Probability and Counting
I have just picked up a text on discrete math and its been ages since I have done this so if anyone can show me with steps to correct my fault, that would be so great.
Repair facility has 25 failed keyboards, 6 of which have electrical defects and 19 which have mechanical defects.
(1) How many ways are there to randomly select 5 of these keyboards for a thorough inspection (order does not matter)? 25 choose 5
(2) IN how many ways can a sample of 5 keyboards be selected so that exactly tow have an electrical defect? (25 choose 5)/(6 choose 5)
(3) If a sample of 5 keyboards is randomly selected, what is the probability that at least 4 of these will have a mechanical defect? (5 choose 4)....?
Am I correct in my thinking logic or way off?
probability discrete-mathematics
asked Sep 18 '13 at 4:35
hherklj kljkljkljhherklj kljkljklj
5624
add a comment |
I have just picked up a text on discrete math and its been ages since I have done this so if anyone can show me with steps to correct my fault, that would be so great.
Repair facility has 25 failed keyboards, 6 of which have electrical defects and 19 which have mechanical defects.
(1) How many ways are there to randomly select 5 of these keyboards for a thorough inspection (order does not matter)? 25 choose 5
(2) IN how many ways can a sample of 5 keyboards be selected so that exactly tow have an electrical defect? (25 choose 5)/(6 choose 5)
(3) If a sample of 5 keyboards is randomly selected, what is the probability that at least 4 of these will have a mechanical defect? (5 choose 4)....?
Am I correct in my thinking logic or way off?
probability discrete-mathematics
asked Sep 18 '13 at 4:35
hherklj kljkljkljhherklj kljkljklj
5624
add a comment |
I have just picked up a text on discrete math and its been ages since I have done this so if anyone can show me with steps to correct my fault, that would be so great.
Repair facility has 25 failed keyboards, 6 of which have electrical defects and 19 which have mechanical defects.
(1) How many ways are there to randomly select 5 of these keyboards for a thorough inspection (order does not matter)? 25 choose 5
(2) IN how many ways can a sample of 5 keyboards be selected so that exactly tow have an electrical defect? (25 choose 5)/(6 choose 5)
(3) If a sample of 5 keyboards is randomly selected, what is the probability that at least 4 of these will have a mechanical defect? (5 choose 4)....?
Am I correct in my thinking logic or way off?
probability discrete-mathematics
asked Sep 18 '13 at 4:35
hherklj kljkljkljhherklj kljkljklj
5624
I have just picked up a text on discrete math and its been ages since I have done this so if anyone can show me with steps to correct my fault, that would be so great.
Repair facility has 25 failed keyboards, 6 of which have electrical defects and 19 which have mechanical defects.
(1) How many ways are there to randomly select 5 of these keyboards for a thorough inspection (order does not matter)? 25 choose 5
(2) IN how many ways can a sample of 5 keyboards be selected so that exactly tow have an electrical defect? (25 choose 5)/(6 choose 5)
(3) If a sample of 5 keyboards is randomly selected, what is the probability that at least 4 of these will have a mechanical defect? (5 choose 4)....?
Am I correct in my thinking logic or way off?
probability discrete-mathematics
probability discrete-mathematics
asked Sep 18 '13 at 4:35
hherklj kljkljkljhherklj kljkljklj
5624
asked Sep 18 '13 at 4:35
hherklj kljkljkljhherklj kljkljklj
5624
edited Sep 18 '13 at 4:42
hherklj kljkljklj
asked Sep 18 '13 at 4:35
hherklj kljkljkljhherklj kljkljklj
5624
asked Sep 18 '13 at 4:35
hherklj kljkljkljhherklj kljkljklj
5624
asked Sep 18 '13 at 4:35
hherklj kljkljkljhherklj kljkljklj
5624
5624
add a comment |
add a comment |
1 Answer
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For part 2, you need exactly two out of five so there are ${6 choose 2}$ ways to do that leaving ${19choose 3}$ for the remaining mechanically failed devices. So in total there are ${6choose 2}{19choose 3}$ total ways. Now for part three, think of what it means to say "at least" and use the same process. So you need a sample with exactly 4 plus a sample with 5....
For cases where we are sampling without replacement, a hyper geometric distribution is used. Say you want the probability of $K$ out $N$ successes with a sample size $n$. Then the mass function looks like
$$P(X<k)=sum_{i=0}^k{frac{{Kchoose i}{N-K choose n-i}}{N choose n}}$$
answered Sep 18 '13 at 4:42
Eleven-ElevenEleven-Eleven
5,40572659
So the for 3, by "at least", would it be (5 choose 4)*(19 choose 1)? Or am I thinking of "at least" from the wrong perspective?
– hherklj kljkljklj
Sep 18 '13 at 4:49
2
Wrong perspective... You have 19 mechanically failed keyboards and 6 electrically failed keyboards. So it should be ${6choose 1}{19choose 4}$, but at least means you can have 4 or 5 mechanically defective keyboards in your sample.... Can you finish it then?
– Eleven-Eleven
Sep 18 '13 at 4:55
So then if I am leaning in the right direction, would it be ((6 choose 1)*(19 choose 4))/(5 choose 4) ?
– hherklj kljkljklj
Sep 18 '13 at 4:58
No...you have a new criterion to meet with the second case of 5 mechanically failed keyboards so that is ${19choose 5}$. Now you must sum the two cases. This is counting , not probability so there should be no division in your method. If you were looking for the probabilities of these cases you divide by ${25choose 5}$. This is sampling without replacement and calls for the hyper geometric distribution.
– Eleven-Eleven
Sep 18 '13 at 5:04
Okay so I saw part 3 calls for probability so you should have $frac{{19choose 4}{6choose 1}+{19choose 5}{6choose 0}}{{25choose 5}}$
– Eleven-Eleven
Sep 18 '13 at 5:08
|
show 1 more comment
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1 Answer
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For part 2, you need exactly two out of five so there are ${6 choose 2}$ ways to do that leaving ${19choose 3}$ for the remaining mechanically failed devices. So in total there are ${6choose 2}{19choose 3}$ total ways. Now for part three, think of what it means to say "at least" and use the same process. So you need a sample with exactly 4 plus a sample with 5....
For cases where we are sampling without replacement, a hyper geometric distribution is used. Say you want the probability of $K$ out $N$ successes with a sample size $n$. Then the mass function looks like
$$P(X<k)=sum_{i=0}^k{frac{{Kchoose i}{N-K choose n-i}}{N choose n}}$$
answered Sep 18 '13 at 4:42
Eleven-ElevenEleven-Eleven
5,40572659
So the for 3, by "at least", would it be (5 choose 4)*(19 choose 1)? Or am I thinking of "at least" from the wrong perspective?
– hherklj kljkljklj
Sep 18 '13 at 4:49
2
Wrong perspective... You have 19 mechanically failed keyboards and 6 electrically failed keyboards. So it should be ${6choose 1}{19choose 4}$, but at least means you can have 4 or 5 mechanically defective keyboards in your sample.... Can you finish it then?
– Eleven-Eleven
Sep 18 '13 at 4:55
So then if I am leaning in the right direction, would it be ((6 choose 1)*(19 choose 4))/(5 choose 4) ?
– hherklj kljkljklj
Sep 18 '13 at 4:58
No...you have a new criterion to meet with the second case of 5 mechanically failed keyboards so that is ${19choose 5}$. Now you must sum the two cases. This is counting , not probability so there should be no division in your method. If you were looking for the probabilities of these cases you divide by ${25choose 5}$. This is sampling without replacement and calls for the hyper geometric distribution.
– Eleven-Eleven
Sep 18 '13 at 5:04
Okay so I saw part 3 calls for probability so you should have $frac{{19choose 4}{6choose 1}+{19choose 5}{6choose 0}}{{25choose 5}}$
– Eleven-Eleven
Sep 18 '13 at 5:08
|
show 1 more comment
For part 2, you need exactly two out of five so there are ${6 choose 2}$ ways to do that leaving ${19choose 3}$ for the remaining mechanically failed devices. So in total there are ${6choose 2}{19choose 3}$ total ways. Now for part three, think of what it means to say "at least" and use the same process. So you need a sample with exactly 4 plus a sample with 5....
For cases where we are sampling without replacement, a hyper geometric distribution is used. Say you want the probability of $K$ out $N$ successes with a sample size $n$. Then the mass function looks like
$$P(X<k)=sum_{i=0}^k{frac{{Kchoose i}{N-K choose n-i}}{N choose n}}$$
answered Sep 18 '13 at 4:42
Eleven-ElevenEleven-Eleven
5,40572659
So the for 3, by "at least", would it be (5 choose 4)*(19 choose 1)? Or am I thinking of "at least" from the wrong perspective?
– hherklj kljkljklj
Sep 18 '13 at 4:49
2
Wrong perspective... You have 19 mechanically failed keyboards and 6 electrically failed keyboards. So it should be ${6choose 1}{19choose 4}$, but at least means you can have 4 or 5 mechanically defective keyboards in your sample.... Can you finish it then?
– Eleven-Eleven
Sep 18 '13 at 4:55
So then if I am leaning in the right direction, would it be ((6 choose 1)*(19 choose 4))/(5 choose 4) ?
– hherklj kljkljklj
Sep 18 '13 at 4:58
No...you have a new criterion to meet with the second case of 5 mechanically failed keyboards so that is ${19choose 5}$. Now you must sum the two cases. This is counting , not probability so there should be no division in your method. If you were looking for the probabilities of these cases you divide by ${25choose 5}$. This is sampling without replacement and calls for the hyper geometric distribution.
– Eleven-Eleven
Sep 18 '13 at 5:04
Okay so I saw part 3 calls for probability so you should have $frac{{19choose 4}{6choose 1}+{19choose 5}{6choose 0}}{{25choose 5}}$
– Eleven-Eleven
Sep 18 '13 at 5:08
|
show 1 more comment
For part 2, you need exactly two out of five so there are ${6 choose 2}$ ways to do that leaving ${19choose 3}$ for the remaining mechanically failed devices. So in total there are ${6choose 2}{19choose 3}$ total ways. Now for part three, think of what it means to say "at least" and use the same process. So you need a sample with exactly 4 plus a sample with 5....
For cases where we are sampling without replacement, a hyper geometric distribution is used. Say you want the probability of $K$ out $N$ successes with a sample size $n$. Then the mass function looks like
$$P(X<k)=sum_{i=0}^k{frac{{Kchoose i}{N-K choose n-i}}{N choose n}}$$
answered Sep 18 '13 at 4:42
Eleven-ElevenEleven-Eleven
5,40572659
For part 2, you need exactly two out of five so there are ${6 choose 2}$ ways to do that leaving ${19choose 3}$ for the remaining mechanically failed devices. So in total there are ${6choose 2}{19choose 3}$ total ways. Now for part three, think of what it means to say "at least" and use the same process. So you need a sample with exactly 4 plus a sample with 5....
For cases where we are sampling without replacement, a hyper geometric distribution is used. Say you want the probability of $K$ out $N$ successes with a sample size $n$. Then the mass function looks like
$$P(X<k)=sum_{i=0}^k{frac{{Kchoose i}{N-K choose n-i}}{N choose n}}$$
answered Sep 18 '13 at 4:42
Eleven-ElevenEleven-Eleven
5,40572659
edited Sep 18 '13 at 11:26
answered Sep 18 '13 at 4:42
Eleven-ElevenEleven-Eleven
5,40572659
answered Sep 18 '13 at 4:42
Eleven-ElevenEleven-Eleven
5,40572659
answered Sep 18 '13 at 4:42
Eleven-ElevenEleven-Eleven
5,40572659
5,40572659
So the for 3, by "at least", would it be (5 choose 4)*(19 choose 1)? Or am I thinking of "at least" from the wrong perspective?
– hherklj kljkljklj
Sep 18 '13 at 4:49
2
Wrong perspective... You have 19 mechanically failed keyboards and 6 electrically failed keyboards. So it should be ${6choose 1}{19choose 4}$, but at least means you can have 4 or 5 mechanically defective keyboards in your sample.... Can you finish it then?
– Eleven-Eleven
Sep 18 '13 at 4:55
So then if I am leaning in the right direction, would it be ((6 choose 1)*(19 choose 4))/(5 choose 4) ?
– hherklj kljkljklj
Sep 18 '13 at 4:58
No...you have a new criterion to meet with the second case of 5 mechanically failed keyboards so that is ${19choose 5}$. Now you must sum the two cases. This is counting , not probability so there should be no division in your method. If you were looking for the probabilities of these cases you divide by ${25choose 5}$. This is sampling without replacement and calls for the hyper geometric distribution.
– Eleven-Eleven
Sep 18 '13 at 5:04
Okay so I saw part 3 calls for probability so you should have $frac{{19choose 4}{6choose 1}+{19choose 5}{6choose 0}}{{25choose 5}}$
– Eleven-Eleven
Sep 18 '13 at 5:08
|
show 1 more comment
So the for 3, by "at least", would it be (5 choose 4)*(19 choose 1)? Or am I thinking of "at least" from the wrong perspective?
– hherklj kljkljklj
Sep 18 '13 at 4:49
2
Wrong perspective... You have 19 mechanically failed keyboards and 6 electrically failed keyboards. So it should be ${6choose 1}{19choose 4}$, but at least means you can have 4 or 5 mechanically defective keyboards in your sample.... Can you finish it then?
– Eleven-Eleven
Sep 18 '13 at 4:55
So then if I am leaning in the right direction, would it be ((6 choose 1)*(19 choose 4))/(5 choose 4) ?
– hherklj kljkljklj
Sep 18 '13 at 4:58
No...you have a new criterion to meet with the second case of 5 mechanically failed keyboards so that is ${19choose 5}$. Now you must sum the two cases. This is counting , not probability so there should be no division in your method. If you were looking for the probabilities of these cases you divide by ${25choose 5}$. This is sampling without replacement and calls for the hyper geometric distribution.
– Eleven-Eleven
Sep 18 '13 at 5:04
Okay so I saw part 3 calls for probability so you should have $frac{{19choose 4}{6choose 1}+{19choose 5}{6choose 0}}{{25choose 5}}$
– Eleven-Eleven
Sep 18 '13 at 5:08
So the for 3, by "at least", would it be (5 choose 4)*(19 choose 1)? Or am I thinking of "at least" from the wrong perspective?
– hherklj kljkljklj
Sep 18 '13 at 4:49
So the for 3, by "at least", would it be (5 choose 4)*(19 choose 1)? Or am I thinking of "at least" from the wrong perspective?
– hherklj kljkljklj
Sep 18 '13 at 4:49
2
2
Wrong perspective... You have 19 mechanically failed keyboards and 6 electrically failed keyboards. So it should be ${6choose 1}{19choose 4}$, but at least means you can have 4 or 5 mechanically defective keyboards in your sample.... Can you finish it then?
– Eleven-Eleven
Sep 18 '13 at 4:55
Wrong perspective... You have 19 mechanically failed keyboards and 6 electrically failed keyboards. So it should be ${6choose 1}{19choose 4}$, but at least means you can have 4 or 5 mechanically defective keyboards in your sample.... Can you finish it then?
– Eleven-Eleven
Sep 18 '13 at 4:55
So then if I am leaning in the right direction, would it be ((6 choose 1)*(19 choose 4))/(5 choose 4) ?
– hherklj kljkljklj
Sep 18 '13 at 4:58
So then if I am leaning in the right direction, would it be ((6 choose 1)*(19 choose 4))/(5 choose 4) ?
– hherklj kljkljklj
Sep 18 '13 at 4:58
No...you have a new criterion to meet with the second case of 5 mechanically failed keyboards so that is ${19choose 5}$. Now you must sum the two cases. This is counting , not probability so there should be no division in your method. If you were looking for the probabilities of these cases you divide by ${25choose 5}$. This is sampling without replacement and calls for the hyper geometric distribution.
– Eleven-Eleven
Sep 18 '13 at 5:04
No...you have a new criterion to meet with the second case of 5 mechanically failed keyboards so that is ${19choose 5}$. Now you must sum the two cases. This is counting , not probability so there should be no division in your method. If you were looking for the probabilities of these cases you divide by ${25choose 5}$. This is sampling without replacement and calls for the hyper geometric distribution.
– Eleven-Eleven
Sep 18 '13 at 5:04
Okay so I saw part 3 calls for probability so you should have $frac{{19choose 4}{6choose 1}+{19choose 5}{6choose 0}}{{25choose 5}}$
– Eleven-Eleven
Sep 18 '13 at 5:08
Okay so I saw part 3 calls for probability so you should have $frac{{19choose 4}{6choose 1}+{19choose 5}{6choose 0}}{{25choose 5}}$
– Eleven-Eleven
Sep 18 '13 at 5:08
|
show 1 more comment
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