Anyone good at probability?

[Kevin26385]

I am stuck on 3 homework problems for a statistics class I am taking this summer. I was curious if anyone could explain how to run through these problems:

Suppose 50% of the students at State U. are enrolled in a math course, and 60% of the students are enrolled in an English course. 38% are enrolled in both types of courses. What is the probability a randomly selected student is enrolled in an English course but NOT a math course?

Suppose there are 9 Democrats and 11 Republicans from which to choose a group of 4. What is the probability the group contains at least one Republican? In addition to calculating the probability

What is the probability a randomly selected male student lives on campus?
Male 73 (commutes) 58 (on-campus)
Female 67 (commutes) 66 (on-campus)

[youvalss]

Re: Anyone good at probability?

What is the probability that the 2nd one will get this thread locked

[bgeno]

Re: Anyone good at probability?

I am stuck on 3 homework problems for a statistics class I am taking this summer. I was curious if anyone could explain how to run through these problems:

Suppose 50% of the students at State U. are enrolled in a math course, and 60% of the students are enrolled in an English course. 38% are enrolled in both types of courses. What is the probability a randomly selected student is enrolled in an English course but NOT a math course?

Suppose there are 9 Democrats and 11 Republicans from which to choose a group of 4. What is the probability the group contains at least one Republican? In addition to calculating the probability

What is the probability a randomly selected male student lives on campus?
Male 73 (commutes) 58 (on-campus)
Female 67 (commutes) 66 (on-campus)

I don't know about the first two. I would have known two years ago when I took probability, but I'm not sure how to do it now.

I THINK for the third one you add up all the male students and divide the total number of male students by the students who live on campus.

[snepp]

Re: Anyone good at probability?

Probably not?

[GAMEC0CK2002]

Re: Anyone good at probability?

I don't know about the first two. I would have known two years ago when I took probability, but I'm not sure how to do it now.

I THINK for the third one you add up all the male students and divide the total number of male students by the students who live on campus.

Agreed. Part (on campus males)/Whole (on campus + commuting males)

I'll just take a stab at the Dem/GOP question. Answer 55%

Total selection pool is 20 individuals and 11 are GOP. Not sure that you having 4 chances at picking raises the chances...maybe it does. EDIT: Thinking about it, it probably does b/c each time your selection pool gets smaller.


My guess for the Math/English question. Common sense says it has to be less than 60%....and you know the overlap is 38% So 60-38= 22% ???

[Reaman]

Re: Anyone good at probability?

Not at all!

[Kevin26385]

Re: Anyone good at probability?

I am finding out there a certain formulas for each one of these.

#2 is .2605
#3 is .2197

...I think haha

I can't figure out #1 for the life of me though, thanks for all the suggestions so far guys

[Speedy]

Re: Anyone good at probability?

My guess for the Math/English question. Common sense says it has to be less than 60%....and you know the overlap is 38% So 60-38= 22% ???

Yeh, logic should be the answer to this one. If there's 100 kids at a school and 50 take math, 60 take english and 38 take both classes...then obviously 22 take english but no math. The answer I would assume must be 22%.

[NoDakHusker]

Re: Anyone good at probability?

I don't know about the first two. I would have known two years ago when I took probability, but I'm not sure how to do it now.

I THINK for the third one you add up all the male students and divide the total number of male students by the students who live on campus.

Correct. I took a prob & stats class...in my senior year of HS and that was almost 2 years ago. I barely remember any of it.(not like I've had any real world application of it)
Really the only statistics I'm good with is sports lol.

[GAMEC0CK2002]

Re: Anyone good at probability?

I am finding out there a certain formulas for each one of these.

#2 is .2605
#3 is .2197

...I think haha

I can't figure out #1 for the life of me though, thanks for all the suggestions so far guys

Either of those results seems low for the Democrat-Republican question. 9 vs. 11 is essentially a 50/50 shot like flipping a coin. Your answer is basically saying you are WAY WAY more likely to pick a Democrat even though they make up a SMALLER portion of the selection pool.

[Blzer]

Re: Anyone good at probability?

I'll take a shot at them, but I don't have a calculator handy so you'll have to do the math.

Suppose 50% of the students at State U. are enrolled in a math course, and 60% of the students are enrolled in an English course. 38% are enrolled in both types of courses. What is the probability a randomly selected student is enrolled in an English course but NOT a math course?

In this problem, think of a Venn diagram. This is a situation where the classes taken are not both independent probabilities, like heads and tails are on a coin toss. So if you wanted to figure out the probability that there is a student in an English course but not in a math course, it should be exactly that... take the 60% in the English course and subtract the 38% that are in both types of courses. That would mean that 22% are exclusively in the English course. This, again, makes sense because these are not independent probabilities, so that the 38% of people taking both courses are inclusive of the 60% of people that are taking English courses. I do hope that makes sense.

Suppose there are 9 Democrats and 11 Republicans from which to choose a group of 4. What is the probability the group contains at least one Republican?

This is one of those binomial distribution problems. If we're finding out the probability that the group contains at least one Republican (that is, one or two or three or four Republicans), it is the same as 1 minus the probability that there are exactly zero Republicans. The normal probability of selecting a Republican is 11/(11+9) = 11/20 = .55, so we'll keep that number saved for later.

What you do is you take the total population of the selected group which is 4, and we're going to choose zero of the 4 for the Republican's sake. So this is a (4)-Choose-(0) problem, which is [4!]/[(0!)(4!)] = 1.

Anyway, once you have that number (which is just 1), you will multiply it by the probability that you have a Republican (.55) and raise it to the 0th power as you're selecting zero Republicans, which is still 1. Multiply that by (1-.55) = .45, raise that to the (4-0) power, which is (.45)^4. I do not know that number, but once you get it, take 1 and subtract it from that.

In short, your number will be 1 - (.45)^4, which should be a high number probably above .9 or 90%. If you think about it, they're almost asking it like this... if you flip a coin four times, what is the probability that you get a heads at least once? Well, think about it as what is not the probability that you never flip a heads? That's a rather high probability. I'm just not going to calculate that number.

In addition to calculating the probability

What is the probability a randomly selected male student lives on campus?
Male 73 (commutes) 58 (on-campus)
Female 67 (commutes) 66 (on-campus)

This is a "probability x given y" problem, where basically we are omitting the females completely. We are asking "what is the problem a student lives on campus, given that it is a male?" So I think it's simply 58/(58+73) = 58/131, whatever that number is.




Hope I answered them correctly.

[Blzer]

Re: Anyone good at probability?

I didn't have a calculator, but I whipped out a piece of paper and did #2 on it. My camera doesn't exactly take stills the best for this kind of thing, so I made a "video" that shows my work, panning across the steps that I did. There is no audio.


<object width="640" height="385"><param name="movie" value="http://www.youtube.com/v/sCFngTnNHPg&hl=en_US&fs=1&"></param><param name="allowFullScreen" value="true"></param><param name="allowscriptaccess" value="always"></param><embed src="http://www.youtube.com/v/sCFngTnNHPg&hl=en_US&fs=1&" type="application/x-shockwave-flash" allowscriptaccess="always" allowfullscreen="true" width="640" height="385"></embed></object>


Hope this helps!

[deaduck]

Re: Anyone good at probability?

I'll take a shot at them, but I don't have a calculator handy so you'll have to do the math.


In this problem, think of a Venn diagram. This is a situation where the classes taken are not both independent probabilities, like heads and tails are on a coin toss. So if you wanted to figure out the probability that there is a student in an English course but not in a math course, it should be exactly that... take the 60% in the English course and subtract the 38% that are in both types of courses. That would mean that 22% are exclusively in the English course. This, again, makes sense because these are not independent probabilities, so that the 38% of people taking both courses are inclusive of the 60% of people that are taking English courses. I do hope that makes sense.


This is one of those binomial distribution problems. If we're finding out the probability that the group contains at least one Republican (that is, one or two or three or four Republicans), it is the same as 1 minus the probability that there are exactly zero Republicans. The normal probability of selecting a Republican is 11/(11+9) = 11/20 = .55, so we'll keep that number saved for later.

What you do is you take the total population of the selected group which is 4, and we're going to choose zero of the 4 for the Republican's sake. So this is a (4)-Choose-(0) problem, which is [4!]/[(0!)(4!)] = 1.

Anyway, once you have that number (which is just 1), you will multiply it by the probability that you have a Republican (.55) and raise it to the 0th power as you're selecting zero Republicans, which is still 1. Multiply that by (1-.55) = .45, raise that to the (4-0) power, which is (.45)^4. I do not know that number, but once you get it, take 1 and subtract it from that.

In short, your number will be 1 - (.45)^4, which should be a high number probably above .9 or 90%. If you think about it, they're almost asking it like this... if you flip a coin four times, what is the probability that you get a heads at least once? Well, think about it as what is not the probability that you never flip a heads? That's a rather high probability. I'm just not going to calculate that number.



This is a "probability x given y" problem, where basically we are omitting the females completely. We are asking "what is the problem a student lives on campus, given that it is a male?" So I think it's simply 58/(58+73) = 58/131, whatever that number is.




Hope I answered them correctly.

I took a probability class decades ago and also an abstract math class and the two have constantly battled each other in my head ...making math a never ending war.

I concur with your thinking on Question 1.

Not sure I agree with Question 2's reasoning because, and I think is the abstract math speaking, but I always get locked on this... because each member of the four person group has 55% chance of being Republican. This odds does not fluxate in regards to the next person in the group chosen to join the four...still just 55% chance it's also Republican....and so on. To calculate any other odds is adding up variables that have ceased to exist outside the individual chance.

And I think Question 3 is probably worded poorly. I think the maleness of the subject isn't a critera for a random draw. Otherwides why even inclue the female numbers?

[mKoz26]

Re: Anyone good at probability?

And I think Question 3 is probably worded poorly. I think the maleness of the subject isn't a critera for a random draw. Otherwides why even inclue the female numbers?

Just to throw you off. They do this all the time in math courses.

[GAMEC0CK2002]

Re: Anyone good at probability?

The highest possible chance you could select a Republican would be on the 4th draw assuming you picked Democrats on the 3 proceeding selections.

Part/Total --> 11 (Republicans)/17 (Republicans + Remaining Democrats) = roughly 65%

Not sure what the original question is looking for. You have to start making assumptions to get anything other than 55%.