[vubuntu]

For me both problem 2 and problem 3 should also be 1/2 .

The variances are due to how people interpret the outcome space .

It is right that probability = favaroble out comes / total possible outcomes.

In problem 2, in my opinion the possible outcomes are not how it was suggested in the post but as below. When a family has 2 kids , the below are the only possible outcomes

1) Potential Outcome 1: Both are boys 2) Potential Outcome 2: One boy and one girl 3) Potential outcome 3: Both are girls

The 3rd one is not a legal potential outcome in our particular constraint of problem 2, since problem #2 statement already states 'at least one is a boy'

So Total possible legal outcomes = 2 Favorable out comes for our event (both boys) = 1

So probability for Problem 2 = 1/2

Similarly for problem #3, I think the post unnecessarily complicates the calculation of problem space. The fact that 'Tuesday' is mentioned is irrelevant in my opinon, if you state the problem #3 in a different way that is more clearly understood.

There are 14 baskets labelled as follows "Sunday Boy", "Sunday Girl", "Monday Boy", "Monday Girl",....."Saturday Boy", "Saturday Girl". A stork came and dropped 2 babies. One baby was dropped in "Tuesday Boy" basket. What is the probability that both are boys?

Now the total outcomes and favarable outcomes are :

Total possible outcomes = Number of ways second baby could have been dropped = 14 possible baskets = 14

Favorable outcomes = second baby dropped in 'boy' basket = 7 possible baskets = 7

Probability that both are boys = 7 / 14 = 1/2

[bo1024]

If you understand problem 2, then you will get problem 3 as well, so let's focus on 2.

> It is right that probability = favaroble out comes / total possible outcomes.

No, not actually: It is only right if all outcomes are equally likely! (There's an old joke about the guy who has a 50% chance of winning the lottery, since either he will win it or he won't.)

In particular, you make that mistake here:

> 1) Potential Outcome 1: Both are boys 2) Potential Outcome 2: One boy and one girl 3) Potential outcome 3: Both are girls

These three outcomes are not all equally likely. Outcome 1 has probability 1/4, outcome 2 has probability 1/2, and outcome 3 has probability 1/4. (This is if you assume that each child has a half chance each of being a boy/girl.)

Norvig gets rid of this problem by listing out all four possible outcomes, which are all equally likely.

1. First child boy, Second child boy

2. First child boy, Second child girl

3. First child girl, Second child boy

4. First child girl, Second child girl

[mamon]

Again, both Norvig and you are messing with ordering of events. If you list 4 events like this, assuming that ordering of boys and girls matters, then you should stick with that. So, if ordering mattered when children were born it should also matter when you are doing "checks". So you shouldn't formulate problem as "one of the children is boy", you should formulate problem as "first child is boy" (with ordering in place), which eliminates possibility 3. Otherwise you are "solving" problem by listing sample space of completely different problem.

[bo1024]

The event "at least one child is a boy" is well-defined on the 4-state sample space. It is the set of events (first boy/second boy, first boy/second girl, first girl / second boy). It has probability 3/4.

[norvig]

When you say "1) Potential Outcome 1: Both are boys 2) Potential Outcome 2: One boy and one girl 3) Potential outcome 3: Both are girls" you are right that those are three potential outcomes, but they are not equiprobable outcomes. One boy and one girl is twice as probable each of the others.