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80% of the items produced are non-defective. if three items are examined, what is the probability that at most one is non- defective? use the binomial probability function to answer this question.

Respuesta :

mathmate
Binomial distribution can be used because the situation satisfies all the following conditions:
1. Number of trials is known and remains constant (n)
2. Each trial is Bernoulli (i.e. exactly two possible outcomes) (success/failure)
3. Probability is known and remains constant throughout the trials (p)
4. All trials are random and independent of the others
The number of successes, x, is then given by
[tex]P(x)=C(n,x)p^x(1-p)^{n-x}[/tex]
where
[tex]C(n,x)=\frac{n!}{x!(n-x)!}[/tex]

Here given
p=0.8   (success=non-defective)
n=3
x=0 or 1
Then
[tex]P(0)+P(1)[/tex]
[tex]=\sum_{x=0}^1C(n,x)p^x(1-p)^{n-x}[/tex]
[tex]=C(3,0)0.8^0(1-0.8)^{3-0}+C(3,1)0.8^1(1-0.8)^{3-1}[/tex]
[tex]=1(1)0.08+3(0.8)(0.04)[/tex]
[tex]=0.08+0.096[/tex]
[tex]=0.104[/tex]

Answer: probability of at most one non-defective out of a random sample of three is 0.104
frika

The probability that produced item is non-defective is [tex]p=0.8[/tex] and the probability that  produced item is defective is [tex]q=1-p=1-0.8=0.2.[/tex]

1. Find the probability all produced items are defective:

[tex]Pr_1=q^3=(0.2)^3=0.008.[/tex]

2. Find the probability that one produced item is non-defective and 2 are defective:

[tex]Pr_2=C_3^1p^1q^2=\dfrac{3!}{1!(3-1)!}\cdot 0.8\cdot (0.2)^2=\dfrac{2\cdot 3}{2}\cdot 0.8\cdot 0.04=3\cdot 0.032=0.096.[/tex]

3. The probability that at most one produced is non- defective is

[tex]Pr=Pr_1+Pr_2=0.008+0.096=0.104.[/tex]

Answer: 0.104.