Hi Jason:

My thinking went along this way:

For 17 pennies, 19 nickels, and 21 dimes, P(draw 2p, 2n, & 2d, order not important):

Let's see, to reach in and pull out ANY six coins out of the bag, there is a 100% probability.

Since there are nearly an EQUAL number of each of three different kinds of coins, maybe I had an nearly even or nearly 50% chance to end up with two of each?

But if I pulled them out one at a time, two pennies, then two nickels, and then two dimes, IN ORDER, then my chances would be (17/57)*(16/56)*(19/55)*(18/54)*(21/53)*(20/52) = 0.15% chance, which is a very tiny chance to do that, which didn't seem at all in the ball park of what I thought it should be.

So then I thought about 2 pennies, 1 nickel, and 1 dime, and thought about how to get one of each if I draw 3, which is a 50% chance, which was misleading me again.

But then I didn't really think too much about 2 pennies, 2 nickels, and 2 dimes, and one of each, which now that I "make a table" of possibilities, shows that it's a 40% chance to get one of each. So maybe if I show the students the chances to get them "in order", vs the table of possibilities for 2 of each, order doesn't matter, and the chance to get one of each, then they might see that the probability should be between 0.15% and 40% for at least somewhat of an estimated range?

P1P2N1

P1P2N2

P1P2D1

P1P2D2

P1N1N2

P1N1D1

P1N1D2

P1N2D1

P1N2D2

P1D1D2

P2N1N2

P2N1D1

P2N1D2

P2N2D1

P2N2D2

P2D1D2

N1N2D1

N1N2D2

N1D1D2

N2D1D2

8/20 ways to get one of each type out of 2 pennies, 2 nickels, and 2 dimes. Or C(2,1)*C(2,1)*C(2,1)/C(6,3) = .4

Then that might convince them of the true nature of the situation. Plus, if I do it that way, I think I can demonstrate the link between the C(n,r) formula and the table of values for the larger main problem.

Does that sound like a reasonable approach?

Sincerely,

Michael

> --- In probability@yahoogroups.com, Michael L Werth <mwerth@b...>

wrote:

> Thanks Pedro!

>

> I'm very happy that you could confirm what I've been trying. Now I

need to figure out a way to explain this to high school freshmen!

>

> Talk to you later.

>

> Sincerely,

>

> Michael

Perhaps you should investigate why you, yourself, thought the

probability should be close to 50%. If you made that mistake,

certainly some of them will, too. How will you explain to them where

they made their mistake and why they're wrong?

Maybe it's similar to how some people often believe in the false

statement: "If I flip a coin 100 times, the probability that I get 50

heads is high." Actually, as you know, the probability is very low.

Or maybe that's not it. But whatever it is, it may be worth thinking

about. Good luck.

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