today's fun with probability

[eub]

1) You meet a man at the bowling alley [this is how it was posed, honest]. He has two children, bowling on lane 17. You see one girl there; the other child doesn't happen to be visible.

2) Man, bowling, two children. You ask if he has any daughters. He says yes.

In each case (any difference?), what's the probability that he has two daughters?

Assume male and female are mutually exclusive and 50-50 and independent between children. Make any further assumptions you need.

[edit: spoilers in comments, natch]

Comments

Re: do not taunt happy fun mathchick

[thatmathchick.livejournal.com]

For a moment I almost psyched myself out. I recently ended up discussing the problem:

Q: You roll 2 dice. What is the probability that you get at least one 6?

A: The answer is not 2/6- which would be derived from the notion that on each roll you have a 1/6 chance of rolling a 6. It is actually (2/6 - 1/36). If you simply add the probabilities, you double count the case where you roll 6 twice.

simpler example:

Q: You pick a 2 bit number. What is the probability that it has a 1 in it?

A: If you think of adding the probabilities, you would get 1/2 for the first digit and 1/2 for the second digit. Which would make it a 100% chance of at least one 1. HUH???

The possible choices are 10, 11, 00, 01. 3/4 possibilities have a 1 in them. Again, you take 1/2 + 1/2 - 1/4 = 3/4.

In general, for 2 independent events A and B, the problems can be stated:

I) Probability (A or B) = Probability (A) + Probability (B) - Probability (A and B)

II) Probability (A given B) = Probability (A).

The bowling, loonie - twoonie, and your 2 bit number examples are all problem II. The roll at least one 6 and my 2 bit number examples are problem I.


(Huh- apparently I'm also not sure how to spell the term for Canadian $2 coin.)

Re: do not taunt happy fun mathchick

[eub.livejournal.com]

I've lost track of what A and B bind to...

In my (4) scenario, I'd want to talk about the events "2^0 bit is high" and "2^1 bit is high" for x uniform from {1, 2, 3}. Call those C and D. C and D are not independent, right? P(C) = P(D) = 2/3; P(C and D) = 1/3.

A and B, to be independent, would be those bit highnesses for x from {0, 1, 2, 3}, maybe? I don't remember exactly how this machinery works -- do we then talk about the independence (the lack thereof) of "A and not x=0" and "B and not x=0", or what?