Binomial Distribution Part III

Example

For a family of size 8 we want to find the coefficient for families with seven ‘successes’ and one ‘failure’. Let Aa be a ‘success’ and aa be a ‘failure’.
Then n = 8, r = 7, n-r = 1.
Now:

n!	=	8!
r!(n-r)!		7!1!

8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 40320
7! = 7 x 6 x 5 x 4 x 3 x 2 x 1 = 5040
1! = 1

8!	=	40320	=8
7!1!		(5040)(1)

To find the probability of seven Aa and one aa individuals in family of size eight, ignoring the order of progeny:

n!	(x)r (y)n - r
r!(n - r)!

Let:
x = probability of ‘success’ for a single event;
y = probability of ‘failure’ for a single event.

In our example of Aa x aa 1/2 Aa: 1/2 aa, then x = 1/2 and y = 1/2.