Probability Defective Items Binomial Distribution
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Suppose that a large lot with 10000 manufactured items has 30 percent defective items and 70 percent nondefective. You choose a subset of 10 items to test.
(a) What is the probability that at most 1 of the 10 test items is defective?
(b) Approximate the previous answer using the binomial distribution.
I am getting for (a) that $P(text{at most 1 def item}) = 0.7^{10} + {10choose 1} cdot 0.3^1 cdot 0.7^9$
I do not understand what is meant by (b), since the answer for (a) already uses binomial distribution?
probability binomial-distribution
edited Jul 21 '15 at 3:19
Conrado Costa
4,4421032
asked Jul 21 '15 at 3:14
Dee ChantelleDee Chantelle
707
$endgroup$
add a comment |
$begingroup$
Suppose that a large lot with 10000 manufactured items has 30 percent defective items and 70 percent nondefective. You choose a subset of 10 items to test.
(a) What is the probability that at most 1 of the 10 test items is defective?
(b) Approximate the previous answer using the binomial distribution.
I am getting for (a) that $P(text{at most 1 def item}) = 0.7^{10} + {10choose 1} cdot 0.3^1 cdot 0.7^9$
I do not understand what is meant by (b), since the answer for (a) already uses binomial distribution?
probability binomial-distribution
edited Jul 21 '15 at 3:19
Conrado Costa
4,4421032
asked Jul 21 '15 at 3:14
Dee ChantelleDee Chantelle
707
$endgroup$
add a comment |
$begingroup$
Suppose that a large lot with 10000 manufactured items has 30 percent defective items and 70 percent nondefective. You choose a subset of 10 items to test.
(a) What is the probability that at most 1 of the 10 test items is defective?
(b) Approximate the previous answer using the binomial distribution.
I am getting for (a) that $P(text{at most 1 def item}) = 0.7^{10} + {10choose 1} cdot 0.3^1 cdot 0.7^9$
I do not understand what is meant by (b), since the answer for (a) already uses binomial distribution?
probability binomial-distribution
edited Jul 21 '15 at 3:19
Conrado Costa
4,4421032
asked Jul 21 '15 at 3:14
Dee ChantelleDee Chantelle
707
$endgroup$
Suppose that a large lot with 10000 manufactured items has 30 percent defective items and 70 percent nondefective. You choose a subset of 10 items to test.
(a) What is the probability that at most 1 of the 10 test items is defective?
(b) Approximate the previous answer using the binomial distribution.
I am getting for (a) that $P(text{at most 1 def item}) = 0.7^{10} + {10choose 1} cdot 0.3^1 cdot 0.7^9$
I do not understand what is meant by (b), since the answer for (a) already uses binomial distribution?
probability binomial-distribution
probability binomial-distribution
edited Jul 21 '15 at 3:19
Conrado Costa
4,4421032
asked Jul 21 '15 at 3:14
Dee ChantelleDee Chantelle
707
edited Jul 21 '15 at 3:19
Conrado Costa
4,4421032
asked Jul 21 '15 at 3:14
Dee ChantelleDee Chantelle
707
edited Jul 21 '15 at 3:19
Conrado Costa
4,4421032
edited Jul 21 '15 at 3:19
Conrado Costa
4,4421032
edited Jul 21 '15 at 3:19
Conrado Costa
4,4421032
4,4421032
asked Jul 21 '15 at 3:14
Dee ChantelleDee Chantelle
707
asked Jul 21 '15 at 3:14
Dee ChantelleDee Chantelle
707
asked Jul 21 '15 at 3:14
Dee ChantelleDee Chantelle
707
707
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The $10$ items are not chosen independently since they are chosen without replacement. If it were with replacement, with a tiny chance that the same item might be chosen more than once, then the binomial distribution would be exact rather than a very close approximation. As it is, part (a) must use a hypergeometric distribution. The answer you've written for part (a) is in fact a correct answer to part (b).
answered Jul 21 '15 at 3:21
Michael HardyMichael Hardy
1
$endgroup$
$begingroup$
so the answer for (a) would be 3000C0 * 7000 C 10 / 10000 C 10 +3000 C1 * 7000 C9 /10000 C 10 ?
$endgroup$
– Dee Chantelle
Jul 21 '15 at 3:22
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@DeeChantelle Yes.
$endgroup$
– Graham Kemp
Jul 21 '15 at 3:25
$begingroup$
The hypergeometric distribution has the same expected value as the binomial distribution and a very slightly smaller variance. If instead of 10000 there were only 40, then the variance of the hypergeometric distribution would be far smaller, but the variance of the binomial distribution would still be the same.
$endgroup$
– Michael Hardy
Jul 21 '15 at 3:28
add a comment |
$begingroup$
The binomial distribution is only an approximation
More explicitly
$P(text{at most 1 defective }) = P(text{ no defective item}) + P(text{1 defective item}) \= frac{7000}{10000}frac{6999}{9999}ldots frac{6991}{9991} + 10 frac{3000}{10000}frac{7000}{9999}frac{6999}{9998} ldots frac{6992}{9991} $
answered Jul 21 '15 at 3:30
Conrado CostaConrado Costa
4,4421032
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The $10$ items are not chosen independently since they are chosen without replacement. If it were with replacement, with a tiny chance that the same item might be chosen more than once, then the binomial distribution would be exact rather than a very close approximation. As it is, part (a) must use a hypergeometric distribution. The answer you've written for part (a) is in fact a correct answer to part (b).
answered Jul 21 '15 at 3:21
Michael HardyMichael Hardy
1
$endgroup$
$begingroup$
so the answer for (a) would be 3000C0 * 7000 C 10 / 10000 C 10 +3000 C1 * 7000 C9 /10000 C 10 ?
$endgroup$
– Dee Chantelle
Jul 21 '15 at 3:22
$begingroup$
@DeeChantelle Yes.
$endgroup$
– Graham Kemp
Jul 21 '15 at 3:25
$begingroup$
The hypergeometric distribution has the same expected value as the binomial distribution and a very slightly smaller variance. If instead of 10000 there were only 40, then the variance of the hypergeometric distribution would be far smaller, but the variance of the binomial distribution would still be the same.
$endgroup$
– Michael Hardy
Jul 21 '15 at 3:28
add a comment |
$begingroup$
The $10$ items are not chosen independently since they are chosen without replacement. If it were with replacement, with a tiny chance that the same item might be chosen more than once, then the binomial distribution would be exact rather than a very close approximation. As it is, part (a) must use a hypergeometric distribution. The answer you've written for part (a) is in fact a correct answer to part (b).
answered Jul 21 '15 at 3:21
Michael HardyMichael Hardy
1
$endgroup$
$begingroup$
so the answer for (a) would be 3000C0 * 7000 C 10 / 10000 C 10 +3000 C1 * 7000 C9 /10000 C 10 ?
$endgroup$
– Dee Chantelle
Jul 21 '15 at 3:22
$begingroup$
@DeeChantelle Yes.
$endgroup$
– Graham Kemp
Jul 21 '15 at 3:25
$begingroup$
The hypergeometric distribution has the same expected value as the binomial distribution and a very slightly smaller variance. If instead of 10000 there were only 40, then the variance of the hypergeometric distribution would be far smaller, but the variance of the binomial distribution would still be the same.
$endgroup$
– Michael Hardy
Jul 21 '15 at 3:28
add a comment |
$begingroup$
The $10$ items are not chosen independently since they are chosen without replacement. If it were with replacement, with a tiny chance that the same item might be chosen more than once, then the binomial distribution would be exact rather than a very close approximation. As it is, part (a) must use a hypergeometric distribution. The answer you've written for part (a) is in fact a correct answer to part (b).
answered Jul 21 '15 at 3:21
Michael HardyMichael Hardy
1
$endgroup$
The $10$ items are not chosen independently since they are chosen without replacement. If it were with replacement, with a tiny chance that the same item might be chosen more than once, then the binomial distribution would be exact rather than a very close approximation. As it is, part (a) must use a hypergeometric distribution. The answer you've written for part (a) is in fact a correct answer to part (b).
answered Jul 21 '15 at 3:21
Michael HardyMichael Hardy
1
edited Jul 21 '15 at 3:29
answered Jul 21 '15 at 3:21
Michael HardyMichael Hardy
1
answered Jul 21 '15 at 3:21
Michael HardyMichael Hardy
1
answered Jul 21 '15 at 3:21
Michael HardyMichael Hardy
1
1
$begingroup$
so the answer for (a) would be 3000C0 * 7000 C 10 / 10000 C 10 +3000 C1 * 7000 C9 /10000 C 10 ?
$endgroup$
– Dee Chantelle
Jul 21 '15 at 3:22
$begingroup$
@DeeChantelle Yes.
$endgroup$
– Graham Kemp
Jul 21 '15 at 3:25
$begingroup$
The hypergeometric distribution has the same expected value as the binomial distribution and a very slightly smaller variance. If instead of 10000 there were only 40, then the variance of the hypergeometric distribution would be far smaller, but the variance of the binomial distribution would still be the same.
$endgroup$
– Michael Hardy
Jul 21 '15 at 3:28
add a comment |
$begingroup$
so the answer for (a) would be 3000C0 * 7000 C 10 / 10000 C 10 +3000 C1 * 7000 C9 /10000 C 10 ?
$endgroup$
– Dee Chantelle
Jul 21 '15 at 3:22
$begingroup$
@DeeChantelle Yes.
$endgroup$
– Graham Kemp
Jul 21 '15 at 3:25
$begingroup$
The hypergeometric distribution has the same expected value as the binomial distribution and a very slightly smaller variance. If instead of 10000 there were only 40, then the variance of the hypergeometric distribution would be far smaller, but the variance of the binomial distribution would still be the same.
$endgroup$
– Michael Hardy
Jul 21 '15 at 3:28
$begingroup$
so the answer for (a) would be 3000C0 * 7000 C 10 / 10000 C 10 +3000 C1 * 7000 C9 /10000 C 10 ?
$endgroup$
– Dee Chantelle
Jul 21 '15 at 3:22
$begingroup$
so the answer for (a) would be 3000C0 * 7000 C 10 / 10000 C 10 +3000 C1 * 7000 C9 /10000 C 10 ?
$endgroup$
– Dee Chantelle
Jul 21 '15 at 3:22
$begingroup$
@DeeChantelle Yes.
$endgroup$
– Graham Kemp
Jul 21 '15 at 3:25
$begingroup$
@DeeChantelle Yes.
$endgroup$
– Graham Kemp
Jul 21 '15 at 3:25
$begingroup$
The hypergeometric distribution has the same expected value as the binomial distribution and a very slightly smaller variance. If instead of 10000 there were only 40, then the variance of the hypergeometric distribution would be far smaller, but the variance of the binomial distribution would still be the same.
$endgroup$
– Michael Hardy
Jul 21 '15 at 3:28
$begingroup$
The hypergeometric distribution has the same expected value as the binomial distribution and a very slightly smaller variance. If instead of 10000 there were only 40, then the variance of the hypergeometric distribution would be far smaller, but the variance of the binomial distribution would still be the same.
$endgroup$
– Michael Hardy
Jul 21 '15 at 3:28
add a comment |
$begingroup$
The binomial distribution is only an approximation
More explicitly
$P(text{at most 1 defective }) = P(text{ no defective item}) + P(text{1 defective item}) \= frac{7000}{10000}frac{6999}{9999}ldots frac{6991}{9991} + 10 frac{3000}{10000}frac{7000}{9999}frac{6999}{9998} ldots frac{6992}{9991} $
answered Jul 21 '15 at 3:30
Conrado CostaConrado Costa
4,4421032
$endgroup$
add a comment |
$begingroup$
The binomial distribution is only an approximation
More explicitly
$P(text{at most 1 defective }) = P(text{ no defective item}) + P(text{1 defective item}) \= frac{7000}{10000}frac{6999}{9999}ldots frac{6991}{9991} + 10 frac{3000}{10000}frac{7000}{9999}frac{6999}{9998} ldots frac{6992}{9991} $
answered Jul 21 '15 at 3:30
Conrado CostaConrado Costa
4,4421032
$endgroup$
add a comment |
$begingroup$
The binomial distribution is only an approximation
More explicitly
$P(text{at most 1 defective }) = P(text{ no defective item}) + P(text{1 defective item}) \= frac{7000}{10000}frac{6999}{9999}ldots frac{6991}{9991} + 10 frac{3000}{10000}frac{7000}{9999}frac{6999}{9998} ldots frac{6992}{9991} $
answered Jul 21 '15 at 3:30
Conrado CostaConrado Costa
4,4421032
$endgroup$
The binomial distribution is only an approximation
More explicitly
$P(text{at most 1 defective }) = P(text{ no defective item}) + P(text{1 defective item}) \= frac{7000}{10000}frac{6999}{9999}ldots frac{6991}{9991} + 10 frac{3000}{10000}frac{7000}{9999}frac{6999}{9998} ldots frac{6992}{9991} $
answered Jul 21 '15 at 3:30
Conrado CostaConrado Costa
4,4421032
answered Jul 21 '15 at 3:30
Conrado CostaConrado Costa
4,4421032
answered Jul 21 '15 at 3:30
Conrado CostaConrado Costa
4,4421032
answered Jul 21 '15 at 3:30
Conrado CostaConrado Costa
4,4421032
4,4421032
add a comment |
add a comment |
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