Question

Probability Question, any idea? Thanks

Consider a production line assembling products for a day, total N products were assembled. There is a possibility P1 for a product has one or more defects, and defects occur randomly and independently. A staff doing check for a proportion P2 of total products, and any defects will be found and fixed, so all these checked products will be free of defects (100% pass check). So at the end of day, the manager will randomly select a product for the final checking, then
1) What is the probability that a randomly chosen product free of defects?
2) Suppose there are 50 products assembled. What is the distribution of the number of products with free of defects?
3) Suppose P1 = 5%, What is the minimum value of the proportion P2 of products needed to be checked so that there is a 50% probability that none of the 50 products have defects?

Answer 1

the probability that a product is still defective after first check is...
(1-P2)*P1
thus probability manager chooses product with no defect is...
1-(1-P2)*P1

Answer 2

thanks @dumbcow , this is what I did as well. However I have no confidence my answer is right. I will keep doing next question, when I finish it, could you please help me to check?

The next question is "Suppose there are 50 products assembled. What is the distribution of the number of products with free of defects?"

Answer 3

sure

Answer 4

@dumbcow I am finished, thanks in advance for checking my answers

n=50 and decide this is Binomial distribution, so y = 0,1,2,3...50 as number of products free of defects. \[P(Y=y) = \left(\begin{matrix}n \\ y\end{matrix}\right) p^{y} (1-p) ^{n-y} | y = 0,1,2 .... 50 \]

\[P(Y=y) = \left(\begin{matrix}50 \\ y\end{matrix}\right) p^{y} (1-p) ^{50-y} | y = 0,1,2 .... 50 \]

and we know from first question the probability of product free of defect is 1 - (1-P2)*P1

so replace p in above formula to 1 - (1-P2)*P1

so the distribution will be \[P(Y=y) = \left(\begin{matrix}50 \\ y\end{matrix}\right) [1-(1-P2)*P1]^{y} * [(1-P2)*P1] ^{50-y} | y = 0,1,2 .... 50 \]

Answer 5

looks good ... i agree :)

Answer 6

That's great, thanks. @dumbcow I will keep doing the last question.
It is "Suppose P1 = 5%, What is the minimum value of the proportion P2 of products needed
to be checked so that there is a 50% probability that none of the 50 products have defects?"

Answer 7

@dumbcow I think I am doing something wrong as I got negative value of P2

first none of the 50 products have defects means y= 0
so \[P(Y=0) = \left(\begin{matrix}50 \\ 0\end{matrix}\right) * [1-(1-P2)*0.05]^{0} * [(1-P2)*0.05]^{50}\]

so \[P(Y=0) = 1 * 1 * (0.05-0.05P2)^{50}\]
then 50% probability means \[(0.05-0.05P2)^{50} = 0.5\]
\[\frac{ 0.05 - \sqrt[50]{0.5} }{ 0.05 } \approx -18.72\] ???

Answer 8

oh thats why....y is 50

"n=50 and decide this is Binomial distribution, so y = 0,1,2,3...50 as number of products free of defects. "

50 products free of defects

Answer 9

basically what you showed correctly is that its impossible for a 50% chance all products have defects :)

Answer 10

@dumbcow Yes I made a silly mistake :) It should be \[P(Y=50) = \left(\begin{matrix}50 \\ 50\end{matrix}\right) * [1-(1-P2)*0.05]^{50} * [(1-P2)*0.05]^{0} = (1-0.05+0.05P2)^{50} \]
\[ (1-0.05+0.05P2)^{50} = 0.5\]

\[P2 = \frac{ \sqrt[50]{0.5}+0.05-1 }{ 0.05 } \approx \frac{ 0.03623 }{ 0.05 } \approx 0.725\]
Thank you for your help

Answer 11

yep , no problem