2 years, 11 months ago
Statistics help
Alright, I'm having a mental block or something, and need some help with a stats problem. Any help would be appreciated.
You have 3 stacks of 100 cables. There are 10 bad cables in stack A, 15 bad in stack B, and 20 bad cables in the last stack(stack C). Four cables are required to secure a chandelier, and they are pulled at random from the three stacks. If two or more bad cables are pulled for a particular chandelier, it will be defective and fail. Assume that the number of cables per stack remains constant
1. What is the probability of a chandelier failing?
2. A safety inspector selects chandeliers at random. 1 in 10 are tested randomly.
a. What is the probability of him finding a defective chandelier?
b. What is the probability of there being a defective chandelier and him not finding it?
Here is what I've done...assume ╣ is Intersection (the upside down U)...
#1(probability of bad chand):
P(D|StackA) = 0.1
P(D|StackB) = 0.15
P(D|StackC) = 0.2
P(D ╣ StackA) = P(D|StackA) * P(StackA)
= 0.1 * (1/3) = 0.0333 = 3.33%
P(D ╣ StackB) = P(D|StackB) * P(StackB)
= 0.15 * (1/3) = 0.05 = 5%
P(D ╣ StackC) = P(D|StackC) * P(StackC)
= 0.2 * (1/3) = 0.0666 = 6.66%
P(D) = P(D ╣ StackA) + P(D ╣ StackB) + P(D ╣ StackC)
= 0.0333 + 0.05 + 0.0666 = 0.1499 = 14.99%
Probability of bad chandelier:
P(D2|D1) = 0.1499 * 0.1499 = 0.0225 = 2.25%
#2
P=0.0225 n=10 (P-1)=0.9775
P(x=0) = / 0!(10-0)! * (0.0225)^0 * (0.9775)^10
= 1 * 1 * 0.7965 = 0.7965
P(x>=1) = 1 - 0.7965 = 0.2035
...and that's where I'm stuck...HELP!
You have 3 stacks of 100 cables. There are 10 bad cables in stack A, 15 bad in stack B, and 20 bad cables in the last stack(stack C). Four cables are required to secure a chandelier, and they are pulled at random from the three stacks. If two or more bad cables are pulled for a particular chandelier, it will be defective and fail. Assume that the number of cables per stack remains constant
1. What is the probability of a chandelier failing?
2. A safety inspector selects chandeliers at random. 1 in 10 are tested randomly.
a. What is the probability of him finding a defective chandelier?
b. What is the probability of there being a defective chandelier and him not finding it?
Here is what I've done...assume ╣ is Intersection (the upside down U)...
#1(probability of bad chand):
P(D|StackA) = 0.1
P(D|StackB) = 0.15
P(D|StackC) = 0.2
P(D ╣ StackA) = P(D|StackA) * P(StackA)
= 0.1 * (1/3) = 0.0333 = 3.33%
P(D ╣ StackB) = P(D|StackB) * P(StackB)
= 0.15 * (1/3) = 0.05 = 5%
P(D ╣ StackC) = P(D|StackC) * P(StackC)
= 0.2 * (1/3) = 0.0666 = 6.66%
P(D) = P(D ╣ StackA) + P(D ╣ StackB) + P(D ╣ StackC)
= 0.0333 + 0.05 + 0.0666 = 0.1499 = 14.99%
Probability of bad chandelier:
P(D2|D1) = 0.1499 * 0.1499 = 0.0225 = 2.25%
#2
P=0.0225 n=10 (P-1)=0.9775
P(x=0) = / 0!(10-0)! * (0.0225)^0 * (0.9775)^10
= 1 * 1 * 0.7965 = 0.7965
P(x>=1) = 1 - 0.7965 = 0.2035
...and that's where I'm stuck...HELP!
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M$3 Answers
Since cables are pulled at random from the three stacks, they can be combined into a single group of 300 cables, 45 (15%) of which are bad. A failed chandelier can be made with two, three, or four bad cables, the probability of which is
p = 0.15^2 + 0.15^3 + 0.15^4 = 0.0225 + 0.003375 + 0.00050625
or about a 2.64% chance. If 1 in 10 chandeliers is tested at random, the chance of finding a defective one depends on the number of chandeliers sampled (a sample size of 10,000 is more likely to turn up a defective chandelier than a sample size of 10). The inspections are dependent events (each inspection changes the probability that the next chandelier sampled will be defective), and so the probability of finding a defective chandelier is
p = 1 - ((probability of complementary event)^(sample size))
If 1 sample is taken from a batch of 10 chandeliers, the probability of finding a defective one is 1 - (1 - 0.0264 (=0.9736))^1 = 0.0264. If 10 samples are taken from a batch of 100, the probability of finding a defective one is 0.235; and for 100 samples out of 1000, the probability of finding a defective one is 0.931. The probability of the inspector not finding a defective chandelier is the complement of this, specifically 0.9736^(sample size).
p = 0.15^2 + 0.15^3 + 0.15^4 = 0.0225 + 0.003375 + 0.00050625
or about a 2.64% chance. If 1 in 10 chandeliers is tested at random, the chance of finding a defective one depends on the number of chandeliers sampled (a sample size of 10,000 is more likely to turn up a defective chandelier than a sample size of 10). The inspections are dependent events (each inspection changes the probability that the next chandelier sampled will be defective), and so the probability of finding a defective chandelier is
p = 1 - ((probability of complementary event)^(sample size))
If 1 sample is taken from a batch of 10 chandeliers, the probability of finding a defective one is 1 - (1 - 0.0264 (=0.9736))^1 = 0.0264. If 10 samples are taken from a batch of 100, the probability of finding a defective one is 0.235; and for 100 samples out of 1000, the probability of finding a defective one is 0.931. The probability of the inspector not finding a defective chandelier is the complement of this, specifically 0.9736^(sample size).
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M$
You need to count every possible combination of two or more bad cables. Since there are two possible outcomes for each cable you choose, this is a binomial probability distribution problem. So, the probability of getting m good cables and n bad cables is:
P(m,n) = B(m,n)*P(G)^m*P(D)^n
where B(m,n) is the binomial coefficient (m+n)!/(m!n!), P(G) is the probability of getting a good cable, and P(D) is the probability of getting a defective one. Then, the probability of a chandelier falling is
P(BAD) = P(2,2) + P(1,3) + P(0,4)
If I'm doing the calculation correctly:
P(2,2) = 0.0975375
P(1,3) = 0.011475
P(0,4) = 0.00050625
Giving P(BAD) = 0.10951875 or almost 11%.
I need to think about question 2, but I think @badaspie is on the right track.
P(m,n) = B(m,n)*P(G)^m*P(D)^n
where B(m,n) is the binomial coefficient (m+n)!/(m!n!), P(G) is the probability of getting a good cable, and P(D) is the probability of getting a defective one. Then, the probability of a chandelier falling is
P(BAD) = P(2,2) + P(1,3) + P(0,4)
If I'm doing the calculation correctly:
P(2,2) = 0.0975375
P(1,3) = 0.011475
P(0,4) = 0.00050625
Giving P(BAD) = 0.10951875 or almost 11%.
I need to think about question 2, but I think @badaspie is on the right track.
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M$
I get the 14.99% for Part I. It asks to assume that the number of cables remain constant, which is key.
You choose randomly from the 3 stacks so P(bad cable)= 45/300 = 15%.
P(4 bad cables) = .15^4
P(exactly 3 bad cables) = .15^3 * (4Combination3) = .15^3 * (4! / (3!1!))
P(exactly 2 bad cables) = .15^2 * (4Combination2) = .15^2 * (4! / (2!2!))
Sum these to get the probability of a defective chandelier.
For Part 2, I think you have to assume that all 300 cables were used to make 75 chandeliers. It should be straightforward from there.
You choose randomly from the 3 stacks so P(bad cable)= 45/300 = 15%.
P(4 bad cables) = .15^4
P(exactly 3 bad cables) = .15^3 * (4Combination3) = .15^3 * (4! / (3!1!))
P(exactly 2 bad cables) = .15^2 * (4Combination2) = .15^2 * (4! / (2!2!))
Sum these to get the probability of a defective chandelier.
For Part 2, I think you have to assume that all 300 cables were used to make 75 chandeliers. It should be straightforward from there.
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M$
Y'know, the more I think about it, the more I think there's something more that needs to be taken into account. The probabilities you calculate assume that 10.95% of the chandeliers is bad. But there's some contingent probability going on here. I'm better at probability than statistics, unfortunately.
What I'm thinking is: for any given number of chandeliers N, there is a probability distribution P(M|N) where 0 <= M <= N that is the probability of having M bad chandeliers. For each M, there is an associated probability that the inspector will find at least one bad chandelier, given the sample size., which is N/10. Then you have to sum over M. Wish I had more time to think about this, and it may actually reduce to your formulation in the end. I'm just not quite sure.
Sorry to throw another wrench in.
I think you're right on question 2 (not the numbers, but the idea), but I don't think you're counting up the combinations properly on question 1. See my answer.
You're right about question 1. The cables are constantly being replenished, and the percentage of bad cables remains constant, so the binomial distribution applies and rickg has the correct answer (10.95%).
Recalculating question 2 using this figure, a sample size of 1 has about an 11% chance of finding a defective chandelier, a sample size of 10 has a 67% chance, and a sample size of 100 has a 99.999% chance.