Question

Statistics 2.34 - 2.35

Answer 1

For problem 2.54 I got 2/5 3/5 3/5 5/5 and 0/5

for the second problem I dont know how to do it

Answer 2

any1?

Answer 3

I'd be interested in hearing about your reasoning in arriving at these results for Problem 2.54, especially for P(A).
P(A OR B) does seem to be P(A) + P(B) = 1, because that exhausts the entire sample space.
I agree with your response to P(A AND B). A and B have no overlap, and thus the probability of them happening together is zero.

Regarding 2.35:

Answer 4

First of all, there are 5 possible outcomes making up the sample space: {a,b,c,d,e}, and their respective (unequal) probabilities are given. Add up these probabilities. What do you get? What do you conclude on the basis of this result?

Answer 5

Ive been trying for the past 20 minutes and I got for each respectively 0.4 0.8 0.6 1.0 0.2 for problem 2.55

Are you saying my 2.54 is correct?

Answer 6

I agree with 3 of your 5 responses to 2.34; I was hoping you'd explain your reasoning in calculating the probability of Event A: P(A).

Also, we need to come to agreement on what {a,b,c} means in this context. Does it mean that outcomes a, b and c occur simultaneously? What is your interpretation? If we know the individual probabilities of outcomes a, b and c, how do we use that info to come up with the probability of Event A: P(A) = P({a,b,c})? I don't pretend that I know the answer to that particular question.

Answer 7

P(A) so there is a and b out of a , b, c, d, e so that must mean 2/5?

Answer 8

Your best bet is to review your instructional material, in search for examples of just what Event A: {a,b,c} means.

Answer 9

you talking about 2.54 or 2.55?

Answer 10

More recently, about 2.55.

Answer 11

oh, ok.. But for 2.54 I thought they all were correct?

Answer 12

you said only 3 of 5 I have correct?

Answer 13

P(A) would have to be 2/5 rigt? so P(B) would be 3/5? and shouldn't P(A') be the same as P(A)?

Answer 14

Elvis, I didn't say that. I'm not comfortable just saying "correct" or "incorrect;" I'd prefer to hear about and understand your reasoning before responding. My agreement with 3 of 5 of your responses does not necessarily mean that the other 2 are incorrect; it means that I myself am unsure and therefore wanted to know your reasoning, as well as to know about any examples that you might find in your instructional material.

Answer 15

cool Im happy to hear that because some people just state answers

Answer 16

and dont really help me

Answer 17

Same here. I'd much prefer not to deal in answers alone.

Answer 18

How do I add you as a friend or something? there is a really hard question I asked about statistics that people havent really explained well and Im just stuck on with an answer I dont know how to get

Answer 19

It may be helpful to look at the example of tossing a single die. There are 6 possible outcomes: {1,2,3,4,5,6}. Each outcome is equally likely and has probability 1/6. Now let me throw you a trick question. What is the probability of Event Z: {1,3}? Suppose you only make one toss.

Then change the rules and suppose you make 2 tosses.

What would be the probabilities that you'd expect?

Answer 20

2.55?

Answer 21

i thought we consider c as one outcome? because its both in A and B

Answer 22

My latest post actually applies to both 2.54 and 2.55. I'm picking your brain as well as mine to come to an understanding of what {1,3} or {a,c} would mean in this context.

Answer 23

why six outcomes? there is only abcde or 5

Answer 24

oh nevermind lol i see u have confused me a little

Answer 25

Back to that example of tossing a die ONCE: Could we end up with event {1,3}?

Remember, I'm talking about an example; my example has 6 possible outcomes, whereas 2.55 has 5. Tossing a die is, I'd assume, more familiar to you than is some event such as E:{a,b,c}.

Answer 26

yes

Answer 27

so 2/6?

Answer 28

I find it very useful to revert to easier examples when I don't understand something well or one of my students doesn't.

But look: Suppose we toss the die just once. Could we obtain the result {1,3}?

Answer 29

if we toss it twice

Answer 30

Right. But I'm restricting both of to tossing it only once.
So, could we, on one toss, obtain {1,3}?

Answer 31

no

Answer 32

My point exactly.
Now supposing we toss the same die twice, each time recording the (single) outcome.
Could we conceivably get {1,3}?

Answer 33

yes first possibility is 1/6 second is 1/5

Answer 34

total 1/30

Answer 35

But each toss is INDEPENDENT, and this is a case of "drawing a card WITH REPLACEMENT." Thus, your second toss is completely independent of your first toss, isn't it?

Answer 36

ohhhhh i forgot

Answer 37

In other words, the probabilities don't change. The probability of getting a digit between 1 and 6 inclusive is 1/6 for {1,2,3,4,5,6}.

Answer 38

oh ok

Answer 39

My first guess (note that I've used the word "guess") is that

IF one die is thrown twice and the (single) outcome recorded each time, and the probability of any outcome in {a,b,c,d,e,f} is 1/6, the probability of obtaining {1,3} is (1/6)(1/6).

Explain why this makes sense (or does not make sense) to you.

Answer 40

Note that I'm multiplying probabilities, not adding them (as you have tried, a few comments earlier). Why multiply instead of add?

Answer 41

because although you obatin a 1 on the first roll doesnt rule out it can be rolled the se3cond

Answer 42

Right; the first roll has no effect whatsoever on the 2nd roll.

Answer 43

is that what my question is saying for 5.54?

Answer 44

I don't know yet. I've explained that you and I need to work together to come to an understanding of what {a,b,c} actually means.

Answer 45

oh ok...istill believe the answer is 2/5 though

Answer 46

for P(A)

Answer 47

5.54

Answer 48

Supposing that in 2.54, a, b, c, d and e are lying in a bowl. You cover your eyes and pick one of them out. The probability of getting a on the first pick is 1/5, right?
Now we have to decide whether we replace that a or do not replace it. If we replace it, the probability of obtaining b on the next pick is 1/5. If we do not replace it, but the first pick produced an a, then the prob. of obtaining b on the next pick is 1/4.

Answer 49

oh ok.. so the answer is then 1/5 x 1/4? i thought it was a simple 2/5

Answer 50

so P(B) is 1/5 x 1/4 x 1/3 or 1/60?

Answer 51

I cannot do this problem without examining the possibilities here and then making assumptions. That's why I've been asking you to look for examples of this kind of problem in the instructional material available to you, whether it be textbook, class handouts, etc.
Elvis: Please remember that whether the "answer" is correct or not depends upon our assumptions. Unless you tell me what you're assuming, I can't tell you whether or not your reasoning is correct or not.

If you tell me that P(B) is (1/5)(1/4)(1/3), then I automatically assume that you are drawing from the bowl WITHOUT REPLACEMENT and that each time, you got what you wanted (a or b or c).

Answer 52

Please list whatever reference material you may have available to you right now.
Do you have a texbook or something the teacher wrote for your study? Are there examples on the teacher's website?

Answer 53

yes ill try

Answer 54

If you're not under pressure of time to complete this assignment, I'd suggest you e-mail your teacher listing the assumptions you/ we / I are making and asking for feedback in regard to whether those assumptions are legitimate or not. I'm sure you want to get this problem done and done correctly, but this kind of problem really ca n't be rushed through.

Answer 55

http://www.acc.umu.se/~marshi/Courses/Kvalitetsteknik%20och%20forsoksplanering/Uppgifter/Uppgift10_42/0470053046Applied%20Statistics%20and%20Probability%20for%20Engineers%20B.pdf This is basically the textbook Im using I am on page 35

Answer 56

I've clicked on your link and am redirected to the new URL, but so far the textbook has not downloaded onto my computer. In the meanteime, would you mind going back to that tough question from earlier (not 2.54 or 2.55)? Perhaps we could together solve it to your satisfaction.

Answer 57

ok definitely

Answer 58

Plastic parts produced by an injection-molding operation are checked for conformance to specifications. Each tool contains 12 cavities in which parts are produced, and these parts fall into a conveyor when the press opens. An inspector chooses 3 parts from among the 12 at random. Two cavities are affected by a temperature malfunction that results in parts that don't conform to specifications.

a) what is the probability that the inspector finds exactly one nonconforming part?
b) what is the probability that the inspector finds at least one nonconforming part?

Answer 59

an answer on yahoo said that

there are 12C3 = 220 ways for the inspector to pick 3 parts to check.
When exactly one of these three is bad: there are 2 ways to pick the bad part and 10C2 = 45 ways to pick good parts. 2*45/220 = 9/22

b) the probability that the inspector checks only good parts is 10C3/12C3 = 120/220 = 6/11
therefore the probability that he finds at least one bad is 1 - 6/11 = 5/11



I reaally dontunderstand the reasoning for using 10 instead of 12 in the combinaations

Answer 60

Before I go into that "10 VS 12": are you familiar with binomial probability? And, if so, have you considered that this problem might be solvable using binomial probability?

Answer 61

i dont know what 10 VS 12 means

Answer 62

You wanted to know "the reasoning for using 10 instead of 12 in the combinaations"

Answer 63

yes

Answer 64

Here are my thots: there are 12 cavities in the mold. 2 of them aren't quite right, so might produce defective parts. Therefore, according to my reasoning, the chances that the inspector will pick a defective part at random from the 12 from the mold is 2/12, or 1/6. I need to know whether or not this makes sense to you.

Answer 65

yeah thats what I thought

Answer 66

OK, at first glance this really does seem to be a problem in binomial probability.
the probability of "success" (which is that a part drawn at random is defective) is 1/6. There are 12 parts to choose from. What is the prob. that the inspector picks exactly 1 defectgive part?

If you have a TI-84 calculator, the proper command for calc. this prob. would be

binompdf(12,1/6,1). Are you familiar with that, and does this make sense to you?

Answer 67

I do not have a calculator or a table of binom. prob. with me, unfort.

Answer 68

i have a calculator dont know how to use it for these questions

Answer 69

but the trick to the questio is the inspector must pick 3 parts at once where one is defective

Answer 70

Whoa. The problem statement says that the inspector chooses 3 parts from the 12. I'd say the chances that one of these is defective is still 1/6; would you?

Then n=3, prob. = 1/6, and x = 1

I assume you're calc. binom. prob. with the formula that involves nCr; am I correct?

Answer 71

yes

Answer 72

OK, then:

so you'd be calculating 3C1 (1/6)^1(1-1/6)^(3-1). Up for that?

Answer 73

I am but I am not sure what that means

Answer 74

That's the formula for calculating binomial probability when n=3, p = 1/6 and x = 1.

Answer 75

oh ok

Answer 76

I trust you know how to calculate that. 3C1 = (3!)/[1!(3-1)!] = 1/2, unless I'm sorely mistaken.

Answer 77

3C1 is 2 that times 1/6 equals 1/3

Answer 78

i mwan 3

Answer 79

3C1 is 3

Answer 80

right?

Answer 81

I get (3)(1/6)^1*(5/6)^(3-1). Compare. Agreed or not?
Elvis, I feel bad about this but do need to get off my comp. If you'd like to complete our discussion, you may do so either thru openstudy or thru e-mail (wrgnsc@rit.edu).

Answer 82

Yes, I evaluate 3C1 as 3.

Answer 83

ok ill email you abput this and other things

Answer 84

when will you be available to discuss?

Answer 85

I believe the correct result is 1/72, but that depends upon wehther your/my/our assumptions are correct. Great working with you! Best to e-mail me; then I can respond immed. when available again, and/or set up a specific time to work on OpenStudy.

Answer 86

ok thanks

Answer 87

Cool. Over and out!

Answer 88

I've glanced through a few sections of your online prob/stat textbook. There's a section where this book compares binomial and hypergeometric probabilities. Somehow I'm not comfortable with my own suggestion that if the probability of getting a non-conforming result is 1/6, then the same probability applies to the case where 12 moldings are made but only 3 of those 12 are chosen.

Although I've tried solving a few problems involving hypergeometric probability in the past, I am by no means an expert at it. Either we'll both have to read up on hypergeom. prob.., or you'll need to ask through OpenStudy for someone familiar with this sub-topic.