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Sagot :
Answer:
0.7061 = 70.61% probability she will have her first crash within the first 30 races she runs this season
Step-by-step explanation:
For each race, there are only two possible outcomes. Either the person has a crash, or the person does not. The probability of having a crash during a race is independent of whether there was a crash in any other race. 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.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
A certain performer has an independent .04 probability of a crash in each race.
This means that [tex]p = 0.04[/tex]
a) What is the probability she will have her first crash within the first 30 races she runs this season
This is:
[tex]P(X \geq 1) = 1 - P(X = 0)[/tex]
When [tex]n = 30[/tex]
We have that:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{30,0}.(0.04)^{0}.(0.96)^{30} = 0.2939[/tex]
[tex]P(X \geq 1) = 1 - P(X = 0) = 1 - 0.2939 = 0.7061[/tex]
0.7061 = 70.61% probability she will have her first crash within the first 30 races she runs this season
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