Question 1026994
 
Question:
If a family is planning to have 6 Children, what is the probability of having:
 1) 5 boys, 1 girl
 2) 3 girls, 3 boys
 
Solution:
Assume equal probabilities of having girls and boys. 
This problem can be solved by drawing a binary probability tree with 64 ultimate branches, and the required answer by counting.
Alternative to the tree, it could be a table 6 rows (usually columns, but here I put them in rows to make typing easier).
BGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBG
BBGGBBGGBBGGBBGGBBGGBBGGBBGGBBGGBBGGBBGGBBGGBBGGBBGGBBGGBBGGBBGG
BBBBGGGGBBBBGGGGBBBBGGGGBBBBGGGGBBBBGGGGBBBBGGGGBBBBGGGGBBBBGGGG
BBBBBBBBGGGGGGGGBBBBBBBBGGGGGGGGBBBBBBBBGGGGGGGGBBBBBBBBGGGGGGGG
BBBBBBBBBBBBBBBBGGGGGGGGGGGGGGGGBBBBBBBBBBBBBBBBGGGGGGGGGGGGGGGG
BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG
From the above table, seek out the cases where there are 5 boys and 1 girl, or 3 girls and 3 boys.  (Each column represents one case, 64 columns in all).
 
If you find this method to straining to your eyes, you can always try the binomial probability distribution, which says that
P(x,n,p)={{{C(n,x)*p^x*(1-p)^(n-x)}}}
where
x is the number of boys
n is the total number of children (6)
p is the probability of having a boy (0.5)
and
C(n,x) is the binomial coefficient of taking x objects out of n, and equals
n!/(x!(n-x)!)
 
Hence
(1) P(5 boys, 6,0.5) = {{{C(6,5)*(0.5^5)*(0.5^1) = 6*(1/32)(1/2)=6/64=3/32}}}
(2) P(3 boys, 6,0.5) = {{{C(6,3)*(0.5^3)*(0.5^3) = 20(1/8)(1/8)=20/64=5/16}}}