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Sagot :
Answer:
0.8746 = 87.46% probability of at least 6 failures in 7 trials.
Step-by-step explanation:
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.
7 trials
This means that [tex]n = 7[/tex]
The probability of success in any trial is 9%?
So the probability of a failure is 100 - 9 = 91%, which means that [tex]p = 0.91[/tex]
Probability of at least 6 failures in 7 trials?
This is:
[tex]P(X \geq 6) = P(X = 6) + P(X = 7)[/tex]
In which
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 6) = C_{7,6}.(0.91)^{6}.(0.09)^{1} = 0.3578[/tex]
[tex]P(X = 7) = C_{7,7}.(0.91)^{7}.(0.09)^{0} = 0.5168[/tex]
[tex]P(X \geq 6) = P(X = 6) + P(X = 7) = 0.3578 + 0.5168 = 0.8746[/tex]
0.8746 = 87.46% probability of at least 6 failures in 7 trials.
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