Respuesta :
Answer:
The probability that of the two chips selected both are defective is 0.1089.
Step-by-step explanation:
Let X = number of defective chips.
It is provided that there are 2 defective chips among 6 chips.
The probability of selecting a defective chip is:
[tex]P(X)=p=\frac{2}{6}=0.33[/tex]
A sample of n = 2 chips are selected.
The random variable X follows a Binomial distribution with parameter n = 2 and p = 0.33.
The probability function of a Binomial distribution is:
[tex]P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0, 1, 2, ...[/tex]
Compute the probability that of the two chips selected both are defective as follows:
[tex]P(X=2)={2\choose 2}(0.33)^{2}(1-0.33)^{2-2}=1\times 0.1089\times 1=0.1089[/tex]
Thus, the probability that of the two chips selected both are defective is 0.1089.
The sample space of selecting two chips is:
S = (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5)
The probability of both chips are defective is [tex]\frac{1}{9}[/tex] .
Probability:
Given that, Out of six computer chips, two are defective.
Probability of getting one chip is defective ,
[tex]P_{1}=\frac{2}{6}=\frac{1}{3}[/tex]
When two chips are randomly chosen for testing,
Then, probability that both of them are defective is,
[tex]P(E)=\frac{1}{3}*\frac{1}{3}=\frac{1}{9}[/tex]
Learn more about the probability here:
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