Respuesta :
Answer:
[tex]P(X = 1) = C_{6,1}.(0.01)^{1}.(0.99)^{5} = 0.0571[/tex]
Step-by-step explanation:
For each laser pointer, there are only two possible outcomes. Eithey they are defective, or they are not. This means that we use the binomial probability distribution to solve this problem.
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 combinatios 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.
In this problem we have that:
[tex]p = 0.01, n = 6[/tex]
Which one of the following expressions describes the probability that exactly one of the six laser pointers is defective?
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 1) = C_{6,1}.(0.01)^{1}.(0.99)^{5} = 0.0571[/tex]
The probability that exactly one of the six laser pointers is defective can be determine by binomial probability distribution.
The expressions [tex](0.99)^5(0.01)^1[/tex]describes the probability that exactly one of the six laser pointers is defective.
Given:
The probability is [tex]p=\dfrac{1\%}{100}=0.01[/tex].
Number of pointers [tex]n=6[/tex].
Consider the formula for binomial probability distribution.
[tex]P(X=x)=C_{n.x}.p^x.(1-p)^{n-x}[/tex]
Where [tex]C_{n.x}[/tex] is number of different combinations of X objects from a set of n elements.
Substitute the value in above equation.
[tex]P(X=1)=C_{6,1}.(0.01)^1.(0.99)^5[/tex]
Thus, the expressions [tex](0.99)^5(0.01)^1[/tex]describes the probability that exactly one of the six laser pointers is defective.
Learn more about what binomial probability distribution is here:
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