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Subject: OT: Probability Question
Posted by: Ender
- [2131019] Sun, May 07, 2006, 21:35
Forgive the OT post, but this is likely the forum most trafficked by the guys I'm looking for on this one.
I am going to swallow my pride and ask a couple questions that I should already have figured out, but haven't given them enough time. I am tutoring a student in Finite Math tomorrow and my skills are rusty in this area from having spent waaay too much time on 8th grade level math and not enough in keeping higher math skills sharp.
Anyway:
Problem #1:
A lot contains 300 parts of which 30 are defective and 270 are good. Three parts are selected at random (no replacement).
(a) What is the probability that exactly one of the 3 parts is defective?
I started off with (270/300)(269/299)(30/298), but I lack confidence in that response for some reason. I tried to convince myself by finding P(0) + P(1) + P(2) + P(3) and it didn't add up to 1. That's where I lost confidence in the answer. I couldn't find the error in my logic of the response or in the calculation of the sum of the probabilities.
(b) What is the probability that at least one of the three parts is defective.
This is what led me to the P(1) + P(2) + P(3) calculation above. I double checked the P(0) value and it didn't jive.
Problem 2
John Q. Badstudent is taking a math class and a sociology class. He estimates that the probability of passing the math class is .45, the probability of passing sociology is .60, and the probability of passing at least one of these classes is .90 What is the probability of passing both classes?
This one escapes, but I am certain I need consider Union, Intersection, and the like. I just having trouble fitting it all together.
There are enough statisticians and actuaries running around here I figure someone can help me out. I appreciate any of it.
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| 1 | JeffG
Leader
ID: 01584348
Sun, May 07, 2006, 22:57
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Ok, it has been a generation since I took stats, but here is how I remember it.
If you have A unique parts and are choosing B and order does not matter, the the number of combinations is (A B) read "A choose B" which is A! divided by (B! * (A - B)!).
1A If you consider each of the 300 parts unique and order does not matter, there are '300 choose 3' denoted (300 3) combinations or 300!/(3!)*(297)!. That is your denominator. which is (300*299*298)/(3*2*1).
So picking 2 good and 1 bad is (270 2) * (30 1). That is your numerator.
Using the calculator. I get .245
1B. The probability at least one is defective is one minus the probability none are defective.
so that is (270 3)/(300 3). So 1 - .728 is .272.
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| 2 | JeffG
Leader
ID: 01584348
Sun, May 07, 2006, 23:06
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2. I am missing something on this. If they are unrelated events, the probability of event A is .45 and probability of event B is .60. Then isn't the probability of event A AND event B is p(a) * p(b). That is .27.
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| 3 | Sludge
ID: 14411118
Sun, May 07, 2006, 23:10
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I started off with (270/300)(269/299)(30/298), ... I couldn't find the error in my logic of the response or in the calculation of the sum of the probabilities.
Not a bad approach. JeffG's answer is correct, but you can also get the correct answer with your approach as well.
The probability you give is the probability of drawing GGD in that order. What about drawing GDG or DGG? (270/300)*(30/299)*(269/298) and (30/300)*(270/299)*(269/298), respectively. Note that these are each the same as your original, but you need to add those three together (e.g. multiply your original by 3). What do you get? 0.2445.
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| 4 | Sludge
ID: 14411118
Sun, May 07, 2006, 23:12
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Since this seems like homework, I wasn't going to answer #2, but I feel I must now that JeffG has given an incorrect answer.
Look at the general rule of additivity for probabilities.
P(A or B) = P(A) + P(B) - P(A and B)
P(A union B) = P(A) + P(B) - P(A intersect B)
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| 5 | Ender
ID: 285713
Mon, May 08, 2006, 09:19
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Homework? LOL
I'm the teacher in this situation. I am tutoring an adult student to help her prepare for a test. These were 2 study questions. SO any help can be given with a clean conscience :)
Thanks. Problem 1 makes sense now. I new there was something I was leaving out. I didn't divide out the permutations of the 3 parts.
So Problem 2 amounts to P(M)+P(S) - P(M&S) = .45 + .60 - .90 = .15? That's the formula I found that led me to the comment about Unions and INtersections above, but it just doesn't feel right. I know sometimes probability works out that way. I am used to surprising answers in this branch (the birthday problem as an example), but the "at least one class" still leaves me scratching my head. It feels like it should somehow be P(1) + P(Both) and I can reconcile those 2 ideas.
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| 6 | Sludge
ID: 11042612
Mon, May 08, 2006, 11:15
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Ender -
Draw a Venn diagram, and it will make sense.
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| 7 | Ender
ID: 285713
Mon, May 08, 2006, 11:37
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Got it. Thanks again.
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| 8 | holt
ID: 1331253
Tue, May 09, 2006, 08:06
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I'm getting .2472 for 1A and .2767 for 1B but those are rounded off. the actual numbers are pretty long.
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| 9 | weykool
ID: 3645097
Tue, May 09, 2006, 09:20
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When making 3 selections with 2 outcome you will always have 8 possible outcomes:
GGG GGD GDG DGG GDD DGD DDG DDD
None defective-(270/300)(269/299)(268/298)*1 = .7282
One defective--(270/300)(269/299)(30/298) *3 = .2445
Two defective--(270/300)(30/299)(29/298) *3 = .0264
All defective--(30/300)(29/299)(28/298) *1 = .0009
If you add them all up you will get 1.0
1A = .2445
1B = .2718
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| 11 | weykool
ID: 394849
Tue, May 09, 2006, 16:23
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As for problem #2 it doesnt compute.
IF the probability of passing is .45 and .60 then the probablity of failing must be .55 and .40.
The probabilty of failing both must be .55 x .40 = .22
So the probability passing at least 1 would be .78 not .90
The 4 outcomes are:
PP PF FP FF
PP=.45*.60 = .27
PF=.45*.40 = .18
FP=.55*.60 = .33
FF=.55*.40 = .22
Total = 1.00
Passig at least 1 = .78
Passing both = .22
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| 12 | Salt Bandit
ID: 35618614
Tue, May 09, 2006, 16:42
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weykool - Your calculations for problem 2 assume they are independent events, which is not the case for this problem.
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| 13 | weykool
ID: 394849
Tue, May 09, 2006, 18:37
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Salt please explain.
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| 14 | Sludge
ID: 11042612
Tue, May 09, 2006, 19:07
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P(M) = 0.45
P(S) = 0.6
P(M or S) = 0.9
All of the above are given.
P(M and S) = P(M)+P(S)-P(M or S) = 0.45 + 0.6 - 0.9 = 0.15. This is your PP above.
PF = P(M and S') = 0.30.
FP = P(M' and S) = 0.45.
FF = P(M' and S') = P((M or S)') = 0.10.
A' = "A complement" or "not A".
Again, draw a Venn Diagram, and all will be revealed.
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| 15 | Salt Bandit
ID: 810262813
Mon, May 22, 2006, 15:20
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weykool - sorry to take so long for this response.
The probabilty of failing both must be .55 x .40 = .22
The equation you're using P(A and B) = P(A) * P(B) only holds true of A and B are independent.
Events are independent when knowing the result of one event doesn't change the probability of the other event;
P(A|B) = P(A)
More info
Given the information for problem 2, you can conclude the chances of passing math and soc are dependent (because P(A and B) != P(A) * P(B))
Sludge's response in [14] uses more general equations that don't reply on independence.
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