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Sagot :
The probability that at least one will have a defect if if 10 items are chosen at random is 0.65132.
binomial distribution is given as :
[tex]P(X=x) = (_{x} ^{n}) p^{x}q^{n-x} ; x = 0,1,2,3,........n\\q = 1 - p ;[/tex]
where
x = number of times for a specific outcome within n trials
{n x}= number of combinations
p = probability of success on a single trial
q = probability of failure on a single trial
n = number of trials
Let D be a random variable represents the number of defected items out of 10.
i.e., D(0,1,2,3,4,5,6,7,8,9,10)
probability that given items is defected is :
p = 0.1
q = 1 - p
q = 1 - 0.1 = 0.9
where
p = probability of defected items
q = probability of non-defective
then
D- bin(10,0.1)
P(D=0) = [tex]^{10} C_{0}{(0.1)^0}{(0.9)^{10}}[/tex]
[tex]= \frac{10!}{0!(10-0)!} * (0.9)^{10}[/tex]
= 0.3486784401
[tex]P(D\geq 1) = 1-P[/tex]
1 - = 0.3486784401
= 0.65132
The probability that at least one will have a defect if if 10 items are chosen at random is 0.65132.
To know more about distributive probability
https://brainly.com/question/21883407
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