all assigned to the north court.
Item b:
- Binomial distribution with [tex]p = 0.5[/tex], thus all four to the south court is the same probability of 0.0625.
- Then, 2 x 0.0625 = 0.1250 = 12.50% probability that the first four players that arrive are all assigned to the same court.
A similar problem is given at https://brainly.com/question/24881950