SOLUTION: Construct a valid probability distribution for the number of boys a couple could have if they were to have 4 children.

There are 16 ways their 4 children could turn out, all assumed equally likely:

    ways   number of boys
 1. BBBB         4
 2. BBBG         3
 3. BBGB         3
 4. BBGG         2
 5. BGBB         3
 6. BGBG         2
 7. BGGB         2
 8. BGGG         1
 9. GBBB         3
10. GBBG         2
11. GBGB         2
12. GBGG         1
13. GGBB         2
14. GGBG         1
15. GGGB         1
16. GGGG         0

So there are 5 numbers of boys, that X can take on, 0 through 4.

1 way  to have 0 boys out of 16 ways to have 4 children.
4 ways to have 1 boy out of 16 ways to have 4 children.
6 ways to have 2 boys out of 16 ways to have 4 children.
4 ways to have 3 boys out of 16 ways to have 4 children.
1 way  to have 4 boys out of 16 ways to have 4 children.

So the answer is:

    X   p(X)
 1. 0   1/16
 2. 1   4/16 reduces to 1/4
 3. 2   6/16 reduces to 3/8
 4. 3   4/16 reduces to 1/4
 5. 4   1/16

Notice that the sum of all the probabilities is 16/16  = 1.
None are negative.
None are greater than 1.

Edwin