Friday, April 30, 2010

The fun in teaching probability

My last class this semester will be on Monday. Whether you are a student or professor, the end of the semester is always something to look forward to, especially if a four month long summer free of classes awaits. This time, though, I won’t feel the same way – the sentiment is primarily because I've been teaching probability and statistics. Both these topics are so rich and full of startling revelations that I will, for the first time, miss preparing for lectures. The class itself, consisting of ninety undergraduates, was a lot of fun (even if grading was not); I used the personal response system which worked well (I ask students a question and have their answers displayed immediately, enabling instant feedback, like an audience poll in a game show).

The mathematician Steven Strogatz, whose short pieces in the New York Times have thrilled many readers, recently wrote about conditional probability, a notoriously twisted concept. I spent two weeks covering it. Using a common example, I tried to tell my students that “testing positive if you have the disease” is not the same as “having the disease if you test positive”. To some this is mere semantics or subterfuge -- and indeed, much of probability can seem like smoke and mirrors, like the Monty Hall Problem -- but trust me it is not. Strogatz provides another example:
Perhaps the most pulse-quickening topic of all is “conditional probability” — the probability that some event A happens, given (or “conditional” upon) the occurrence of some other event B. It’s a slippery concept, easily conflated with the probability of B given A. They’re not the same, but you have to concentrate to see why. For example, consider the following word problem.

Before going on vacation for a week, you ask your spacey friend to water your ailing plant. Without water, the plant has a 90 percent chance of dying. Even with proper watering, it has a 20 percent chance of dying. And the probability that your friend will forget to water it is 30 percent. (a) What’s the chance that your plant will survive the week? (b) If it’s dead when you return, what’s the chance that your friend forgot to water it? (c) If your friend forgot to water it, what’s the chance it’ll be dead when you return?
I am too tired to give a detailed answer; moreover, conditional probability is just one concept in probability. There are plenty others, and this semester I often felt like I've stumbled upon a treasure trove of delightful ideas. Even this week I was discovering some that I had glossed over as a student. Someday, I will write extended pieces on what I have learned and how they apply to common situations.

Meanwhile, for those of you who would like to be tested, here are a couple of questions I asked in my exams -- up to the challenge? The first one (credit Leonard Mlodinow's The Drunkard's Walk) is easier than the second; it does not need any prior knowledge of probability.
a. Two students were partying in another state the day before their final chemistry exam. They got back only after the exam was over. However, they made up an excuse. They lied to the professor that they had a flat tire while returning and asked if they could take a make-up test. The professor agreed, wrote out a test, and sent the two students to separate rooms to take it. The only question on the test, worth 100 points, was: “Which tire was it?”

What is the probability that both students will give the same answer?

b. Suppose you toss a coin once and roll a die 4 times (these are two independent sets of experiments). Success in a coin-toss is getting a heads, while success in a die roll is getting a 5 or a 6. What is the probability that the number of successes in the coin toss equals the number of successes in the 4 die rolls?


Anonymous said...

what are the answers?
0.25 and 0.33?

Hari said...

Your first answer is correct. For the second I got 0.2962. I calculated it this way: [P(getting 0 successes in the coin toss)* P(getting 0 success in the 4 die rolls)] + [P(getting 1 success in the coin toss)* P(getting exactly 1 success in the 4 die rolls)].

Anonymous said...

so 1/2*2/3*2/3*2/3*2/3 +

1/2*1/3*2/3*2/3 ?

i get 0.1481 (half of 0.2962)

Hari said...

There are four ways to get exactly one success in 4 die rolls -- 4 choose 1. In general, there are "n choose x" ways of getting x successes in n trials (basis of the binomial distribution). So the second term is 1/2*4*2/3*2/3*2/3*1/3.

Anonymous said...

thanks. makes sense.

for 4 die rolls with exactly one success the number of ways is

1/3*2/3*2/3*2/3 + 2/3*1/3*2/3*2/3 +
2/3*2/3*1/3*2/3 + 2/3*2/3*2/3*1/3

i considered only one way

saban_arslan21 said...

Divx online film izlemek harika. ucuz bitkisel zayıflama hapı burada. Anlık bölgesel zayıflama hapı resmi satış sitesi. Kilo vermek sağlıklı zayıflama hapı le daha kolay. cildinizi kozmetik ürünleri ile gençleştirin. Hızlı nöbetçi eczane arama. Haydi stop ter le hayata tutun. Düzleştirin çatlak kremi ile cildinizi. Gücünüze güç katın jinsei kolye ile. Kaliteli sağlık ürünleri resmi sitesi. Hızlı diyet hapı le kilo verme. Yaşamınızı elektronik market ürünleriyle kolaylaştırın.