# Probability Homework help

1. Apr 15, 2013

### Drudge

"Half of the population are men and half are women.

What is the probability that in a 100 random sample, there are 50 women?"

0.5 right!?!??!

Last edited by a moderator: Apr 15, 2013
2. Apr 15, 2013

### phinds

How did you get that? You have to show your work in order to get help on this forum.

3. Apr 15, 2013

### Drudge

No, I´m just figuring out. I mean, to me that is common sense. If there are as many woman as men, then randomly choosing from them should give a probability of 0.5 of either gender.

The answer in my answer sheet is 0.08, which I dont get.

4. Apr 15, 2013

### phinds

Yes, I can see that you don't "get it" BECAUSE you are using "common sense" instead of math. Even common sense tells me .5 is wrong but that's a different discussion.

How would you do it mathematically?

EDIT: also ... read the problem statement. I think you are in a sense solving the wrong problem.

Last edited: Apr 15, 2013
5. Apr 16, 2013

### Drudge

What problem do you think I am instead solving?

I think I get it. It is something like (I dont know really how to demonstrate this in "maths") the probability of EXACTLY 50 is 0.08, because there could be in turn 49, 48, 47, 51, 52, 53 (and so on), which are also close to the expected value, and thus part of the probability (of a total of one).

Also the problem is supposed to be solved like: 100ℂ50*0.5^50*0.5^50
that is "of all the cases where there could be fifty women (100ℂ50), how likely is there to be so much" or something like that.

Another one I cannot figure out.

"There are 4 options in a question. Each question has either one or multiple answers right, which in case means that only if you answer all the options right do you get the question right.

What is the probability of answering a question right?"

=1/4 * 2/4 * 3/4 * 4*4 ?

(there is no answer for this in my records)

Last edited: Apr 16, 2013
6. Apr 16, 2013

### Drudge

I don´t know should I post a new thread or just continue posting here.

Ok, this one has to be a mistake in the book!

"Approximately 0.01% of people suffer from disease X. In a random sample of 10 000, how likely is there to be 2 people with disease X?"

So,

0.01% = 0.0001
0.0001 * 10 000 = 1

( (1^2) / 2! ) * e^-1 = 0.1839397206 (Poisson approximation)

The answer in the book is 1,86 * 10^-40. You can get this answer if you multiply 10 000 with 0.01 (not 0.0001) which is 100.

And:

( (100^2) / 2!) * e^-100 = 1,86 * 10^-40

So what do you think? Is there a mistake in the book. Does it use the percentage 0.01% without converting it (into 0.0001)?

7. Apr 16, 2013

### Bacle2

Why don't you break it down into cases? What if exactly one, exactly two, etc. are the right

8. Apr 16, 2013

### Drudge

You want me to calculate every probability up to 50 and then see if there is 0.08 left, or something like that?

9. Apr 16, 2013

### HallsofIvy

Staff Emeritus
Yes, that is exactly what he means- do the work rather than guessing.

If you are clever, you might see a pattern after the first few so that you don't have to actually do all 50.

10. Apr 16, 2013

### Drudge

Ok, so I was not that clever and I count up to 49 adding every previous on to the sum and got ≈0.46. Because the probabilities are symmetrical, 0.46 * 2 = 0.92. So 1 - 0.92 = 0.08.

So yes, I understand now, thank you very much. What do you think about the two other problems I posted?

EDIT: Concerning the second question, I thought I was counting up the cases (1/4 * 2/4...). it dosen seem right though, and I dont have any where to check. Do you think it is right?

11. Apr 16, 2013

### Ray Vickson

I don't understand why you are adding so many things (if that is what you are doing). The probability of exactly 50 women in the sample is $p_{50} = C(100,50)/2^{100} = 0.079589... \approx 0.08.$

BTW: you should NOT round off to 0.46 and then compute 1-2*-0.46 = 0.08. Whenever the answer is a small number obtained by subtraction of two not-small numbers you can get serious roundoff error effects. You should keep more digits before stating/rounding the final answer. (Of course, you *might* have done that, while just writing rounded intermediate results for the sake of easier typing.)

For your third question, I agree with your answer: $P\{2 \text{ have the disease }\} \approx 0.184.$

12. Apr 17, 2013

### Drudge

Ok, I understand and agree with you completely. However I was just really testing a hypothesis, and not counting real results. So I think its ok to approximate for the sake of argument, no, yes? Maybe not, anyway I think the reason I did not understand first was that I did not understand that they where asking EXACTLY 50.

Thank you! I cant believe this book. Well I guess there has to be some errors somewhere.

13. Apr 17, 2013

### phinds

Which is why I said I thought you were solving the wrong problem.

14. Apr 17, 2013

### Drudge

Anybody know what the probability of a question is that:

"has four options, of which any number of can be right"

I cannot think of anything better than 1/4 * 2/4 * 3/4 * 4/4, but cannot help but think that it is not right (at least I know something).

15. Apr 18, 2013

### haruspex

Very unlikely to get such a question. Seriously, I have no idea what you're asking. Pls rephrase it more clearly, preferably in a new thread.

16. Apr 19, 2013

### Drudge

Ok, I have to admit that I made that question up, but what it has to do with is my university admission exam format.

So, for example:

1) Question

A)x
B)y
C)z
D)q

Each question has four options as above and the only way to get points off a question is to mark ALL the right answers. So the answer to the above could be A, B, C, D or AB, AC, AD, BC, BD.... or ABC, BCD... or ABCD.

And so I was wondering what the chances are of guessing a question right (not that I plan on doing so). Also the exam, aswell as having these kind of option-questions, has alot to do with probability problems and I was annoyed that I couldn't even figure this kind of problem out, when the test is much more harder!

17. Apr 19, 2013

### haruspex

You showed 4 answers, each of which may be chosen independently of the others. You could think of this as a sequence of four 0s and 1s. How many possibilities? If all are equally likely (including ticking none), what is the chance of each one?
Btw, I believe the marking systems apply a penalty based on this, so that if you answered all questions at random your expected average score would be 0.

18. Apr 21, 2013

### Drudge

Ok, not really sure what you meant (by the way ticking none is not an option), but I think I figured it out. You must count up all the cases (A, B, C,....) and then the probability is one out of all the choices.

Thanks anyway.

Hey,

should not the "degrees of freedom" in this picture (link below) be 23 and not 15 (on the principle that in a t-Test: Two-Sample Assuming unequal variances, the degrees of freedom is n(1) + n(2) -2 )

http://imageshack.us/photo/my-images/801/img1791d.jpg

EDIT: This is more statistics I´m afraid, but i hope it is not a problem

19. Apr 21, 2013

### haruspex

Yes, but I was trying to get you to see there's a very easy way of counting them all. Don't worry for the moment whether ticking none is valid. You can represent any set of ticks by a sequence of four bits, yes? Conversely, any sequence of four bits represents a unique set of ticks. How many possibilities does that give?
Wrt the t-test (more properly, Welch's test here?) I believe you are right, but I'm no expert on practical stats.

20. Apr 21, 2013

### Drudge

2 * 2 * 2 * 2 ?