At long last, the solution.

As in the case with flipping two coins, the problem revolves around figuring out what options have been eliminated by the statement of the problem. To review, the statement is that I have two children, one of whom is a born born on a Tuesday. The question is what is the probability that my other child is a boy?

First, I'll do this formally, to show that formal mathematics works just fine to solve problems like this. That is, you don't have to stoop to writing out the various possibilities and then tabulating them, as we did in the coin problem.

There is a formula in probability theory that comes in handy here:

(1) P(X|Y) = P(X and Y) / P(Y)

(The notation P(X|Y) means: the probability that X occurs given that Y has occurred, or in short-hand: the probability of X given Y.)

From (1), by the way, you can get another useful formula:

(2) P(X and Y) = P(Y) P(X|Y)

And since P(X|Y) = P(X) if X and Y are independent, that last formula can be written P(X and Y) = P(X) P(Y) in the case of independence.

Finally, since the "and" operation is commutative (i.e., the order of its operands is irrelevant), we have:

(3) P(Y and X) = P(X) P(Y|X) = P(Y) P(X|Y)

In this case, we want to determine:

P(both my children are boys | one child is a boy born on a Tuesday)

By (1), that's equal to P(both my children are boys and one child is a boy born on a Tuesday) / P(one of my children is a boy born on a Tuesday).

The first probability is clearly equivalent to P(both my children are boys and one of them was born on Tuesday), which by (2) is P(both my children are boys) P(one of my children was born on Tuesday). The first of these is easy to calculate: 1/4. The second is trickier, but the easy way to see it is to imagine that neither was born on Tuesday. Since the likelihood for either not being born on Tuesday is 6/7, the chance that neither was born on Tuesday is just 6/7 squared, or 36/49. That means the chance that at least one was born on Tuesday is 1 - 36/49 = 13/49. So that first factor in our equation above is (1/4)(13/49) = 13/196.

Now let's look at the second factor, P(one of my children is a born born on Tuesday). We can apply similar logic by imagining that neither is. The chance that a single child is a boy born on Tuesday is clearly 1/2 (boy) times 1/7 (born on Tuesday) = 1/14. So the chance that a single child is NOT a boy born on Tuesday is 13/14. Hence the chance that neither of two children is a boy born on Tuesday is 13/14 squared, or 169/196. The chance that at least one child is a boy born on Tuesday is then 1 - 169/196 = 27/196.

Plugging these values back into our original formula, we get the final probability as (13/196) / (27/196) = 13/27. And that is the answer.

Another way to do it (a bit less formally, but still perfectly accurate) is more like the two-coins problem. Let's write B for boy, G for girl, T for Tuesday, and !T for not-Tuesday. Then all possible outcomes for one child can be written like this:

G

BT

B!T

These are not equally likely, of course. P(G) = 1/2, P(BT) = 1/14, and P(B!T) = 3/7. But these add up to 1, and are mutually exclusive, so they must cover every possibility.

So for two children, there are nine possible outcomes:

G,G

G,BT

G,B!T

BT,G

BT,BT

BT,B!T

B!T,G

B!T,BT

B!T,B!T

Again, though, these are not equally likely. But their probabilities are easy to compute since the outcomes are independent. Furthermore, some of these outcomes are not possible since we know that one child is a boy born on Tuesday. So the only outcomes remaining are:

G,BT => (1/2)(1/14) = 1/28

BT,G => (1/14)(1/2) = 1/28

BT,BT => (1/14)(1/14) = 1/196

BT,B!T => (1/14)(3/7) = 3/98

B!T,BT => (3/7)(1/14) = 3/98

The sum of these is 1/28 + 1/28 + 3/98 + 3/98 + 1/196 = 27/196. Two boys show up with probability 3/98 + 3/98 + 1/196 = 13/196. The ratio is 13/27, as before.

## Friday, July 23, 2010

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