PDA

View Full Version : Prob w/Combination of At Least AND Combination

ckbeme
04-21-2006, 05:40 PM
I'm stuck on this. I've had similar probs, but none that combine the At Least with this type of selection. I really hate to ask for help on a problem that I think is probably fairly simple, but I'm really behind in my class and I don't want to waste too much time trying to figure out One Problem.

A bus has 8 students from the1st grade and 10 students from the 2nd grade. If 2 children become ill, what is the probability that at least one will be a 2nd grade student?

If any of you give me a solution, could you try to give me details on how you arrived at the solution - and tell me how you knew what formula/method to use? As always, thanks for any help.

>>>> Ck - Struggling, Starving, Sleepless Student <<<<

JohnM
04-21-2006, 08:49 PM
The probability of "at least 1" is the same as 1 minus the probability of "none."

P(at least 1) = 1 - P(none)

There are 8 first graders and 10 second graders. If 2 get sick, the probability of no 2nd graders is:

(8/18) * (7/17) = 56/306 = 0.183

P(at least 1) = 1 - 0.183 = 0.817

ckbeme
04-22-2006, 04:51 PM
The probability of "at least 1" is the same as 1 minus the probability of "none."

P(at least 1) = 1 - P(none)

There are 8 first graders and 10 second graders. If 2 get sick, the probability of no 2nd graders is:

(8/18) * (7/17) = 56/306 = 0.183

P(at least 1) = 1 - 0.183 = 0.817

Thanks, John. I appreciate the detailed and simplified explanation. I thought the solution would use the complementary event rule, but I couldn't figure out how to calculate the 'no 2nd graders' formula. I was trying to use the Combination rule and couldn't make it work.

I got a solution from another forum that used the combination and complementary event; it's interesting to see how you came up with the solution using multiplcation. Here is the other solution I got:

AT LEAST ONE IS THE OPPOSITE OF NONE!It is just that simple.
18C2 is the total number of ways for two children to become ill. 8C2 is the total number of way for only two first graders to become ill. So the probability that only first graders will be ill is 8C2/18C2.

Thus the probability that not only first graders will be ill is 1-8C2/18C2.
In other words, at least one second grader is ill.

Your solution seems easier to me, but I bet our text would show this type of problem (if only they showed this type of problem!) being worked using combination and complement - they have used combination a lot.

I see how both answers were calculated; I just need to find some similar problems and work them both ways. I'm taking a Web course that has almost zero resources aside from the text, so I need a different source.I've searched for sites with practice problems AND detailed solutions, and have come up with very few.

I'm desperate for a web source. I recently started to use this and one other forum. It's been a tremendous help, but I don't want to abuse the assistance from the forums. Anyway, I need some repetition on certain types of problems. Do you know of any site that might help me?

Sorry this is so long, but I wanted to explain what I really need.

Ck

JohnM
04-22-2006, 05:02 PM
Either way works fine....

My approach uses conditional probability. Using an analogy with red and blue marbles, let's say a box has 8 red and 10 blue marbles.

What's the probability of selecting no blue marbles if we select two at random, without replacement?

On the first draw, P(red) = 8/18. Without putting back the red marble, what's the probability of selecting a red one again? It's 7/17.

So, 1 - [ (8/18) * (7/17) ] = P(at least 1 red) = .817

What other forum have you gotten help from?

ckbeme
04-23-2006, 04:27 AM
Either way works fine....
My approach uses conditional probability.

Yeah, both ways do work fine, but I'd like to work a few probs so I could decide which method seems more logical to me.

I think that's one of my problems with this course - When I start working a problem, sometimes I have trouble figuring out what method to use - conditional probability, addition rule, counting rule, etc. I often find myself getting confused about a fairly easy question with familiar content. I hope that gets better soon.

What other forum have you gotten help from?

I've been using Math Forums @ Math Goodies http://www.mathgoodies.com/forums/ I almost left the site before I discovered the forum. The name Math Goodies just didn't sound like it would have any college level material. I'm glad that I was wrong about the site content. They have a very active forum and the posts I've put up have gotten multiple replies within a fairly short time.

That additional example you sent was helpful. (where you turned the kids into marbles) Speaking of help, did you think of any sites that might work for me? Refreshing your memory---- I've searched for sites with practice problems AND detailed solutions, and have come up with very few. Many of those few that I find are covering a different Statistics discipline. ---- Perhaps you could put on that Blue Baseball Cap - we'll call it your thinking hat - and Voila! you would have a list of hundreds of perfect web sites for me. Perhaps.

Later,

CK

JohnM
04-23-2006, 08:38 AM
Speaking of help, did you think of any sites that might work for me? Refreshing your memory---- I've searched for sites with practice problems AND detailed solutions, and have come up with very few. Many of those few that I find are covering a different Statistics discipline. ---- Perhaps you could put on that Blue Baseball Cap - we'll call it your thinking hat - and Voila! you would have a list of hundreds of perfect web sites for me. Perhaps.

mathgoodies is a great site for getting high-school algebra help - I knew they occasionally assisted with prob/stats....

Actually, I don't know of any sites that list a lot of practice problems and detailed solutions....and if there were, we wouldn't be getting as many visitors as we do, would we?:D

But, there are some really good books that may have what you're looking for - go to Amazon.com and search under Schaum's Outlines - look for beginning statistics or probability...

The problem is, what you're looking for has a price-tag attached - no one is going to produce that much valuable work without compensation....