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Sagot :
Using the binomial distribution, there is a 0.6328 = 63.28% probability that she wins at most 1 prize.
For each box, there are only two possible outcomes, either it has a prize, or it does not. The probability of a box having a prize is independent of any other box, hence, 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:
- She buys 5 boxes, hence [tex]n = 5[/tex]
- 1 in 4 boxes has a prize, hence [tex]p = \frac{1}{4} = 0.25[/tex]
The probability is:
[tex]P(X \leq 1) = P(X = 0) + P(X = 1)[/tex]
Hence:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{5,0}.(0.25)^{0}.(0.75)^{5} = 0.2373[/tex]
[tex]P(X = 1) = C_{5,1}.(0.25)^{1}.(0.75)^{4} = 0.3955[/tex]
Then
[tex]P(X \leq 1) = P(X = 0) + P(X = 1) = 0.2373 + 0.3955 = 0.6328[/tex]
0.6328 = 63.28% probability that she wins at most 1 prize.
A similar problem is given at https://brainly.com/question/24863377
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