Mathematics, 01.11.2020 04:50 rue61

A production process produces 2% defective parts. A sample of five parts from the production is selected what is the probability that the sample contains two or less of the sampled parts are defective

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Answer from: zakiraboo94

Approximately $99.992\%$ , 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 of finding no defective part among the five samples,The probability of finding one defective part among the five samples, andThe probability of finding two defective parts among the five samples.No defective parts

The probability that one chosen part is defective is $2\% = 0.02$ . Therefore, the probability that chosen part is not defective would be (1 - 0.02) .

The probability that all five parts are not defective (assuming that they are independent from one another) would be (1-0.02)^5 .

One defective part

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:

Sample one is defective, which corresponds to a probability of 0.02

, andSample two through five are not defective, which corresponds to a probability of (1 - 0.02)

for each of the (5-1)

samples.

That gives the probability that sample one is the only sample that fails:

$0.02 \times (1 -0.02)^{5-1}$ .

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 $C(n,\, r)$ gives the number of ways to choose out of objects:

$\displaystyle C(n,\, r) = \frac{n!}{r!\, (n-r)!}$

There are $\displaystyle C(5,\, 1) = \frac{5!}{1!\, (5-1)!} = 5$ ways to choose one out of five objects. Hence, the probability that exactly one out of these five samples will be defective would be:

$C(5,\, 1) \times \left(0.02 \times (1-0.02)^{5-1}\right) = 5\times \left(0.02 \times (1 - 0.02)^{5-1}\right)$ .

Two defective parts

There are $\displaystyle C(5,\, 2) = \frac{5!}{2!\, (5-2)!} = 10$ 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:

$(0.02)^2 \times (1 - 0.02)^{5-2}$ .

The probability that exactly one out of these five samples will be defective would be:

$C(5,\, 2) \times (0.02)^2 \times (1 - 0.02)^{5-2} = 10 \times \left(0.02 \times (0.98)^{5-2}\right)$ .

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:

Answer from: Quest

b.-4 units

all you do is count how many times you move

Answer from: Quest

It would take her 20 minutes.

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