Question

Explain why the following simulation fails to model the situation properly. Use a random integer from 0 to 3 to represent the number of girls in a family of 3 children. The answer is : The outcomes are not equally likely. The simulation assumes they are.

Correct Reasoning: Having 0 through 3 girls in a family of 3 children are not equally likely. For example, having a total of 3 girls is less likely than having 2 girls out of 3 children. The simulation assumes they are equally likely.

My question is why aren't the outcomes equally likely?

Answer 1

for each child, it's approximately 50/50 whether it will be male or female. think of a coin labeled male on one side and female on the other. having one kid, you flip the coin once.
having two kids, you flip the coin twice. here are the possible outcomes:

male / male
male / female
female / male
female / female

4 different outcomes:
1 outcome gives you 2 girls
1 outcome gives you 2 boys
2 outcomes give you a boy and a girl

if we extend this to 3 kids:

male/male / male
male/male / female
male/female / male
male/female / female
female/male / male
female/male / female
female/female / male
female/female / female

8 outcomes:
1 outcome gives you 3 girls
1 outcome gives you 3 boys
3 outcomes give you 2 boys and a girl
3 outcomes give you 2 girls and a boy

Answer 2

Look at the sample space for 3 children. Let b = boy, g = girl.

These are the possibilities (in order of age, youngest to oldest):

bbb
bbg
bgb
bgg
gbb
gbg
ggb
ggg

Possibilities
0 girls = 1
1 girl = 3
2 girls = 2
3 girls = 1

Probabilities
P(0 girls) = 1/8
P(1 girl) = 3/8
P(2 girls) = 3/8
P(3 girls) = 1/8

So now you see that the probability of 3 girls is less likely than the probability of 2 girls. In fact, having 2 girls is three times as likely than having 3 girls! Simply because there are more ways for this to happen.

Therefore, the probability of 0 to 3 girls is NOT equally likely.

You might wonder why age is important. Say we did not include it as a factor, then our sample space would look like this if age didn't matter:

bbb
bbg
bgg
ggg

In this case you might think that having 0, 1, 2 or 3 girls would be equally likely. It's like counting the number of times you get 2 tails out of three. You can't just count the times this happens in the 2nd and 3rd flip, you also need to count the number of times it happens on the first. Otherwise, your sample space for coins would look like this:

hhh
hht
htt
ttt

But you know in fact there are other ways to get 2 tails:

tth
tht
htt

So taking into account age exposes all the possibilities.

Make sense?