Find the measure for < MQX 54 59 63 65

A batch of 40 semiconductor chips is inspected by choosing a sample of 5 chips. Assume 10 of the chips do not conform to custome

butalik [34]

Answer:

a) 658008 samples

b) 274050 samples

c) 515502 samples

Step-by-step explanation:

a) How many ways sample of 5 each can be selected from 40 is just a combination problem since the order of selection isn't important.

So, the number of samples = ⁴⁰C₅ = 658008 samples

b) How many samples of 5 contain exactly one nonconforming chip?

There are 10 nonconforming chips in the batch, and 1 nonconforming chip for the sample of 5 be picked from ten in the following number of ways

¹⁰C₁ = 10 ways

then the remaining 4 conforming chips in a sample of 5 can be picked from the remaining 30 total conforming chips in the following number of ways

³⁰C₄ = 27405 ways

So, total number of samples containing exactly 1 nonconforming chip in a sample of 5 = 10 × 27405 = 274050 samples

c) How many samples of 5 contain at least one nonconforming chip?

The number of samples of 5 that contain at least one nonconforming chip = (Total number of samples) - (Number of samples with no nonconforming chip in them)

Number of samples with no nonconforming chip in them = ³⁰C₅ = 142506 samples

Total number of samples = 658008

The number of samples of 5 that contain at least one nonconforming chip = 658008 - 142506 = 515502 samples

6 0

3 years ago

I need the answer for this problem please

Savatey [412]

Answer:

B

Step-by-step explanation:

Find all probabilities:

A. False

$Pr(\text{red shirt}|\text{large shirt})=\dfrac{\text{number red large shirts}}{\text{number large shirts}}=\dfrac{42}{77}=\dfrac{6}{11}\\ \\Pr(\text{large shirt})=\dfrac{\text{number large shirts}}{\text{number shirts}}=\dfrac{77}{165}=\dfrac{7}{15}$

B. True

$Pr(\text{blue shirt}|\text{large shirt})=\dfrac{\text{number blue large shirts}}{\text{number large shirts}}=\dfrac{35}{77}=\dfrac{5}{11}\\ \\Pr(\text{blue shirt})=\dfrac{\text{number blue shirts}}{\text{number shirts}}=\dfrac{75}{165}=\dfrac{5}{11}$

C. False

$Pr(\text{shirt is medium and blue})=\dfrac{\text{number medium and blue shirts}}{\text{number shirts}}=\dfrac{48}{165}=\dfrac{16}{55}\\ \\Pr(\text{medium shirt})=\dfrac{\text{number medium shirts}}{\text{number shirts}}=\dfrac{88}{165}=\dfrac{8}{15}$

D. False

$Pr(\text{large shirt}|\text{red shirt})=\dfrac{\text{number red large shirts}}{\text{number red shirts}}=\dfrac{42}{90}=\dfrac{7}{15}\\ \\Pr(\text{red shirt})=\dfrac{\text{number red shirts}}{\text{number shirts}}=\dfrac{90}{165}=\dfrac{6}{11}$