SOA #132 is defective

[David.Han]

You can solve that problem using the following two approaches, which yield to different answers. If we modify this question as follows:

(1) A store has 800 modems in its inventory, 300 coming from Source A and the remainder from Source B. Of the modems from Source A, 20% are defective. Of the modems from Source B, 8% are defective. Calculate the probability that exactly two out of a random sample of five modems from the store's inventory are defective.

(2) A store has 8000 modems in its inventory, 3000 coming from Source A and the remainder from Source B. Of the modems from Source A, 20% are defective. Of the modems from Source B, 8% are defective. Calculate the probability that exactly two out of a random sample of five modems from the store's inventory are defective.

As you can see, the answer based on method A.1 approaches to that based on method A.2.

[Colymbosathon ecplecticos]

I really doubt that there is anything wrong with the problem. Do you have a question that you'd like to ask?

[David.Han]

Do you find anything wrong with the second method? The second method shows correct answer is D, but the first method supports answer C.

I really doubt that there is anything wrong with the problem. Do you have a question that you'd like to ask?

[jraven]

The problem states that a specific percentage of the modems from each source are defective; thus there are definitely 6 defective modems from A and 4 defective modems from B, for a total of 10. This is not an expected amount, it is guaranteed by the phrasing of the question. So then the question simply boils down to "Given that there are 10 defective modems out of 80, what is the probability that exactly two out of a sample of 5 are defective?" Which is what the SOA solution computes.

If the problem were stated as "There is a 20% probability that a modem produced by A is defective, and an 8% probability that a modem produced by B is defective." then things would be significantly more complicated -- but the answer wouldn't follow your method A.2. The problem is that when looking at one modem, the probability that it will be defective is 0.125, as you compute -- however the chance that the a second modem will be defective, given your knowledge about the first modem, will not be the same.

[At the extreme, suppose there is only 1 modem from A and 79 from B, and that A and B have defect probabilities of 100% and 0% respectively; then the probability that a randomly chosen modem is defective is 1/80, but the probability that exactly two out of a sample of 5 are defective is exactly zero (and not the binomial value calculated by method A.2). In order for method A.2 to work you'd need an infinite collection of A and B modems, with a 3/8 and 5/8 probability that any randomly selected modem is from A or B respectively.]

[David.Han]

The problem states that a specific percentage of the modems from each source are defective; thus there are definitely 6 defective modems from A and 4 defective modems from B, for a total of 10. This is not an expected amount, it is guaranteed by the phrasing of the question. So then the question simply boils down to "Given that there are 10 defective modems out of 80, what is the probability that exactly two out of a sample of 5 are defective?" Which is what the SOA solution computes.

If the problem were stated as "There is a 20% probability that a modem produced by A is defective, and an 8% probability that a modem produced by B is defective." then things would be significantly more complicated -- but the answer wouldn't follow your method A.2. The problem is that when looking at one modem, the probability that it will be defective is 0.125, as you compute -- however the chance that the a second modem will be defective, given your knowledge about the first modem, will not be the same.

[At the extreme, suppose there is only 1 modem from A and 79 from B, and that A and B have defect probabilities of 100% and 0% respectively; then the probability that a randomly chosen modem is defective is 1/80, but the probability that exactly two out of a sample of 5 are defective is exactly zero (and not the binomial value calculated by method A.2). In order for method A.2 to work you'd need an infinite collection of A and B modems, with a 3/8 and 5/8 probability that any randomly selected modem is from A or B respectively.]

You pointed out two very interesting points here:

In the context of Exam P, it is usually implied that it refers to probability when we say a percentage of something. We can find numerous examples in SOA 142.

The second point you mentioned is about conditional probability. First, here it is not a conditional probability problem (select simultaneously, not sequentially). Second, the order of action does not matter. For example, if you are drawing a ball out of 50 balls which are numbered from 1 to 50. You do not place it back after drawing. What is the probability that your fifth draw is the ball numbered 25? The probability is 1/50. Basically it means here we do not have advantage of first mover.

[jraven]

You pointed out two very interesting points here:

In the context of Exam P, it is usually implied that it refers to probability when we say a percentage of something. We can find numerous examples in SOA 142.

Then those problems may be defective; this one is not. [Though it is unusual to know the exact amount of defective products before sampling.]

The second point you mentioned is about conditional probability. First, here it is not a conditional probability problem (select simultaneously, not sequentially). Second, the order of action does not matter. For example, if you are drawing a ball out of 50 balls which are numbered from 1 to 50. You do not place it back after drawing. What is the probability that your fifth draw is the ball numbered 25? The probability is 1/50. Basically it means here we do not have advantage of first mover.

Whether or not the balls are selected simultaneously or not has no bearing on whether you need conditional probability -- if you want to justify the use of the binomial distribution you need to enumerate the cases which meet the condition (exactly two defects), identify a probability for each case and add them up. However the probability of getting, say (Defective, OK, OK, Defective, OK) is NOT p (1-p) (1-p) p (1-p); as a result when you add up the probabilities of the \left(\overset{5}{2}\right) ways of getting exactly two items you do NOT end up with a probability of
\left(\overset{5}{2}\right) p^2 (1 - p)^3
as your method 2 asserts.

I will simply repeat my example -- Suppose there is one modem of type A and 79 of type B. All of type A are defective, none of type B are. What is the probability that exactly two modems from a sample of five are defective?

Your second approach to the problem would compute
p = 1 \,\cdot\, \left(\frac{1}{80}\right) + 0 \,\cdot\, \left(\frac{79}{80}\right) = \frac{1}{80}
and would then conclude that the probability that exactly two modems from a sample of five are defective is
\left(\overset{5}{2}\right) p^2 (1 - p)^3 > 0

On the other hand since there's only ONE defective modem, it's obvious that the probability of finding two in the sample must be zero.

[David.Han]

Then those problems may be defective; this one is not. [Though it is unusual to know the exact amount of defective products before sampling.]



Whether or not the balls are selected simultaneously or not has no bearing on whether you need conditional probability -- if you want to justify the use of the binomial distribution you need to enumerate the cases which meet the condition (exactly two defects), identify a probability for each case and add them up. However the probability of getting, say (Defective, OK, OK, Defective, OK) is NOT p (1-p) (1-p) p (1-p); as a result when you add up the probabilities of the \left(\overset{5}{2}\right) ways of getting exactly two items you do NOT end up with a probability of
\left(\overset{5}{2}\right) p^2 (1 - p)^3
as your method 2 asserts.

I will simply repeat my example -- Suppose there is one modem of type A and 79 of type B. All of type A are defective, none of type B are. What is the probability that exactly two modems from a sample of five are defective?

Your second approach to the problem would compute
p = 1 \,\cdot\, \left(\frac{1}{80}\right) + 0 \,\cdot\, \left(\frac{79}{80}\right) = \frac{1}{80}
and would then conclude that the probability that exactly two modems from a sample of five are defective is
\left(\overset{5}{2}\right) p^2 (1 - p)^3 > 0

On the other hand since there's only ONE defective modem, it's obvious that the probability of finding two in the sample must be zero.

I can agree with your example more. The basic difference in understanding this question is whether those percentages are some probability or a certain proportion. In your example, you know for sure all modems from A are defective and none of B is defective. If this is the case, all arguments that follow do make sense and no doubt about it.

[jraven]

I can agree with your example more. The basic difference in understanding this question is whether those percentages are some probability or a certain proportion. In your example, you know for sure all modems from A are defective and none of B is defective. If this is the case, all arguments that follow do make sense and no doubt about it.

The fact that the probabilities imply known percentages is irrelevant to the matter of whether your method 2 works when interpreting the percents as probabilities; they were merely chosen so that the correct answer would be obvious. Your method 2 just doesn't work, even if you interpret the defect percents in the problem as probabilities. [The only time it works is when you assume you have an infinite number of modems of each type to draw your sample from, and that you instead have certain probabilities of getting either an A or a B modem.]

Here's another example then. There is one modem of type A, one modem of type B. There is a 40% probability that a type A modem is defective and a 10% probability that a type B modem is defective. What is the probability that exactly one modem from a two modem sample is defective?

Note unlike my previous example, here we don't know a priori how many modems are actually defective.

Your method 2 would first find that the probability of a randomly selected modem from the whole population being defective is
p = 0.4 \,\cdot\, \left(\frac{1}{2}\right) + 0.1 \,\cdot\, \left({\frac{1}{2}\right) = 0.25
and then conclude that the probability of exact one defect in a two-element sample from the population is
\left(\overset{2}{1}\right) p (1-p) = 2 (0.25) (0.75) = 0.375

On the other hand because the two-element sample is actually the whole population, the correct answer must be
Prob(A modem defective, B modem OK) + Prob(A modem OK, B modem defective) = (0.4) (0.9) + (0.6) (0.1) = 0.42

The issue here is the relative size of the sample population to the whole population -- if the sample population represents a notable portion of the whole population then your method 2 will result in a significant error. I don't know whether the 5-element sample from a population of 80 described in the original problem would cause there to be a significant difference between what your method 2 would produce and the correct answer if you were to interpret the percents as probabilities, but the method is nonetheless incorrect. [Calculating the correct answer when you interpret the percents as probabilities looks messy.]

[JavaGeek]

Answer
A.1 Assumes the problem is solved without replacement (can't select same unit twice)
Here's another way to write the solution:
\frac{5!}{2!\times 3!} \frac{10\times 9\times 70 \times69 \times 68}{80\times 79\times 78\times 77\times 76}=0.1024666539

A.2 Assumes that the items are replaced (you can pick the same unit twice)
\frac{5!}{2!\times 3!} \frac{10\times 10\times 70 \times 70 \times 70}{80^5}=0.1046752930

[HarryPotter]

The Poisson distribution wasn't named after a fish -- it was named after a man ... who was named after a fish.

seems like we get an evolutionist here :slug:

[Actuarialsuck]

seems like we get an evolutionist here :slug:

I think you missed the point of jraven's signature...

[Wolf Follower]

A.1 Assumes the problem is solved without replacement (can't select same unit twice)

A.2 Assumes that the items are replaced (you can pick the same unit twice)

This is the reason why the two answers are different and why answer 1 is the correct one. I made the mistake of solving it the second way the first time I saw a problem like this.

WF

[jraven]

This is the reason why the two answers are different and why answer 1 is the correct one. I made the mistake of solving it the second way the first time I saw a problem like this.


David.Han's issue is with the interpretation of the defect percents -- he feels that they should be viewed as probabilities. Thus instead of knowing for certain that there are 10 defects in the 80-element population, you only know that the expected number of defects is 10. That changes the problem significantly.

Under that interpretation he is correct that the SOA's solution would in fact be wrong. My points were that (1) that's not how the problem should be interpreted and (2) under that interpretation neither of his solution methods is correct.

[HarryPotter]

I think you missed the point of jraven's signature...

think you :slug:http://a10.idata.over-blog.com/300x304/0/18/68/25/photos/drole-poisson.jpg

[Wolf Follower]

David.Han's issue is with the interpretation of the defect percents -- he feels that they should be viewed as probabilities. Thus instead of knowing for certain that there are 10 defects in the 80-element population, you only know that the expected number of defects is 10. That changes the problem significantly.

Under that interpretation he is correct that the SOA's solution would in fact be wrong. My points were that (1) that's not how the problem should be interpreted and (2) under that interpretation neither of his solution methods is correct.

I won't argue with that. However, I think the thing to take away here is that this problem needs to be set up as desired outcomes/total outcomes rather than as a binomial random variable.

WF

[JavaGeek]

The correct solution following the assumption that the # of defects is in fact a random variable following a binomial distribution. [Just for fun]
\left(\overset{30}{0}\right) 0.2^{0}0.8^{30} \sum_{i=2}^{50} \left(\overset{50}{i}\right) 0.08^{i}0.92^{(50-i)}\frac{\left(\overset{i}{2}\right) \left(\overset{80-i}{3}\right)}{\left(\overset{80}{5}\right) }
+ \left(\overset{30}{1}\right) 0.2^{1}0.8^{29} \sum_{i=1}^{50} \left(\overset{50}{i}\right) 0.08^{i}0.92^{(50-i)}\frac{\left(\overset{i+1}{2}\right) \left(\overset{80-i-1}{3}\right)}{\left(\overset{80}{5}\right) }
+ \sum_{j=2}^{27}\sum_{i=0}^{50} \left(\overset{30}{j}\right) 0.2^{j}0.8^{(30-j)} \left(\overset{50}{i}\right) 0.08^{i}0.92^{(50-i)}\frac{\left(\overset{i+j}{2}\right) \left(\overset{80-i-j}{3}\right)}{\left(\overset{80}{5}\right) }
+\left(\overset{30}{28}\right) 0.2^{28}0.8^{2} \sum_{i=0}^{49} \left(\overset{50}{i}\right) 0.08^{i}0.92^{(50-i)}\frac{\left(\overset{28+i}{2}\right) \left(\overset{80-i-28}{3}\right)}{\left(\overset{80}{5}\right) }
+\left(\overset{30}{29}\right) 0.2^{29}0.8^{1} \sum_{i=0}^{48} \left(\overset{50}{i}\right) 0.08^{i}0.92^{(50-i)}\frac{\left(\overset{29+i}{2}\right) \left(\overset{80-i-29}{3}\right)}{\left(\overset{80}{5}\right) }
+\left(\overset{30}{30}\right) 0.2^{30}0.8^{0} \sum_{i=0}^{47} \left(\overset{50}{i}\right) 0.08^{i}0.92^{(50-i)}\frac{\left(\overset{30+i}{2}\right) \left(\overset{80-i-30}{3}\right)}{\left(\overset{80}{5}\right) }

=2.6\times10^{-5}+0.0003+0.1043+2.7\times10^{-18}+4.6\times10^{-20} +3.8\times10^{-22} =0.1046

[optimizer]

The correct solution following the assumption that the # of defects is in fact a random variable following a binomial distribution. [Just for fun]
\left(\overset{30}{0}\right) 0.2^{0}0.8^{30} \sum_{i=2}^{50} \left(\overset{50}{i}\right) 0.08^{i}0.92^{(50-i)}\frac{\left(\overset{i}{2}\right) \left(\overset{80-i}{3}\right)}{\left(\overset{80}{5}\right) }
+ \left(\overset{30}{1}\right) 0.2^{1}0.8^{29} \sum_{i=1}^{50} \left(\overset{50}{i}\right) 0.08^{i}0.92^{(50-i)}\frac{\left(\overset{i+1}{2}\right) \left(\overset{80-i-1}{3}\right)}{\left(\overset{80}{5}\right) }
+ \sum_{j=2}^{27}\sum_{i=0}^{50} \left(\overset{30}{j}\right) 0.2^{j}0.8^{(30-j)} \left(\overset{50}{i}\right) 0.08^{i}0.92^{(50-i)}\frac{\left(\overset{i+j}{2}\right) \left(\overset{80-i-j}{3}\right)}{\left(\overset{80}{5}\right) }
+\left(\overset{30}{28}\right) 0.2^{28}0.8^{2} \sum_{i=0}^{49} \left(\overset{50}{i}\right) 0.08^{i}0.92^{(50-i)}\frac{\left(\overset{28+i}{2}\right) \left(\overset{80-i-28}{3}\right)}{\left(\overset{80}{5}\right) }
+\left(\overset{30}{29}\right) 0.2^{29}0.8^{1} \sum_{i=0}^{48} \left(\overset{50}{i}\right) 0.08^{i}0.92^{(50-i)}\frac{\left(\overset{29+i}{2}\right) \left(\overset{80-i-29}{3}\right)}{\left(\overset{80}{5}\right) }
+\left(\overset{30}{30}\right) 0.2^{30}0.8^{0} \sum_{i=0}^{47} \left(\overset{50}{i}\right) 0.08^{i}0.92^{(50-i)}\frac{\left(\overset{30+i}{2}\right) \left(\overset{80-i-30}{3}\right)}{\left(\overset{80}{5}\right) }

=2.6\times10^{-5}+0.0003+0.1043+2.7\times10^{-18}+4.6\times10^{-20} +3.8\times10^{-22} =0.1046

:geek: A little bored tonight, are we? :)