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After $10$ inspections with no defect, whats the prob of that number of inspection is no more than $20$?
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I have a problem which I'm not sure how to solve. It goes as follows:
A production line has a $5$% defective rate, and its products are inspected one by one until the first defect is found. Given that the first $10$ inspections do not find any defect, what is the probability that the number of inspections is no more than $20$?
I tried the following approach:
Given that this is a Bernoulli process, I thought that if there were no defective parts encountered in the first 10 inspected products, the next 10 products (till $X_20$) will be independent from the past, which means a new Bernoulli process of 10 trials. So, if $S$ is the sum of each $X_i$, then I could calculate its PMF with parameters $p=.05$ and $t=10$, where $t$ is the number of trials:
$p_s(1) = binom101(0.05)(0.95)^9$
$p_s(1) = .3151$
Yet this is wrong. I'm not sure if I had to use a conditional approach (because the second round of inspection occurs in the conditional universe where there were no defects in the first $10$) so as to say that:
Event A : one is defective in $11$ through $20$
Event B : none is defective in the first $10$
$P(A cap B) = P(A) P(B)$
In which case:
$P(A) = p_s(1) = binom101(0.05)(0.95)^9$
$P(B) = p_s(0) = binom100(0.95)^10$
probability
asked Jul 18 at 14:11
Frank Pinto
165
add a comment |Â
up vote
1
down vote
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I have a problem which I'm not sure how to solve. It goes as follows:
A production line has a $5$% defective rate, and its products are inspected one by one until the first defect is found. Given that the first $10$ inspections do not find any defect, what is the probability that the number of inspections is no more than $20$?
I tried the following approach:
Given that this is a Bernoulli process, I thought that if there were no defective parts encountered in the first 10 inspected products, the next 10 products (till $X_20$) will be independent from the past, which means a new Bernoulli process of 10 trials. So, if $S$ is the sum of each $X_i$, then I could calculate its PMF with parameters $p=.05$ and $t=10$, where $t$ is the number of trials:
$p_s(1) = binom101(0.05)(0.95)^9$
$p_s(1) = .3151$
Yet this is wrong. I'm not sure if I had to use a conditional approach (because the second round of inspection occurs in the conditional universe where there were no defects in the first $10$) so as to say that:
Event A : one is defective in $11$ through $20$
Event B : none is defective in the first $10$
$P(A cap B) = P(A) P(B)$
In which case:
$P(A) = p_s(1) = binom101(0.05)(0.95)^9$
$P(B) = p_s(0) = binom100(0.95)^10$
probability
asked Jul 18 at 14:11
Frank Pinto
165
This might help: en.wikipedia.org/wiki/Geometric_distribution
– Theoretical Economist
Jul 18 at 15:03
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have a problem which I'm not sure how to solve. It goes as follows:
A production line has a $5$% defective rate, and its products are inspected one by one until the first defect is found. Given that the first $10$ inspections do not find any defect, what is the probability that the number of inspections is no more than $20$?
I tried the following approach:
Given that this is a Bernoulli process, I thought that if there were no defective parts encountered in the first 10 inspected products, the next 10 products (till $X_20$) will be independent from the past, which means a new Bernoulli process of 10 trials. So, if $S$ is the sum of each $X_i$, then I could calculate its PMF with parameters $p=.05$ and $t=10$, where $t$ is the number of trials:
$p_s(1) = binom101(0.05)(0.95)^9$
$p_s(1) = .3151$
Yet this is wrong. I'm not sure if I had to use a conditional approach (because the second round of inspection occurs in the conditional universe where there were no defects in the first $10$) so as to say that:
Event A : one is defective in $11$ through $20$
Event B : none is defective in the first $10$
$P(A cap B) = P(A) P(B)$
In which case:
$P(A) = p_s(1) = binom101(0.05)(0.95)^9$
$P(B) = p_s(0) = binom100(0.95)^10$
probability
asked Jul 18 at 14:11
Frank Pinto
165
I have a problem which I'm not sure how to solve. It goes as follows:
A production line has a $5$% defective rate, and its products are inspected one by one until the first defect is found. Given that the first $10$ inspections do not find any defect, what is the probability that the number of inspections is no more than $20$?
I tried the following approach:
Given that this is a Bernoulli process, I thought that if there were no defective parts encountered in the first 10 inspected products, the next 10 products (till $X_20$) will be independent from the past, which means a new Bernoulli process of 10 trials. So, if $S$ is the sum of each $X_i$, then I could calculate its PMF with parameters $p=.05$ and $t=10$, where $t$ is the number of trials:
$p_s(1) = binom101(0.05)(0.95)^9$
$p_s(1) = .3151$
Yet this is wrong. I'm not sure if I had to use a conditional approach (because the second round of inspection occurs in the conditional universe where there were no defects in the first $10$) so as to say that:
Event A : one is defective in $11$ through $20$
Event B : none is defective in the first $10$
$P(A cap B) = P(A) P(B)$
In which case:
$P(A) = p_s(1) = binom101(0.05)(0.95)^9$
$P(B) = p_s(0) = binom100(0.95)^10$
probability
asked Jul 18 at 14:11
Frank Pinto
165
edited Jul 18 at 14:55
asked Jul 18 at 14:11
Frank Pinto
165
asked Jul 18 at 14:11
Frank Pinto
165
asked Jul 18 at 14:11
Frank Pinto
165
165
This might help: en.wikipedia.org/wiki/Geometric_distribution
– Theoretical Economist
Jul 18 at 15:03
add a comment |Â
This might help: en.wikipedia.org/wiki/Geometric_distribution
– Theoretical Economist
Jul 18 at 15:03
This might help: en.wikipedia.org/wiki/Geometric_distribution
– Theoretical Economist
Jul 18 at 15:03
This might help: en.wikipedia.org/wiki/Geometric_distribution
– Theoretical Economist
Jul 18 at 15:03
add a comment |Â
1 Answer
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1
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accepted
Suppose defects are iid. There are more than $20$ inspections if the next $10$ parts to be inspected are not defective. This occurs with probability $0.95^10$.
Hence, the probability that there are no more than $20$ inspections is $1 - 0.95^10$.
answered Jul 18 at 15:06
Theoretical Economist
3,1852628
Thank you. I understand the logic and I accept the answer. One final question: why is it that if I calculate the probability for one defect as my first attempt does, the answer is wrong?
– Frank Pinto
Jul 18 at 15:15
@FrankPinto the probability that you calculated is the probability of exactly one defect in ten inspections. This is different from the probability of there being at least one defect in ten inspections, which is essentially what the problem is asking you to find.
– Theoretical Economist
Jul 18 at 16:02
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Suppose defects are iid. There are more than $20$ inspections if the next $10$ parts to be inspected are not defective. This occurs with probability $0.95^10$.
Hence, the probability that there are no more than $20$ inspections is $1 - 0.95^10$.
answered Jul 18 at 15:06
Theoretical Economist
3,1852628
Thank you. I understand the logic and I accept the answer. One final question: why is it that if I calculate the probability for one defect as my first attempt does, the answer is wrong?
– Frank Pinto
Jul 18 at 15:15
@FrankPinto the probability that you calculated is the probability of exactly one defect in ten inspections. This is different from the probability of there being at least one defect in ten inspections, which is essentially what the problem is asking you to find.
– Theoretical Economist
Jul 18 at 16:02
add a comment |Â
up vote
1
down vote
accepted
Suppose defects are iid. There are more than $20$ inspections if the next $10$ parts to be inspected are not defective. This occurs with probability $0.95^10$.
Hence, the probability that there are no more than $20$ inspections is $1 - 0.95^10$.
answered Jul 18 at 15:06
Theoretical Economist
3,1852628
Thank you. I understand the logic and I accept the answer. One final question: why is it that if I calculate the probability for one defect as my first attempt does, the answer is wrong?
– Frank Pinto
Jul 18 at 15:15
@FrankPinto the probability that you calculated is the probability of exactly one defect in ten inspections. This is different from the probability of there being at least one defect in ten inspections, which is essentially what the problem is asking you to find.
– Theoretical Economist
Jul 18 at 16:02
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Suppose defects are iid. There are more than $20$ inspections if the next $10$ parts to be inspected are not defective. This occurs with probability $0.95^10$.
Hence, the probability that there are no more than $20$ inspections is $1 - 0.95^10$.
answered Jul 18 at 15:06
Theoretical Economist
3,1852628
Suppose defects are iid. There are more than $20$ inspections if the next $10$ parts to be inspected are not defective. This occurs with probability $0.95^10$.
Hence, the probability that there are no more than $20$ inspections is $1 - 0.95^10$.
answered Jul 18 at 15:06
Theoretical Economist
3,1852628
answered Jul 18 at 15:06
Theoretical Economist
3,1852628
answered Jul 18 at 15:06
Theoretical Economist
3,1852628
answered Jul 18 at 15:06
Theoretical Economist
3,1852628
3,1852628
Thank you. I understand the logic and I accept the answer. One final question: why is it that if I calculate the probability for one defect as my first attempt does, the answer is wrong?
– Frank Pinto
Jul 18 at 15:15
@FrankPinto the probability that you calculated is the probability of exactly one defect in ten inspections. This is different from the probability of there being at least one defect in ten inspections, which is essentially what the problem is asking you to find.
– Theoretical Economist
Jul 18 at 16:02
add a comment |Â
Thank you. I understand the logic and I accept the answer. One final question: why is it that if I calculate the probability for one defect as my first attempt does, the answer is wrong?
– Frank Pinto
Jul 18 at 15:15
@FrankPinto the probability that you calculated is the probability of exactly one defect in ten inspections. This is different from the probability of there being at least one defect in ten inspections, which is essentially what the problem is asking you to find.
– Theoretical Economist
Jul 18 at 16:02
Thank you. I understand the logic and I accept the answer. One final question: why is it that if I calculate the probability for one defect as my first attempt does, the answer is wrong?
– Frank Pinto
Jul 18 at 15:15
Thank you. I understand the logic and I accept the answer. One final question: why is it that if I calculate the probability for one defect as my first attempt does, the answer is wrong?
– Frank Pinto
Jul 18 at 15:15
@FrankPinto the probability that you calculated is the probability of exactly one defect in ten inspections. This is different from the probability of there being at least one defect in ten inspections, which is essentially what the problem is asking you to find.
– Theoretical Economist
Jul 18 at 16:02
@FrankPinto the probability that you calculated is the probability of exactly one defect in ten inspections. This is different from the probability of there being at least one defect in ten inspections, which is essentially what the problem is asking you to find.
– Theoretical Economist
Jul 18 at 16:02
add a comment |Â
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This might help: en.wikipedia.org/wiki/Geometric_distribution
– Theoretical Economist
Jul 18 at 15:03