Probability of exactly one defective unit
up vote
2
down vote
favorite
Assume 5 out of 100 units are defective. We pick 3 out of the 100 units at random.
What is the probability that exactly one unit is defective?
My answer would be
$P(text{Defect}=1) = P(text{Defect})times P(text{Not defect})times P(text{Not defect}) = 5/100 times 95/99 times 94/98$
However, I am not sure whether or not this is correct or not. Can someone verify?
probability
edited Dec 9 at 13:33
amWhy
191k28223439
asked Dec 9 at 10:45
CruZ
416
add a comment |
up vote
2
down vote
favorite
Assume 5 out of 100 units are defective. We pick 3 out of the 100 units at random.
What is the probability that exactly one unit is defective?
My answer would be
$P(text{Defect}=1) = P(text{Defect})times P(text{Not defect})times P(text{Not defect}) = 5/100 times 95/99 times 94/98$
However, I am not sure whether or not this is correct or not. Can someone verify?
probability
edited Dec 9 at 13:33
amWhy
191k28223439
asked Dec 9 at 10:45
CruZ
416
1
You can choose the defective in three different ways. Also the number of non-defective units is $95$ not $94$.
– Thomas Shelby
Dec 9 at 10:49
1
Sorry, my mistake, I meant 95/99 and 94/98. But then are you saying that I have to calculate like this: P(D=1) = {DDN}U{DND}U{NDD}, where D is af defective unit and N is a non-defective unit?
– CruZ
Dec 9 at 10:53
2
Yes. You will get the same answer mentioned below.
– Thomas Shelby
Dec 9 at 11:03
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Assume 5 out of 100 units are defective. We pick 3 out of the 100 units at random.
What is the probability that exactly one unit is defective?
My answer would be
$P(text{Defect}=1) = P(text{Defect})times P(text{Not defect})times P(text{Not defect}) = 5/100 times 95/99 times 94/98$
However, I am not sure whether or not this is correct or not. Can someone verify?
probability
edited Dec 9 at 13:33
amWhy
191k28223439
asked Dec 9 at 10:45
CruZ
416
Assume 5 out of 100 units are defective. We pick 3 out of the 100 units at random.
What is the probability that exactly one unit is defective?
My answer would be
$P(text{Defect}=1) = P(text{Defect})times P(text{Not defect})times P(text{Not defect}) = 5/100 times 95/99 times 94/98$
However, I am not sure whether or not this is correct or not. Can someone verify?
probability
probability
edited Dec 9 at 13:33
amWhy
191k28223439
asked Dec 9 at 10:45
CruZ
416
edited Dec 9 at 13:33
amWhy
191k28223439
asked Dec 9 at 10:45
CruZ
416
edited Dec 9 at 13:33
amWhy
191k28223439
edited Dec 9 at 13:33
amWhy
191k28223439
edited Dec 9 at 13:33
amWhy
191k28223439
191k28223439
asked Dec 9 at 10:45
CruZ
416
asked Dec 9 at 10:45
CruZ
416
asked Dec 9 at 10:45
CruZ
416
416
1
You can choose the defective in three different ways. Also the number of non-defective units is $95$ not $94$.
– Thomas Shelby
Dec 9 at 10:49
1
Sorry, my mistake, I meant 95/99 and 94/98. But then are you saying that I have to calculate like this: P(D=1) = {DDN}U{DND}U{NDD}, where D is af defective unit and N is a non-defective unit?
– CruZ
Dec 9 at 10:53
2
Yes. You will get the same answer mentioned below.
– Thomas Shelby
Dec 9 at 11:03
add a comment |
1
You can choose the defective in three different ways. Also the number of non-defective units is $95$ not $94$.
– Thomas Shelby
Dec 9 at 10:49
1
Sorry, my mistake, I meant 95/99 and 94/98. But then are you saying that I have to calculate like this: P(D=1) = {DDN}U{DND}U{NDD}, where D is af defective unit and N is a non-defective unit?
– CruZ
Dec 9 at 10:53
2
Yes. You will get the same answer mentioned below.
– Thomas Shelby
Dec 9 at 11:03
1
1
You can choose the defective in three different ways. Also the number of non-defective units is $95$ not $94$.
– Thomas Shelby
Dec 9 at 10:49
You can choose the defective in three different ways. Also the number of non-defective units is $95$ not $94$.
– Thomas Shelby
Dec 9 at 10:49
1
1
Sorry, my mistake, I meant 95/99 and 94/98. But then are you saying that I have to calculate like this: P(D=1) = {DDN}U{DND}U{NDD}, where D is af defective unit and N is a non-defective unit?
– CruZ
Dec 9 at 10:53
Sorry, my mistake, I meant 95/99 and 94/98. But then are you saying that I have to calculate like this: P(D=1) = {DDN}U{DND}U{NDD}, where D is af defective unit and N is a non-defective unit?
– CruZ
Dec 9 at 10:53
2
2
Yes. You will get the same answer mentioned below.
– Thomas Shelby
Dec 9 at 11:03
Yes. You will get the same answer mentioned below.
– Thomas Shelby
Dec 9 at 11:03
add a comment |
2 Answers
2
active
oldest
votes
up vote
7
down vote
accepted
Your answer should be $$frac{binom{5}{1}binom{95}{2}}{binom{100}{3}}$$ Since we want the total number of ways to choose 3 meeting the criteria over the total number of ways to choose 3 out of the 100.
answered Dec 9 at 10:56
BelowAverageIntelligence
5221213
I don't understand why it's 500 instead of 100? But I didn't think of the binomial distribution, thank you!
– CruZ
Dec 9 at 10:59
1
sorry, the 500 was a typo...
– BelowAverageIntelligence
Dec 9 at 11:00
1
@CruZ This isn't the binomial distribution, it's actually called the hypergeometric distribution.
– Quintec
Dec 9 at 14:59
Ah thank you for clearing that up, my mistake! Cheers!
– CruZ
Dec 9 at 18:07
add a comment |
up vote
7
down vote
Here is a suggestion how to proceed as ordering does not play a role
- Choose one defective item: $binom{5}{1}$
- Choose two non-defective ones: $binom{95}{2}$
- Chose any three: $binom{100}{3}$
$$P(mbox{"exactly 1 defective"}) = frac{binom{5}{1}cdot binom{95}{2}}{binom{100}{3}}$$
answered Dec 9 at 10:55
trancelocation
8,8951521
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
Your answer should be $$frac{binom{5}{1}binom{95}{2}}{binom{100}{3}}$$ Since we want the total number of ways to choose 3 meeting the criteria over the total number of ways to choose 3 out of the 100.
answered Dec 9 at 10:56
BelowAverageIntelligence
5221213
I don't understand why it's 500 instead of 100? But I didn't think of the binomial distribution, thank you!
– CruZ
Dec 9 at 10:59
1
sorry, the 500 was a typo...
– BelowAverageIntelligence
Dec 9 at 11:00
1
@CruZ This isn't the binomial distribution, it's actually called the hypergeometric distribution.
– Quintec
Dec 9 at 14:59
Ah thank you for clearing that up, my mistake! Cheers!
– CruZ
Dec 9 at 18:07
add a comment |
up vote
7
down vote
accepted
Your answer should be $$frac{binom{5}{1}binom{95}{2}}{binom{100}{3}}$$ Since we want the total number of ways to choose 3 meeting the criteria over the total number of ways to choose 3 out of the 100.
answered Dec 9 at 10:56
BelowAverageIntelligence
5221213
I don't understand why it's 500 instead of 100? But I didn't think of the binomial distribution, thank you!
– CruZ
Dec 9 at 10:59
1
sorry, the 500 was a typo...
– BelowAverageIntelligence
Dec 9 at 11:00
1
@CruZ This isn't the binomial distribution, it's actually called the hypergeometric distribution.
– Quintec
Dec 9 at 14:59
Ah thank you for clearing that up, my mistake! Cheers!
– CruZ
Dec 9 at 18:07
add a comment |
up vote
7
down vote
accepted
up vote
7
down vote
accepted
Your answer should be $$frac{binom{5}{1}binom{95}{2}}{binom{100}{3}}$$ Since we want the total number of ways to choose 3 meeting the criteria over the total number of ways to choose 3 out of the 100.
answered Dec 9 at 10:56
BelowAverageIntelligence
5221213
Your answer should be $$frac{binom{5}{1}binom{95}{2}}{binom{100}{3}}$$ Since we want the total number of ways to choose 3 meeting the criteria over the total number of ways to choose 3 out of the 100.
answered Dec 9 at 10:56
BelowAverageIntelligence
5221213
answered Dec 9 at 10:56
BelowAverageIntelligence
5221213
answered Dec 9 at 10:56
BelowAverageIntelligence
5221213
answered Dec 9 at 10:56
BelowAverageIntelligence
5221213
5221213
I don't understand why it's 500 instead of 100? But I didn't think of the binomial distribution, thank you!
– CruZ
Dec 9 at 10:59
1
sorry, the 500 was a typo...
– BelowAverageIntelligence
Dec 9 at 11:00
1
@CruZ This isn't the binomial distribution, it's actually called the hypergeometric distribution.
– Quintec
Dec 9 at 14:59
Ah thank you for clearing that up, my mistake! Cheers!
– CruZ
Dec 9 at 18:07
add a comment |
I don't understand why it's 500 instead of 100? But I didn't think of the binomial distribution, thank you!
– CruZ
Dec 9 at 10:59
1
sorry, the 500 was a typo...
– BelowAverageIntelligence
Dec 9 at 11:00
1
@CruZ This isn't the binomial distribution, it's actually called the hypergeometric distribution.
– Quintec
Dec 9 at 14:59
Ah thank you for clearing that up, my mistake! Cheers!
– CruZ
Dec 9 at 18:07
I don't understand why it's 500 instead of 100? But I didn't think of the binomial distribution, thank you!
– CruZ
Dec 9 at 10:59
I don't understand why it's 500 instead of 100? But I didn't think of the binomial distribution, thank you!
– CruZ
Dec 9 at 10:59
1
1
sorry, the 500 was a typo...
– BelowAverageIntelligence
Dec 9 at 11:00
sorry, the 500 was a typo...
– BelowAverageIntelligence
Dec 9 at 11:00
1
1
@CruZ This isn't the binomial distribution, it's actually called the hypergeometric distribution.
– Quintec
Dec 9 at 14:59
@CruZ This isn't the binomial distribution, it's actually called the hypergeometric distribution.
– Quintec
Dec 9 at 14:59
Ah thank you for clearing that up, my mistake! Cheers!
– CruZ
Dec 9 at 18:07
Ah thank you for clearing that up, my mistake! Cheers!
– CruZ
Dec 9 at 18:07
add a comment |
up vote
7
down vote
Here is a suggestion how to proceed as ordering does not play a role
- Choose one defective item: $binom{5}{1}$
- Choose two non-defective ones: $binom{95}{2}$
- Chose any three: $binom{100}{3}$
$$P(mbox{"exactly 1 defective"}) = frac{binom{5}{1}cdot binom{95}{2}}{binom{100}{3}}$$
answered Dec 9 at 10:55
trancelocation
8,8951521
add a comment |
up vote
7
down vote
Here is a suggestion how to proceed as ordering does not play a role
- Choose one defective item: $binom{5}{1}$
- Choose two non-defective ones: $binom{95}{2}$
- Chose any three: $binom{100}{3}$
$$P(mbox{"exactly 1 defective"}) = frac{binom{5}{1}cdot binom{95}{2}}{binom{100}{3}}$$
answered Dec 9 at 10:55
trancelocation
8,8951521
add a comment |
up vote
7
down vote
up vote
7
down vote
Here is a suggestion how to proceed as ordering does not play a role
- Choose one defective item: $binom{5}{1}$
- Choose two non-defective ones: $binom{95}{2}$
- Chose any three: $binom{100}{3}$
$$P(mbox{"exactly 1 defective"}) = frac{binom{5}{1}cdot binom{95}{2}}{binom{100}{3}}$$
answered Dec 9 at 10:55
trancelocation
8,8951521
Here is a suggestion how to proceed as ordering does not play a role
- Choose one defective item: $binom{5}{1}$
- Choose two non-defective ones: $binom{95}{2}$
- Chose any three: $binom{100}{3}$
$$P(mbox{"exactly 1 defective"}) = frac{binom{5}{1}cdot binom{95}{2}}{binom{100}{3}}$$
answered Dec 9 at 10:55
trancelocation
8,8951521
answered Dec 9 at 10:55
trancelocation
8,8951521
answered Dec 9 at 10:55
trancelocation
8,8951521
answered Dec 9 at 10:55
trancelocation
8,8951521
8,8951521
add a comment |
add a comment |
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1
You can choose the defective in three different ways. Also the number of non-defective units is $95$ not $94$.
– Thomas Shelby
Dec 9 at 10:49
1
Sorry, my mistake, I meant 95/99 and 94/98. But then are you saying that I have to calculate like this: P(D=1) = {DDN}U{DND}U{NDD}, where D is af defective unit and N is a non-defective unit?
– CruZ
Dec 9 at 10:53
2
Yes. You will get the same answer mentioned below.
– Thomas Shelby
Dec 9 at 11:03