Respuesta :
Answer:
the probability that the contractor keeps the transistors is 37.58%
Step-by-step explanation:
Since each transistor is equally likely to the chosen , then the random variable X= transistors that have defects from a sample of 10 , has a binomial distribution . Thus its probability is calculated through:
P(X=x) = n!/((n-x)!*x!)*p^x*(1-p)^(n-x)
where
n= sample size = 10
p= probability for any transistor to be defective = 0.2 (20%)
x= number of transistors that are defective
P(X=x) = probability that there are x transistors that are defective
thus replacing values and knowing that only 0 or 1 can be defective in order to have at least 9 working transistors:
P(X≤1)= P(X=1)+P(X=0)
this is equivalent to
P(X≤1)= F(X=1)
where F(X) is the cummulative binomial probability distribution. From tables can be found that
P(X≤1)= F(X=1)= 0.3758 (37.58%)
thus the probability that the contractor keeps the transistors is 37.58%
The probability that the contractor keeps the transistors is [tex]37.58[/tex] %
Binomial distribution :
The probability that there are x transistors that are defective is given by,
[tex]P(X=x)=^{n}C_{x}p^{x} q^{n-x}[/tex]
where
- [tex]n[/tex] is sample size
- [tex]p[/tex] is probability for any transistor to be defective
- [tex]x[/tex] is number of transistors that are defective
Given that, [tex]n=10,p=0.2,q=1-0.2=0.8[/tex]
if at least 9 of the 10 are in working condition
the probability the contractor will keep all the transistors is given as,
[tex]P(x\leq 1)= P(x=1)+P(x=0)\\\\P(x\leq 1)=^{10}C_{1}(0.2)^{1}(0.8)^{9} +^{10}C_{0}(0.2)^{0}(0.8)^{10} \\\\P(x\leq 1)=0.3758[/tex]
Thus, the probability that the contractor keeps the transistors is [tex]37.58[/tex] %
Learn more about the probability here:
https://brainly.com/question/25870256
