Monday, July 28, 2008

Taking Two or More At a Time- What is the Probability You Don't Like These!

Only a few days left for your July MathAnagram. Either it's harder than I thought or no one is paying attention!

Look here for details.

Probability questions will forever addle the minds of students and adults alike. If all problems could involve selecting one object randomly, life would be good. Unfortunately, selecting 'more than one' is becoming common on standardized tests these days. Taking two or more objects immediately ratchets up the difficulty:

Does order count?
Multiple solution paths
Making a list
Combinations? Permutations? Multiplication Principle? Using "rules" of probability?

In what course do students receive sufficient instruction in this important area? Algebra 1? Algebra 2? Precalculus? A probability/statistics/discrete math class? AP Stat? IMO, the lack of standardization in secondary curriculum can lead to some topics getting short shrift.

I've come to the conclusion that middle schoolers should devote more time to some of this, since 4th graders are generally expected (in most states' standards) to solve the "select one object" type. What do you think? Since most readers enjoy the math challenges and not this kind of curriculum discussion, here are our offerings for today...

A bag contains eight coins: two each of pennies, nickels, dimes and quarters.

Question 1: If two coins are randomly selected, show that the probability that the two coins will total at least 20 cents in value is 1/2.

Question 2: If 4 coins are randomly selected, show that the probability of getting exactly one of each kind of coin is 8/35. (At least two methods please!)
Note: This result implies that the chances of getting at least one matching pair of coins among the 4 coins is greater than 75%!!

Question 3: Invent your own!

There are many problems of this type on this blog (see probability, combinatorial math in the index). Further, there are many other excellent blogs and web sites that address these topics and provide wonderful challenges and explanations. Two of the best are Jonathan's over at jd2718 and Isabel over at God Plays Dice.


jd2718 said...

What happens when you throw an extra dime in?

Oh, and I get 384/1680.


Eric Jablow said...

Students can solve problem 1 in a few ways. However, they all depend on the student realizing that the condition is a) one of the drawn coins is a quarter, or b) both drawn coins were dimes.


To find the probability that either drawn coin is a quarter, it's better to find the probability neither is, and that's C(6, 2)/C(8, 2) = 15/28. So, the probability that at least one coin is a quarter is 13/28.

The probability that both coins are dimes is 1/C(8,2) = 1/28. The total is 1/2.

Set theory (inclusion-exclusion):

Let A be the set of draws (taken in order) where the first coin is a quarter, and let B be the set of draws where the second coin is a quarter. Then, |A∪B| = |A| + |B| - |A∩B|. |A| = |B| = 2×7. |A∩B| = 2, and so |A∪B| = 26. The draws with two dimes give 2.
So, the result is (26 + 2) /(8×7) = 1/2.

Problem 2:

You've done this problem before, actually. By permutations, the probability is 8∙6∙4∙2/(8∙7∙6∙5).

jd2718 said...

Or by conditional probability, (8/8)(6/7)(4/6)(2/5).


Dave Marain said...

I like your analysis of the 20 cents problem. Most students would enumerate the possibilities (and probably miss a couple!). Yes, I agree, the 2nd question has appeared before!

I like the 3 dimes variation.
For the P(at least 20 cents), I still get 1/2 as follows: The original ratio was 14/28. The extra dime adds two more dime-dime combinations and 2 more dime-quarter combos. However, it adds 8 more combinations with all the coins. Thus the new ratio is (14+4)/(28+8) = 1/2. Interestingly, I would think that as we add additional dimes, the probability would start to increase since we're increasing the chances of a dime-dime or dime-quarter combo.

For the 2nd question with 3 dimes, I obtained:
P(one of each coin) = (2)(2)(3)(2)/9C4 = 24/126 = 4/21 which is less than the original 8/35. This result makes sense to me since we now have an imbalance of coins which decreases the chances of getting 'different' outcomes. BTW, I chose not to use the conditional probability argument even though it worked well in the original question. I would like you to verify this result (4/21) for me.

Asking students to predict, intuitively, whether the probabilities will increase or decrease seems like a good question.

As always, thanks to Eric and Jonathan for their insights.

Eric Jablow said...

That's easy to check, Dave. Suppose you have 2 pennies, 2 nickels, 2 quarters, and n dimes, for n+6 coins in total.. Let's try the inclusion/exclusion method.

Let A be the set of cases where the first number is a quarter, and B be the set of cases where the second draw is a quarter. Let C be the set of cases where both draws are dimes.

|A| = 2 (n+5) = |B|. |A∩B|= 2, so |A∪B| = 4n + 18.

The number of cases where both draws are dimes, |C|, is n (n-1).

|A∪B∪C| = n² + 3n + 18. The total number of draws is (n + 6)(n + 5) = n² + 11n + 30. Certainly, the quotient (n² + 3n + 18)/(n² + 11n + 30) tends to 1. Is the expression increasing for n>2? It's time to drag out another old mathematical tool, partial fractions.

Since the denominator is the product (n+6)(n+5), the quotient can be written as A + B/(n+6) + C/(n+5) for some A, B, and C. Taking the quotient gives that A = 1.

(n² + 3n + 18)/(n² + 11n + 30) = 1 + B/(n+6) + C/(n + 5).

(-8n-12)/(n² + 11n + 30) = B/(n+6) + C/(n+5).

Multiply by (n+6) (n+5):

(-8n-12) = B(n+5) + C(n+6).

Let n = -6. This gives 36 = -B, or B = -36.

Let n = -5. This gives 28 = C.

The quotient is p(n) = 1 -36/(n + 6) + 28/(n+5). Where is this increasing? Let's avoid calculus, and find Δp(n) = p(n+1) - p(n).

p(n+1) = 1 - 36/(n+7) - 28/(n+6).

Δp(n) = 36/[(n+6)(n+7)] - 28/[(n+5)(n+6)].

This is greater than 0 when

36/[(n+6)(n+7)] > 28/[(n+5)(n+6)]

36/(n+7) > 28/(n+5),

36(n+5) > 28 (n+7),

36n+180 > 28n + 196,

8n > 16,

n > 2.

So, the probability increases.

Dave Marain said...

Wonderful analysis, Eric (but of course!).

Using partial fractions and straight "difference" methods in lieu of calculus is highly instructive for students, although they would probably look at a table of values or the graph on the calculator first these days! (Not that it would verify anything).

I also did a "delta' calculation on your rational function from the beginning. It's messier algebra but it leads to an easy quadratic and n>2 as well.


Eric Jablow said...

No one teaches partial fractions any more? The situation is worse than I thought.

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Dave Marain said...

From what I've observed, partial fractions are taught in some Algebra 2 curricula (rarely), Precalculus classes (occasionally) and, of course, in AP Calculus (has been a required topic for some time). However, many students who have not taken calculus might well have missed this technique altogether. I will leave it to others currently teaching to comment on this further.

I feel it is an important tool. However, since many see it as simply needed for integration and such techniques are currently viewed as less important, it may fall out of favor. I see it as much more than that. It introduces students to the "method of undetermined coefficients" and has utility beyond integrating rational functions. For polynomial functions, it also reinforces the Fundamental Theorem of Algebra (factoring any polynomial into powers of linear and irreducible quadratics). I devoted considerable attention to it.

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Eric Jablow said...

The line

p(n+1) = 1 - 36/(n+7) - 28/(n+6).

should have read

p(n+1) = 1 - 36/(n+7) + 28/(n+6).

jd2718 said...

Dave, I'm late.

With 2 pennies, 2 nickels, 3 dimes, and 2 quarters, the probability that 4 randomly selected coins are one of each can be calculated as:






which are equivalent to 4/21

Eric, partial fractions I got 2nd term of senior year calculus, in order to handle some integrals. In my school we've just stuck them into pre-calc, which seems to me to be a good idea.