kitchen table math, the sequel: help desk - probability

Friday, August 6, 2010

help desk - probability

I'm having trouble with some problems in Mary Dolciani's Algebra and Trigonometry: Structure and Method Book 2.

p 749
11.  A bag contains 2 red, 4 yellow, and 6 blue marbles. Two marbles are drawn at random. Find the probability of each event.

a. Both are red.
b. Both are yellow.
c. Both are blue
d. One is red and one is yellow.
e. Neither is red.
f. Neither is blue.

For a, b, and c, I can solve this problem using two methods and arrive at the same answer:

a. Both are red.

2/12 x 1/11 = 1/66

OR, using the combination formula, nCr:

sample space (number of ways any 2 marbles can be picked out of 12):
12C2 = 12! ÷ 2!(12-2)!
          = 12 x 11 ÷ 2
          = 66

ways to pick 2 red out of 2 red:
2C2 = 1

probability of picking 2 red:
1 ÷ 66 or 1/66


Trouble is: when I use both approaches to solve d, I get 2 different answers, and I don't understand why.

d. One is red and one is yellow.

2/12 x 4/11 = 1/6 x 4/11 = 4/66 = 2/33

OR, using combination formula nCr:

12C2 = 66 (number of ways any 2 marbles can be drawn from 12)

2C1 = 2 (number of ways 1 red marble can be drawn from 2 red marbles)
4C1 = 4 (number of ways 1 blue marble can be drawn from 4 blue marbles)

so:

2 x 4 ÷ 66 = 4/33 (probability of drawing 1 red & 1 yellow)

2/33 ≠ 4/33

What am I missing here?

Thank you!

Algebra and Trigonometry: Structure and Method Book 2

18 comments:

Kevin said...

Do you have to draw the red on the first draw and the yellow on the second draw? or can you draw them in either order? Your two methods compute these two answers and when the marbles are different, they aren't the same problem.

Dan Anderson said...

Hello!

Your first answer is incorrect only because you have only accounted for drawing a red marble first and a yellow second, and you haven't added the possibility of a yellow first and a red second. I.e. P(red then yellow) + p(yellow then red) = 2/33 + 2/33 = 4/33

Allison said...

Probability problems are very confusing, and worse when badly written.

"You draw two" marbles makes it sound as if there is no "first marble" and no "second marble". In the case where the answers are "both of the same color", the order doesn't matter.

But if "draw two" means "first you draw one marble, and then you draw another", then order does matter. Which one came first?

Kevin and Dan are rephrasing the problem to mean "first you draw one, then another", and no novice really knows why they'd do this.

But the first way you solve the problem assumes the lack of simultaneity. When you solve the problem with "2/12 * 1/11" you are really NOT solving the problem by drawing two.

You are solving it by drawing first one and then the other: your 2/12 is "draw one red" and your * then is "and then, draw another red".

(do you know why you are multiplying here? Tthese events aren't independent)

The answer for the combinations is different. In that case, the state space is different. That is, the possible universe of answers from which you are choosing two simultaneous marbles is different, because you really are considering "I drew two at the same time", and then all the way you could have drawn two at the same time. In that case, you're acting as if the two aren't distinguishable: you're saying you don't care which is first and which is second.

This is really subtle. I'll talk more about what the factorials mean in another post.

Hainish said...

I would have taken the problem to mean that you draw a marble, put it back in the bag, and then draw again. (Replacement, I think it's referred to.)

Allison said...

Hainish,

This is why the problem is terribly written. NO NO NO NO, this is NOT with replacement! Unless this problem is even more exceptionally badly written, problems with replacement ALWAYS state such (because the combinations answers are just *wrong* in that case.) But you're right, the phrasing could lead someone to think that. But as a heuristic, absolutely do not assume replacement when solving problems like this.

btw, do you know that you're getting the right answers? Did you get the right answers to part b?

Catherine Johnson said...

Oh thank you so much!

In fact, I kept asking myself: what if the order is reversed?

What if you draw yellow first & then red?

But I didn't know what to do with that question.

Is there a way to use the 'simple' method of calculating probability of drawing 1 red without replacement followed by a second red to calculate 'draw 1 red & 1 yellow with order unspecified'?

I'm very grateful ---- !

Allison said...

Yes. The answer is that you add the probabilities together of the two cases.

The question (given all the other bits of the problem) is find the probability of event E=One is red and one is yellow. That is, find P(E), where E = Draw two, one red and one yellow.

Event E is actually the sum of two different events, where those two events are mutually exclusive.
P(E) = P(E1) + P(E2). That's because E is really E1 OR E2.
So P(E1 OR E2) is P(E1) + P(E2) - P(E1 and E2). Because they are mutually exclusive, you can just add these probabilities together, since P(E1 and E2) = 0.

E1 = First one drawn red, second one drawn yellow.
E2 = First one drawn yellow, second one drawn red.

P(E1) is then computed the way you computed it. P(E2) is computed the same way. Then you add them together.

lgm said...

I was quite surpised that this material is now part of Int. Alg. I. My son's teacher gave up - poor lady only had 5 class periods allowed and the 8th grade teacher had omitted the unit and on top of that she had to teach how to punch it in on the calculator...

The Arlington Algebra Project Units 11&12 might be helpful as might the JMAP resources:
http://www.jmap.org/htmltopics/PROBABILITY/JMAP_BY_TOPIC_PROBABILITY.htm

Be sure you know the difference between
a. independent & dependent event
b. replacement & no replacement
c. combination & permutation
before beginning to work problems and moving on to multiple events.

Catherine Johnson said...

The answer is that you add the probabilities together of the two cases.

That's what I was wondering!

I should have persisted before asking.

Catherine Johnson said...

It's definitely not a replacement problem....that much I worked out for myself though reading and through checking to see what would happen if I worked it as a replacement problem.

I'm still having trouble figuring out when to add mutually exclusive probabilities, obviously. (Having some trouble even detecting 'mutually exclusive.')

I've made progress on that front, but not enough.

lgm said...

Hainish, I had the Dolciani series independently from 7th grade on. The problem isn't confusing at all b/c the experience of completing the coursework has trained the student to understand not to add any more to the words than are there. Unlike modern texts, Dolciani didn't go for adding in unnecessary details or vague wording to deliberately confuse the student. 'Draw two marbles' means just that. Careful examination of the sections in the text will show the difference in wording between 'draw two' with replacement and without replacement. If I had my text, with me, I'd scan some pages in..but I'll do that later.

lgm said...

>>I'm still having trouble figuring out when to add mutually exclusive probabilities, obviously. (Having some trouble even detecting 'mutually exclusive.')

The Arlington Algebra Project, Unit 12 L5 & L6 may be helpful.

Allison said...

The best thing to do is practice writing out these composite events in terms of little events. Don't bother working out the probabilities until you first have a handle on the events.

The little events should be the things you're told you'd do: draw a marble, toss a die. In general, a very good way to decompose the events is in terms of "first toss, second toss, kth toss." That's because the second toss is independent *conditional on the first toss*. And this is true also the other way: the *first toss is independent conditional on the second toss.*.

Probability is NOT about causality. It isn't asking what depended PHYSICALLY on each other. This is the biggest confusion people have.
The above stuff is really hard to get used to. It means you can calculate the probability of event E_2, something that happens on the second toss, when you've not yet calculated E_1, what happened on the first. After a while, this makes sense, but it takes lots of practice.

Sometimes, first toss and second toss isn't enough. Your composite is the "and" or "or" of first and seconds. If the event is "no ball red", that means "first ball not red AND second ball not red". It also means "first ball blue OR yellow AND second ball blue OR yellow."

So, figure out how your composite event is written in terms of those little events. So for statements like probs e and f, find the little statements. If E is "neither toss is blue", then write down in terms of first and second toss. After you're good at that, you might notice statements about "neither" are saying something about "both". "neither toss is blue" means "not the case that both tosses are blue." the event "both tosses are blue" should be easy to calculate.

Catherine Johnson said...

I don't think I've ever even heard of the Arlington Algebra project.

Thanks for the recommendation.

I LOVE the Dolciani text, btw.

Allison said...

--"neither toss is blue" means "not the case that both tosses are blue." the event "both tosses are blue" should be easy to calculate.

arrgh. where's the edit button?

"neither toss is blue" does not mean that. it means "both tosses not blue." " Not both blue" means first could be blue, and second could be blue, or both could be not blue.

Catherine Johnson said...

Nope- turns out I not only have heard of Arlington Algebra, I've downloaded all the chapters.

I wonder what else I've got stashed away on my hard drive.

Lsquared said...

If you're still thinking about this problem, you might find this page useful. It's not an explanation, per se, but there's lots of examples if you just follow the links, and it's intended for students who are almost certainly less mathematically sophisticated than you are.

http://www.langfordmath.com/126notes/Probability/ProbPrac3-07.html

Catherine Johnson said...

I printed out all of the relevant Arlington Algebra sheets, which look tremendously helpful.

Unfortunately, there is no answer key.

So...I'm going to be keeping you guys busy.

heh