Answer:
Approximately [tex]99.992\%[/tex], assuming that the parts are independent from one another.
Step-by-step explanation:
The probability that the sample of five parts contains at most two defective parts is equal to the sum of:
The probability that one chosen part is defective is [tex]2\% = 0.02[/tex]. Therefore, the probability that chosen part is not defective would be [tex](1 - 0.02)[/tex].
The probability that all five parts are not defective (assuming that they are independent from one another) would be [tex](1-0.02)^5[/tex].
Assume that the five samples are numbered from one to five. If sample number one is the only defective sample among the five, it must be true that:
That gives the probability that sample one is the only sample that fails:
[tex]0.02 \times (1 -0.02)^{5-1}[/tex].
However, the event that there is one defective part is true as long as exactly one out of all five samples is defective. (The defective sample might not necessarily be sample number one.) The combination notation [tex]C(n,\, r)[/tex] gives the number of ways to choose [tex]r[/tex] out of [tex]n[/tex] objects:
[tex]\displaystyle C(n,\, r) = \frac{n!}{r!\, (n-r)!}[/tex]
There are [tex]\displaystyle C(5,\, 1) = \frac{5!}{1!\, (5-1)!} = 5[/tex] ways to choose one out of five objects. Hence, the probability that exactly one out of these five samples will be defective would be:
[tex]C(5,\, 1) \times \left(0.02 \times (1-0.02)^{5-1}\right) = 5\times \left(0.02 \times (1 - 0.02)^{5-1}\right)[/tex].
There are [tex]\displaystyle C(5,\, 2) = \frac{5!}{2!\, (5-2)!} = 10[/tex] ways to choose two defective parts out of a sample of five parts.
The probability that the two chosen parts are the only two parts that are defective among the sample of five would be:
[tex](0.02)^2 \times (1 - 0.02)^{5-2}[/tex].
The probability that exactly one out of these five samples will be defective would be:
[tex]C(5,\, 2) \times (0.02)^2 \times (1 - 0.02)^{5-2} = 10 \times \left(0.02 \times (0.98)^{5-2}\right)[/tex].
Take the sum of the probabilities found in these three sections to find the probability that the sample of five contains no more than two defective parts:
