Question 1203658
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There are two possible outcomes on each birth, boy (B) or girl (G).  So for 4 children, the number of total possible outcomes is {{{2^4}}} = {{{16}}}

This is small enough to enumerate all of them.  After that, you
should be able to easily complete the problems (I'll do the first
part as a template).

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

(1) P(two boys and two girls) = 6/16 = 3/8 
To see this, notice the number of outcomes with 2B and 2G is 6 (lines 4,6,7,10,11, and 13).  The probability is the count of the outcomes of interest, divided by the total number of possible outcomes.total possible outcomes.

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As an Algebraic solution, you would use combinations.  The probability of 2 boys and 2 girls is equivalent to the probability of 2 boys out of 4 children total: 
    Using nCr = (n!)/((n-r)!r!)  ("n choose r")  
  P(2 boys and 2 girls) = P(2 boys out of 4 children) = 4C2 / 16 = 6/16 = 3/8
==

The other parts are solved similarly, just adjust how you select the outcomes of interest (or compute the combinations).