## Ted Baxter and the Binomial Distribution

There are many hard things about teaching, although I appreciate that since I’m in mathematics I have advantages over many other fields. For example, students come in with the assumption that there are certainly right and certainly wrong answers to questions. I’m generally spared the problem of convincing students that I have authority to rule some answers in or out. There’s actually a lot of discretion and judgement and opinion involved, but most of that comes in when one is doing research. In an introductory course, there are some techniques that have gotten so well-established and useful we could fairly well pretend there isn’t any judgement left.

But one hard part is probably common to all fields: how closely to guide a student working out something. This case comes from office hours, as I tried getting a student to work out a problem in binomial distributions. Binomial distributions come up in studying the case where there are many attempts at something; and each attempt has a certain, fixed, chance of succeeding; and you want to know the chance of there being exactly some particular number of successes out of all those tries. For example, imagine rolling four dice, and being interested in getting exactly two 6’s on the four dice.

To work it out, you need the number of attempts, and the number of successes you’re interested in, and the chance of each attempt at something succeeding, and the chance of each attempt failing. For the four-dice problem, each attempt is the rolling of one die; there are four attempts at rolling die; we’re interested in finding two successful rolls of 6; the chance of successfully getting a 6 on any roll is ^{1}/_{6}; and the chance of failure on any one roll is —

Well, this is the binomial part: each attempt either succeeds or fails. Since exactly one of those outcomes has to happen, the probability of succeeding plus the probability of failing has to equal *1* exactly. So the chance here of one die failing to come up ~~heads~~ a six is *1 – ^{1}/_{6}* or

^{5}/

_{6}. And we put all this together in a formula based on the number of attempts, the number of successes we’re interested in, the chance of success, and the chance of failure.

I was guiding a student through such a word problem, extracting from it the bits of information needed. The number of trials and number of successes was obvious, as was the probability of success, for this problem, 0.20. I also asked the probability of failure. There’s a couple reasons; one is that we need that written down later anyway, and I like to think writing out what it is and what it meas earlier makes it clearer what’s there and why it’s there. And another is that I like tossing out the simple arithmetic type problems for students, since they break up my monologues, and they’re usually straightforward enough the student can answer it, and I like to think that getting a “yes, that’s right” in the midst of a complicated problem encourages students to not give up.

The student, though, wasn’t getting the 1 minus 0.20. It may have been nerves. It may have been not getting the relative sizes right. The student guessed 19, and 0.19, and wasn’t showing signs of getting nearer, even when I pointed out that the probability of success and the probability of failure had to add up to 1.

Part of me was tempted to just say the answer and write off this as an inconsequential side problem. Part of me answered that the student had to do exactly this sort of calculation so should practice. Meanwhile we were getting lost in something I just didn’t care about — even during an exam, I allow the students to use pocket calculators, so the calculator could figure out 1 minus 0.20 — and losing all forward momentum.

Then I thought of a way to reframe the problem: “You buy something for 20 cents. You give the cashier a dollar. How much change do you get?” The student got 80 cents immediately, saving me from the embarrassment of trying to think of what you could buy these days that was 20 cents. (I think a couple loose candies from Wawa would fit the, er, bill.)

Sometimes whether a calculation is possible is wholly dependent on how the problem is phrased.

Later, I realized that I had just re-created a throwaway gag from The Mary Tyler Moore Show.

## Chiaroscuro 7:44 am

onSaturday, 10 December, 2011 Permalink |So the chance here of one die failing to come up heads is 1 – 1/6 or 5/6I think the chance of most dice failing to come up heads is quite a bit higher.

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## nebusresearch 4:29 pm

onSaturday, 10 December, 2011 Permalink |Gr. Yeah.

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## Joe Fix It 2:18 am

onMonday, 12 December, 2011 Permalink |Ted would have liked it

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## nebusresearch 11:09 am

onMonday, 12 December, 2011 Permalink |Liking is one thing. Calculating right another. When they can get together at once, I’m reasonably content.

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