Mixing
Probability How many samples of five contian exactly one chip
$begingroup$
Hey guys I hav a task which I cannot solve, can u help me ?
140 semiconductor chips are inspected by choosing a sample of five chips. Assume 10 of the chips do not conform to customer requirements.
How many diffrent samples are possible -->140/5=28 am i right ?
and
How many samples of five contain exactly one nonconforming chip ?
probability
asked Mar 13 '15 at 9:58
Bartek KordalskiBartek Kordalski
104
$endgroup$
add a comment |
$begingroup$
Hey guys I hav a task which I cannot solve, can u help me ?
140 semiconductor chips are inspected by choosing a sample of five chips. Assume 10 of the chips do not conform to customer requirements.
How many diffrent samples are possible -->140/5=28 am i right ?
and
How many samples of five contain exactly one nonconforming chip ?
probability
asked Mar 13 '15 at 9:58
Bartek KordalskiBartek Kordalski
104
$endgroup$
$begingroup$
You're answering the question of how many samples there will be if we continue to take five-chip samples from the original set without replacement. That is, if we take the $140$ chips and start putting them in boxes, five chips in each box, we'll fill $28$ boxes with no chips left over. But this result won't help you find the probability that your first sample contains one of the nonconforming chips, which I strongly suspect is the point of this exercise. The answer $416,965,528$ (below) is more applicable to that probability.
$endgroup$
– David K
Mar 13 '15 at 13:55
add a comment |
$begingroup$
Hey guys I hav a task which I cannot solve, can u help me ?
140 semiconductor chips are inspected by choosing a sample of five chips. Assume 10 of the chips do not conform to customer requirements.
How many diffrent samples are possible -->140/5=28 am i right ?
and
How many samples of five contain exactly one nonconforming chip ?
probability
asked Mar 13 '15 at 9:58
Bartek KordalskiBartek Kordalski
104
$endgroup$
Hey guys I hav a task which I cannot solve, can u help me ?
140 semiconductor chips are inspected by choosing a sample of five chips. Assume 10 of the chips do not conform to customer requirements.
How many diffrent samples are possible -->140/5=28 am i right ?
and
How many samples of five contain exactly one nonconforming chip ?
probability
probability
asked Mar 13 '15 at 9:58
Bartek KordalskiBartek Kordalski
104
asked Mar 13 '15 at 9:58
Bartek KordalskiBartek Kordalski
104
asked Mar 13 '15 at 9:58
Bartek KordalskiBartek Kordalski
104
asked Mar 13 '15 at 9:58
Bartek KordalskiBartek Kordalski
104
asked Mar 13 '15 at 9:58
Bartek KordalskiBartek Kordalski
104
104
$begingroup$
You're answering the question of how many samples there will be if we continue to take five-chip samples from the original set without replacement. That is, if we take the $140$ chips and start putting them in boxes, five chips in each box, we'll fill $28$ boxes with no chips left over. But this result won't help you find the probability that your first sample contains one of the nonconforming chips, which I strongly suspect is the point of this exercise. The answer $416,965,528$ (below) is more applicable to that probability.
$endgroup$
– David K
Mar 13 '15 at 13:55
add a comment |
$begingroup$
You're answering the question of how many samples there will be if we continue to take five-chip samples from the original set without replacement. That is, if we take the $140$ chips and start putting them in boxes, five chips in each box, we'll fill $28$ boxes with no chips left over. But this result won't help you find the probability that your first sample contains one of the nonconforming chips, which I strongly suspect is the point of this exercise. The answer $416,965,528$ (below) is more applicable to that probability.
$endgroup$
– David K
Mar 13 '15 at 13:55
$begingroup$
You're answering the question of how many samples there will be if we continue to take five-chip samples from the original set without replacement. That is, if we take the $140$ chips and start putting them in boxes, five chips in each box, we'll fill $28$ boxes with no chips left over. But this result won't help you find the probability that your first sample contains one of the nonconforming chips, which I strongly suspect is the point of this exercise. The answer $416,965,528$ (below) is more applicable to that probability.
$endgroup$
– David K
Mar 13 '15 at 13:55
$begingroup$
You're answering the question of how many samples there will be if we continue to take five-chip samples from the original set without replacement. That is, if we take the $140$ chips and start putting them in boxes, five chips in each box, we'll fill $28$ boxes with no chips left over. But this result won't help you find the probability that your first sample contains one of the nonconforming chips, which I strongly suspect is the point of this exercise. The answer $416,965,528$ (below) is more applicable to that probability.
$endgroup$
– David K
Mar 13 '15 at 13:55
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
In general if you have $n$ objects, and you want to choose $k$ of them, there are $binom{n}{k}$ ways to do so, where
$$ binom{n}{k} = frac{n!}{k! (n-k)!}.$$
So in your case $n= 140$ is the number of chips, and we want to take samples of size $k=5$, so there are $binom{140}{5} = 416,965,528$ ways to sample the chips.
To count the number of samples (of size 5) containing exactly one of the 10 faulty chips: first select the faulty chip, there are $binom{10}{1}=10$ ways to do this. Now choose any four chips from the remaining 130 working chips, there are $binom{130}{4}$ ways for this, so in total there are
$$ binom{10}{1} binom{130}{4} = 10 * 11358880 = 113,588,800$$
possible choices.
edited Feb 2 at 5:21
Michael Servilla
105
answered Mar 13 '15 at 13:46
owen88owen88
3,6741021
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
In general if you have $n$ objects, and you want to choose $k$ of them, there are $binom{n}{k}$ ways to do so, where
$$ binom{n}{k} = frac{n!}{k! (n-k)!}.$$
So in your case $n= 140$ is the number of chips, and we want to take samples of size $k=5$, so there are $binom{140}{5} = 416,965,528$ ways to sample the chips.
To count the number of samples (of size 5) containing exactly one of the 10 faulty chips: first select the faulty chip, there are $binom{10}{1}=10$ ways to do this. Now choose any four chips from the remaining 130 working chips, there are $binom{130}{4}$ ways for this, so in total there are
$$ binom{10}{1} binom{130}{4} = 10 * 11358880 = 113,588,800$$
possible choices.
edited Feb 2 at 5:21
Michael Servilla
105
answered Mar 13 '15 at 13:46
owen88owen88
3,6741021
$endgroup$
add a comment |
$begingroup$
In general if you have $n$ objects, and you want to choose $k$ of them, there are $binom{n}{k}$ ways to do so, where
$$ binom{n}{k} = frac{n!}{k! (n-k)!}.$$
So in your case $n= 140$ is the number of chips, and we want to take samples of size $k=5$, so there are $binom{140}{5} = 416,965,528$ ways to sample the chips.
To count the number of samples (of size 5) containing exactly one of the 10 faulty chips: first select the faulty chip, there are $binom{10}{1}=10$ ways to do this. Now choose any four chips from the remaining 130 working chips, there are $binom{130}{4}$ ways for this, so in total there are
$$ binom{10}{1} binom{130}{4} = 10 * 11358880 = 113,588,800$$
possible choices.
edited Feb 2 at 5:21
Michael Servilla
105
answered Mar 13 '15 at 13:46
owen88owen88
3,6741021
$endgroup$
add a comment |
$begingroup$
In general if you have $n$ objects, and you want to choose $k$ of them, there are $binom{n}{k}$ ways to do so, where
$$ binom{n}{k} = frac{n!}{k! (n-k)!}.$$
So in your case $n= 140$ is the number of chips, and we want to take samples of size $k=5$, so there are $binom{140}{5} = 416,965,528$ ways to sample the chips.
To count the number of samples (of size 5) containing exactly one of the 10 faulty chips: first select the faulty chip, there are $binom{10}{1}=10$ ways to do this. Now choose any four chips from the remaining 130 working chips, there are $binom{130}{4}$ ways for this, so in total there are
$$ binom{10}{1} binom{130}{4} = 10 * 11358880 = 113,588,800$$
possible choices.
edited Feb 2 at 5:21
Michael Servilla
105
answered Mar 13 '15 at 13:46
owen88owen88
3,6741021
$endgroup$
In general if you have $n$ objects, and you want to choose $k$ of them, there are $binom{n}{k}$ ways to do so, where
$$ binom{n}{k} = frac{n!}{k! (n-k)!}.$$
So in your case $n= 140$ is the number of chips, and we want to take samples of size $k=5$, so there are $binom{140}{5} = 416,965,528$ ways to sample the chips.
To count the number of samples (of size 5) containing exactly one of the 10 faulty chips: first select the faulty chip, there are $binom{10}{1}=10$ ways to do this. Now choose any four chips from the remaining 130 working chips, there are $binom{130}{4}$ ways for this, so in total there are
$$ binom{10}{1} binom{130}{4} = 10 * 11358880 = 113,588,800$$
possible choices.
edited Feb 2 at 5:21
Michael Servilla
105
answered Mar 13 '15 at 13:46
owen88owen88
3,6741021
edited Feb 2 at 5:21
Michael Servilla
105
edited Feb 2 at 5:21
Michael Servilla
105
edited Feb 2 at 5:21
Michael Servilla
105
105
answered Mar 13 '15 at 13:46
owen88owen88
3,6741021
answered Mar 13 '15 at 13:46
owen88owen88
3,6741021
answered Mar 13 '15 at 13:46
owen88owen88
3,6741021
3,6741021
add a comment |
add a comment |
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$begingroup$
You're answering the question of how many samples there will be if we continue to take five-chip samples from the original set without replacement. That is, if we take the $140$ chips and start putting them in boxes, five chips in each box, we'll fill $28$ boxes with no chips left over. But this result won't help you find the probability that your first sample contains one of the nonconforming chips, which I strongly suspect is the point of this exercise. The answer $416,965,528$ (below) is more applicable to that probability.
$endgroup$
– David K
Mar 13 '15 at 13:55