Question

In any given production run, a manufacturer produces parts with defects at a rate of one out of every ten parts produced. If a customer randomly tests three parts off the production line, what is the probability that no more than one of the parts will have a defect?

Answer 1

Yeah, basically I know the probability of getting a defect is

\[P(D)_{Probability~of~getting~a~defect} = \frac{ 1 }{ 10} \]

so the complement of that, the probability a defect won't happen is

\[1-P(D) = \frac{ 9 }{ 10 }\]

If three parts are randomly selected.
Trying to figure out what's the probability that no more than 1 of the parts will have a defect.

Answer 2

Was thinking that since we're looking for say 1/3 parts will have a defect.

\[\frac{ number~of~desired~outcomes }{ total~possible~outcomes }\]

~\[\frac{ 1 }{ 3 }\]

then I thought this was the chance we could get a part with a defect out of the three, but I knew that the chance had to be much lower than 1/10 so what I did was this.

\[P(B)*P(D) = (\frac{ 1 }{ 10})*(\frac{ 1 }{ 3 }) = (\frac{ 1 }{ 30 })\]

Answer 3

@Michele_Laino are you good with probability?

Answer 4

@Owlcoffee

Answer 5

@mayankdevnani

Answer 6

no, I'm sorry, I'm not good with probability

Answer 7

@Kainui help? @TheSmartOne

Answer 8

@ayonnaleflore

Answer 9

what exactly are you trying to find

Answer 10

If a customer randomly tests three parts off the production line, what is the probability that no more than one of the parts will have a defect?

Answer 11

basically you take 3 parts off the assembly line and say, well what's the probability one of those parts will be defective.

Answer 12

it would be 1 out of 3

Answer 13

simply because only one might be defective and there are three production lines

Answer 14

@Photon336 what are the answer choices?

Answer 15

there aren't any.

Answer 16

okay the answer would be \[\frac{ 3 }{ 9 }\] simplified to \[\frac{ 1 }{ 3 }\]

Answer 17

i'm not sure about the answer

Answer 18

@imqwerty @ParthKohli

Answer 19

If none of them is to have a defect, then the probability is\[\left(\frac{9}{10}\right)^3\]If exactly one of them is to have a defect, then the probability is\[\binom{3}1\cdot \frac{1}{10} \cdot \left(\frac{9}{10}\right)^2\]...I think.

Answer 20

So we just add those two, yeah

Answer 21

yep, i think too like this !

Answer 22

@Photon336 do you know the answer?

Answer 23

Just use binomial probability.

Answer 24

http://www.stat.yale.edu/Courses/1997-98/101/binprob.gif

n=3, p=0.1

You want P(X=<1) and can use \[\large P(X \le 1) = P(X=0)+P(X=1)\]

Answer 25

Wow! i forgot this (my bad)

Answer 26

but what about PARTHKOHLI'S answer ?

Answer 27

Yes, that's correct.

Answer 28

The answer key says 97.2%

Answer 29

If you use binomial probability, you'll get that.

Answer 30

I didn't learn that yet I'll definitely look into that formula

Answer 31

My thinking was that, well if you had a 1/10th chance of getting a defective part, I was thinking to multiply that by

(1/10)*(1/3) but it's wrong.

Answer 32

Are you sure you haven't learned it? Because otherwise you'd just be deriving the binomial probability formula yourself, to solve this.

Answer 33

so essentially there's no way to do this problem without binomial probability formula?

Answer 34

Yes. Because this is a binomial probability situation. What ParthKohli did was really the binomial formula.

Answer 35

I see thank you I'll have to look at this again