Quick answer: No.
Long answer: There are more likely to be 2 or 3 girls than 0 or 5. This is because there are more possible combinations that sum to 2 or 3, than sum to 0 or 5.
Mathematically, this is given by a binomial distribution, but your question asks you to list all the possible outcomes. This is a bit tedious to do fully, but consider each child in order (e.g. youngest to oldest) and find all the combinations:
(G G G G G) is one
(B G G G G), (G B G G G ), (G G B G G), (G G G B G), (G G G G B) are 5 more
(B B G G G), (B G B G G), (B G G B G), (B G G G B), (G B B G G), (G B G B G), (G B G G B), (G G B B G), (G G B G B), (G G G B B) are 10 more
... and if you switch the Gs and Bs, you get all the above patterns as well, so there are 1 + 5 + 10 + 10 + 5 + 1 = 32 possible combinations.
Note that I grouped each according to how many girls were in each? So our table is:
The mean is the sum of all (xP(x)) - so 0 * 1/32 + 1 * 5/32 etc.
The standard deviation is the square root of the average square distance from the mean - so sqrt ( 1/32 * (0-2.5)^2 + 5/32 * (1-2.5)^2 .... etc.)
= 1.118, or 2/sqrt(5) to be exact.
Let me know if any step is unclear (there is of course the 20min time limit on these questions, so it's not possible to go into as much detail as the regular paid questions allow).
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