Quote :

"In a group of 5 people, what is the probability that they all have different birthdays? [Assume that each is equally likely to be born on any of the 365 days of the year,"

1- probability of them all being born on the same day((1/365)*(1/365)*(1/365)*(1/365)*(1/365)) = probability of them all being born on different days

thats what i would do

11/7/2006 8:12:44 AM

I don't think that works. Do you have any other ideas.

11/7/2006 9:35:41 AM

only other thing i could think of was (5/365)*(4/364)*(3/363)*(2/362)*(1/361)

then subtract that answer from 1 and maybe its your answer

my reasoning is that like after you count one persons birthday, there are 364 days for the other 4, then 363 days for the remaining 3, etc.

11/7/2006 9:41:13 AM

I have all of the wooden board game one...

for the first part, do C(44,4)=135751
for the second part, do (22,4)=7315, since 1/2 the pieces are painted white, you take half of 44
for the third part, simplify 7315/135751

------
how did you find the answers to the checkerboard problem? when i try and plug those numbers in to the calc, it says 'overflow'...same thing with the students taking foreign language, i think i could figure it out, but my calc wont do those high numbers, i have a TI-83

[Edited on November 7, 2006 at 10:24 AM. Reason : a]

11/7/2006 10:10:16 AM

I have them all figured out for that homework assignment now.

If you go to the following website, you can use the combination calculator.....

http://condor.wesleyan.edu/jfrugale/wescourses/2004f/math132/01/calculator.html

Let me know if you are still having trouble.

Thanks again.

11/7/2006 12:52:59 PM

i'm still having trouble with the foreign language one and the mp3 player one...

i'll repost the ones i dont have...

[Edited on November 7, 2006 at 1:29 PM. Reason : /]

11/7/2006 1:22:06 PM

A group of 3 boys and 4 girls are going to sit together in a row of 7 theater seats. If they seat themselves randomly,

What is the probability the boys will be seated together and the girls will be seated together?

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In a group of 5 people, what is the probability that they all have different birthdays?

----------

Solve the following questions correct to 3 decimal places. A college has 700 students. A survey has determined that

68 read French
36 read German
39 read Spanish
14 read French and German
15 read French and Spanish
6 read German and Spanish
3 read all three languages

What is the probability a randomly selected student reads exactly 2 of these 3 languages?
What is the probability a randomly selected student reads at least 1 of the 3 languages?
What is the probability a randomly selected student reads exactly 1 of the 3 languages?
What is the probability a randomly selected student does not read any of the 3 languages?

----------

Suppose 6 men enter a restaurant and each of them gives his hat to a receptionist.
Suppose now that only 3 of the 6 men have hats when they enter, and that the receptionist gives the 3 hats back to 3 randomly picked men (from the 6) when they leave. What is the probability that each of the 3 hats will be returned to the correct owner?

----------

A normal checkerboard has 8 rows and 8 columns of alternating colored squares (see picture of antique checkerboard). Suppose you have a 12 by 12 checkerboard with alternating brown and white squares. If 4 pieces are randomly placed on the checkerboard,

what is the probability that they will all be in the same column?

----------

Most mp3 players (like the iPod, for example) have a "shuffle" feature which allows you to randomize the order of songs played from a playlist. If your playlist consists of songs A, B, C, D, and E (in that order), then if "shuffle" is activated the songs might play in this order: B, E, A, D, C. In effect, the way that a "shuffle" works is to create a random permutation of the playlist prior to playing any of the songs.

Other players, however, have a "random" rather than "shuffle" feature. With this mode of operation, random songs are played no matter what has been played before. So for example, if the above playlist were played in a "random" mode, then the first four songs played might be B, E, B, A. (Each time a new song is chosen for play, it is equally likely to be any of the songs. So in this example with 5 songs on the playlist, song B has a 20% chance of being played each time a new song is selected, even if it has already been played.)

Suppose that a playlist has 14 songs (all different) with 3 of the songs being by Laura Cantrell. If you listen to 4 songs in shuffle mode, what is the probability that you will hear at least one song by Laura Cantrell?

Answer (to at least 4 decimal places accuracy) =

If you listen to 4 songs from this playlist on an mp3 player that has a "random" mode, what is the probability that you will hear at least one song by Laura Cantrell?

Answer (to at least 4 decimal places accuracy) =

----------

A bag contains 20 colored blocks.

6 of them are red
7 of them are black, and
7 are green

If you randomly draw 3 blocks out of the bag,

what is the probability that 2 will be one color and the remaining one a different color?

----------

thanks for any help you can give me.

11/7/2006 1:33:15 PM

When I get back from work I will send you the solutions to the boardgame problem

11/7/2006 2:33:44 PM

Quote :

"A group of 3 boys and 4 girls are going to sit together in a row of 7 theater seats. If they seat themselves randomly, What is the probability the boys will be seated together and the girls will be seated together?"

Well, the total number of ways they can be seated is 7! = 5040.

So, you're trying to organize the boys and the girls into groups. The combination of the girls and boys doesn't matter, as long as they're grouped according to gender. So, take the permutation of each little group. P(4,4) = 24 and P(3,3) = 6. Because you both sections to be grouped, multiply the two figures by each other to get 144.

So, I think the answer is 144/5040

-----

Quote :

"In a group of 5 people, what is the probability that they all have different birthdays?"

Find how many potential birthdays exist in the group. So, multiply 365 by itself 5 times (or just take the exponent). 365^5 = some insanely large number A (save this number into your calculator so you can bring it up later)

Then, find the permutation of the 5 different birthdays in one year. I think this is because it doesn't matter what order they go in, just as long as they're different. P(365,5) = some other insanely large number B (save this number into your calculator, too)

Finally, take B and divide it by A. You should get probability with that.

-----

Quote :

"Solve the following questions correct to 3 decimal places. A college has 700 students. A survey has determined that 68 read French 36 read German 39 read Spanish 14 read French and German 15 read French and Spanish 6 read German and Spanish 3 read all three languages What is the probability a randomly selected student reads exactly 2 of these 3 languages? What is the probability a randomly selected student reads at least 1 of the 3 languages? What is the probability a randomly selected student reads exactly 1 of the 3 languages? What is the probability a randomly selected student does not read any of the 3 languages? "

Use the triple venn diagrams. Start in the middle with the 3 people who read all three languages, and then move outwards. Make sure that you include those 3 people in the other groups. For example, the 3 people who read all three languages also are counted in the 6 who read German and Spanish languages. So subtract the 3 from 6, and you've got 3 who only read German and Spanish. Keep doing that as you move outward.

To find the probability, just put the number of language readers over 700 and you're done.

-----

Quote :

"A normal checkerboard has 8 rows and 8 columns of alternating colored squares (see picture of antique checkerboard). Suppose you have a 12 by 12 checkerboard with alternating brown and white squares. If 4 pieces are randomly placed on the checkerboard, what is the probability that they will all be in the same column?"

That one should just be whatever the numerator of the first question was multiplied by 12. Then, divide that number by whatever the denominator of the first question was, and that should be it.

[Edited on November 7, 2006 at 4:34 PM. Reason : checkers!]

11/7/2006 4:09:38 PM

Quote :

" Most mp3 players (like the iPod, for example) have a "shuffle" feature which allows you to randomize the order of songs played from a playlist. If your playlist consists of songs A, B, C, D, and E (in that order), then if "shuffle" is activated the songs might play in this order: B, E, A, D, C. In effect, the way that a "shuffle" works is to create a random permutation of the playlist prior to playing any of the songs. Other players, however, have a "random" rather than "shuffle" feature. With this mode of operation, random songs are played no matter what has been played before. So for example, if the above playlist were played in a "random" mode, then the first four songs played might be B, E, B, A. (Each time a new song is chosen for play, it is equally likely to be any of the songs. So in this example with 5 songs on the playlist, song B has a 20% chance of being played each time a new song is selected, even if it has already been played.) Suppose that a playlist has 14 songs (all different) with 3 of the songs being by Laura Cantrell. If you listen to 4 songs in shuffle mode, what is the probability that you will hear at least one song by Laura Cantrell? Answer (to at least 4 decimal places accuracy) = If you listen to 4 songs from this playlist on an mp3 player that has a "random" mode, what is the probability that you will hear at least one song by Laura Cantrell? Answer (to at least 4 decimal places accuracy) ="

For the first one, find the probability of listening to 4 songs in shuffle mode and NOT hearing any Laura Cantrell songs. Subtract that probability from 1 and you should get the answer.

For the second one, multiply the number of total songs in the playlist by itself four times (or just take the exponent). Then, multiply the number of songs that aren't by Laura Cantrell four times (or just take the exponent). Divide the second number by the first, and subtract that fraction into 1. You should get the probability of hearing at least one Laura Cantrell song that way.

11/7/2006 4:42:02 PM

[quote]For the first one, find the probability of listening to 4 songs in shuffle mode and NOT hearing any Laura Cantrell songs. Subtract that probability from 1 and you should get the answer.

I have tried every way I know how to do the probability but I must be messing the steps up somehow, what would be the exact step for finding out this first part?

11/7/2006 6:32:39 PM

Take the combination of all the songs that could be played in shuffle mode. In this problem, you're considering the organization of the songs that are played, so the order of the songs does matter. Anyway, that should be C(14,4)

Then, take the combination of all the songs that aren't by Laura Cantrell, and could still be played in shuffle mode. Just subtract the number of Laura Cantrell songs from the total number of songs to get the first part of the combination. Once again, order does matter; that's why you're using combinations. It should be C(11,4)

Then, divide C(11,4) by C(14,4). Subtract that number from 1, and that should be the correct answer.

If you want to get over the potential for an error caused by bad "4 decimals places accuracy" rounding, just put down the fraction as your answer. They accepted them when I took the course.

[Edited on November 7, 2006 at 6:49 PM. Reason : decimal places problems?]

11/7/2006 6:44:11 PM

that worked, thanks, can you do figure this one out...?

Suppose 6 men enter a restaurant and each of them gives his hat to a receptionist. If the receptionist randomly passes the 6 hats back to the 6 men when they leave, what is the probability that each man gets the correct hat?

Suppose now that only 3 of the 6 men have hats when they enter, and that the receptionist gives the 3 hats back to 3 randomly picked men (from the 6) when they leave. What is the probability that each of the 3 hats will be returned to the correct owner?
(I got the first part of this question, i just need to last part)

---------- or...

A bag contains 20 colored blocks.

6 of them are red
7 of them are black, and
7 are green

If you randomly draw 3 blocks out of the bag,

what is the probability that 2 will be one color and the remaining one a different color?

----------

thanks

11/7/2006 7:02:45 PM

Quote :

"Suppose 6 men enter a restaurant and each of them gives his hat to a receptionist. If the receptionist randomly passes the 6 hats back to the 6 men when they leave, what is the probability that each man gets the correct hat? Suppose now that only 3 of the 6 men have hats when they enter, and that the receptionist gives the 3 hats back to 3 randomly picked men (from the 6) when they leave. What is the probability that each of the 3 hats will be returned to the correct owner? (I got the first part of this question, i just need to last part)"

I think you take the permutation of the number of hats to the number of men in the restaurant. So, P(6,3)

Put that in as the denominator of the fraction, with 1 as the numerator, and that should be it.

[Edited on November 7, 2006 at 7:31 PM. Reason : -]

11/7/2006 7:29:29 PM

Quote :

"A bag contains 20 colored blocks. 6 of them are red 7 of them are black, and 7 are green If you randomly draw 3 blocks out of the bag, what is the probability that 2 will be one color and the remaining one a different color?"

I know that you should take the combination of the total number of blocks to the number of blocks you take out of the bag. So, C(20,3) will be the denominator.

After that, I think you have to take the combinations of the various arrangments and multiply the pairs. Then, take all of the products and add them up to get the numerator. For example, 2 reds and 1 black will result in C(6,2) times C(7,1). Do the same for the rest and add them up. That might be the way to get it.

[Edited on November 7, 2006 at 7:44 PM. Reason : -]

11/7/2006 7:41:02 PM

Does anyone else need any help on any of the other problems?

11/7/2006 9:14:13 PM

i'm still having trouble getting the last part of question 12

11/7/2006 9:16:27 PM

that shit is easy.. take all the possible probs and they are all OR so add them all together. i wont tell yah how to do it though. ull wanna smack ur self in the head when u get it

11/7/2006 9:19:08 PM

you all are fucking saints..

couldn't figure out half of this stuff...

11/7/2006 10:49:15 PM

I remember this class...never got such an easy A+

11/8/2006 2:30:07 PM

i got a b+ when i shoulda had an a

11/8/2006 2:41:57 PM

so who's ready to work on 5.4??

11/9/2006 6:42:14 PM

well i've already got a question, number 1 part 5, and number 4 part 2 and 5...

If P(A) = 0.4 and P(B) = 0.25 and P(A n B) = 0.1,

Find P(A n B | A)

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If P(A) = 0.4 and P(B) = 0.35 and P(A n B) = 0.09,

Find P(B | A^c)
Find P(A n B | A)

11/9/2006 7:43:58 PM

5.4, question 6, part 3 (I'm doing the homework late). I used the tip sheet and was able to get the first two parts of the problem, however I am having trouble with the third part where it asks:
What is the probability that both officers will be women given that at least one is a woman?

The tip sheet explains that you take the intersection of the probability of at least one being a woman and the probability of both being women divided by the probability of at least one. I can't seem to figure out how to get where these two intersect. Any help would be greatly appreciated, thanks! Oh and I guess I should post the entire problem to give you some background:
Solve the following questions correct to 3 decimal places. There are 4 men and 6 women are on a committee. Beth is one of the women. Two of the 10 people are to be chosen to serve as officers. If the officers are chosen randomly, What is the probability that both officers will be women given that at least one is a woman?

11/11/2006 2:41:02 PM

Just look at the example on the website for the section 5.4 Notes. There is an example in there that is identical.

Let me know if you still have trouble.

11/11/2006 5:14:02 PM

thank you,
I figured that out, finally, but I am having trouble figuring out what Beth's probability is. I am having the same trouble with the "amy" problem. I can get all parts except for the individual's probability, so that I can come up with the end answer. In reviewing the examples, I can't even figure out how the got Beth's probability. I may have to watch the movie again, unless you can point me in the right direction. Thanks for your help, though.

11/11/2006 5:36:45 PM

I still can't figure out how to find Beth's probability, so if you can help, I'd appreciate it, thank you!

11/11/2006 7:14:17 PM

how much do webassigns count for your grade

i think mine was mostly the tests

11/11/2006 9:39:20 PM

For Beth's probability, I think you just need to use a little logical reasoning to get the answer. That's how I got it when I did the webassign (that's how I wrote it down in my notes, at least).

If Beth is an officer, then 5 women are left over out of the remaining 9 people that could be selected. That should be a probability of 5/9. When I did the problem, there were 4 men and 5 women in the committee, but the process for getting the correct answer should be the same.

11/11/2006 10:00:25 PM

^^ web assigns actually count for quite a bit of your grade, almost half, I believe. But this is the online class, so they probably count for more than they do in the on-campus class.

[Edited on November 12, 2006 at 8:51 PM. Reason : ?]

11/12/2006 8:50:21 PM

Has anyone started on Section 5.5 HW?

Thanks.

11/13/2006 10:41:00 AM

i'm looking at it right now...

11/13/2006 11:50:30 PM

meh not yet tom night yes. keep this thread alive for tom. night

11/14/2006 12:30:48 AM

Three percent of the people in a certain city are in a "high risk" group for HIV infection. And 6% of the hish risk group are HIV positive. But only 0.7% of the people not in the high risk group are HIV positive.

What is the probability that a randomly chosen person from this city is HIV positive?

What is the probability that a randomly chosen person from this city is in the high risk group AND is HIV positive?

What is the probability that a randomly chosen person from this city is in the high risk group OR is HIV positive?

If you know that a specific person is HIV positive, what is the probability that person is in the high risk group?

---------------

Gateway and Micron are computer stocks that frequently move together because they represent the same sector of the economy. During a 25 day period, the value of Gateway stock went up 14 days. On 10 of these 14 days, the value of Micron stock also went up.

On the 11 days when Gateway stock did not go up, Micron stock went up on 2 of these days.

Turn all these fractions into probabilities and put all the info into a tree diagram. (For example 14/25 is the probability that Gateway stock goes up, so 11/25 is the probability that it doesn't go up. And 10/14 is the conditional probability that Micron goes up if Gateway goes up, etc.)

What is the probability neither stock goes up?

If Gateway goes up, what is the conditional probability Micron goes up?

If Micron goes up, what is the conditional probability Gateway goes up?

If Micron does not go up, what is the conditional probability Gateway will go up?

---------------

anyone tried these yet? i'm working on them but am having some trouble

11/14/2006 2:24:36 PM

I have done both of them, but I don't really know how to explain them on here.

Have you finished all of the other problems?

11/14/2006 2:49:39 PM

not yet, i have the first few, i only have 34 out of 73, i gotta go to work at 3 tho, so i'll have to get on later and see whats up...

11/14/2006 2:56:54 PM

Does anyone know how to do part 5 on problem #4 for the 5.5 webassign. the problems reads as this---

Gateway and Micron are computer stocks that frequently move together because they represent the same sector of the economy. During a 25 day period, the value of Gateway stock went up 17 days. On 11 of these 17 days, the value of Micron stock also went up.

On the 8 days when Gateway stock did not go up, Micron stock went up on 2 of these days.

Turn all these fractions into probabilities and put all the info into a tree diagram. (For example 17/25 is the probability that Gateway stock goes up, so 8/25 is the probability that it doesn't go up. And 11/17 is the conditional probability that Micron goes up if Gateway goes up, etc.)


If Micron does not go up, what is the conditional probability Gateway will go up?

11/14/2006 4:09:59 PM

I have all of them done. If you need help just let me know.

11/14/2006 4:33:48 PM

I just need help on that part 5 of problem 4 that is listed above

11/14/2006 4:35:50 PM

I need the same. I've finished everything except the final part of the Gateway/Micron question. You'd think you would use the same method as the one right before it, but that doesn't work. Hmmm.

11/14/2006 4:45:23 PM

I drew out a venn diagram and then filled in the corresponding areas.

To get the answer to the fifth part you take P(G|M^C), so therefore you take the # of days that ONLY Gateway goes up (ex. 5/25) and divide it by the # of days that the Micron stock DOES NOT go up (ex. 13/25).

Hope this helps.

11/14/2006 4:47:22 PM

I HAVE BEEN WORKING ON THIS ONE FOR THE LAST 45 min and cant figure it out

Three percent of the people in a certain city are in a "high risk" group for HIV infection. And 7% of the hish risk group are HIV positive. But only 0.6% of the people not in the high risk group are HIV positive.

What is the probability that a randomly chosen person from this city is in the high risk group OR is HIV positive?

11/14/2006 5:41:48 PM

If you draw out the tree diagram, they you just calculate the probabilities of each positive and negative as shown in the example tree diagram in the problem.

once you get the probabilities, then you add the + and - probabilities of the HR side and then add the + side of the not HR side.

Let me know if you get it.

11/14/2006 6:06:09 PM

That helps a lot, thanks! ^^^

11/14/2006 8:40:49 PM

Does anyone know how to do this one??

Suppose A and B are independent events with P(A) = 0.27 and P(B) = 0.49. What is the probabability that at least one of the events occurs?

11/15/2006 9:13:59 PM

I figured out the other one ^^^^^ does anyone have any clue on this one I have been tryin for the last 45 min to get it?????


Tom and Alice work independently in an attempt to solve a certain problem, i.e. whether one of them solves it does not affect the chances that the other will solve it. The probability that Tom solves it is 0.2 and the probability that Alice solves it is 0.4. What is the probability that the problem will be solved?


If the problem is solved, what is the probability that Tom solves it?

11/15/2006 10:12:19 PM

Does anyone know how to approach this problem?

-----------------------------------------------------------------------------------------------------------
The Reds and the Cubs are playing 5 games. Each game the probability that the Reds win is 0.53, and the probability of the Reds winning is not affected by who has won any previous games.

(a) What is the probability the Reds win all 5 games?
(b) What is the probability the Reds win 4 and lose 1?
(c) What is the probability the Cubs win 4 and lose 1?
(d) What is the probability the Cubs win all 5 games?


I know how to do parts a and d, I need help on b and c.

THANKS

11/16/2006 6:16:13 PM

^^^^You have to draw a tree diagram starting off with either the Reds or the Cubs winning and drawing it until you get to the fifth game. Then you look for branches on the tree that have the the Reds winning 4 games and the Cubs winning one. You multiply the probablilities up the tree ex: .53*.53*.53*.47*.53. You then add all the different outcome probablities up?? Does this make sense at all??

11/16/2006 6:48:24 PM

Ok thanks, I was using the right method just forgot one multiplication factor.

I have gotten them all done now if anyone needs any help.

11/16/2006 7:46:44 PM

damn, i was gonna post how to do them but u already figured it out

11/16/2006 7:47:58 PM