Binomial Distribution Part I
Example
Once again we make a testcross Aa x aa ÷ 1/2 Aa + 1/2
aa. We have two classes of progeny with expected proportions
of 1/2. Instead of a family of size three, let us use
a family size = 8.
We can expand the binomial expression
(1/2
Aa + 1/2 aa)8
to get the expected proportions of each type of family.
Again we will combine (add) the separate probabilities
of families that differ only in the order of each type
of offspring.
In a family of size eight we expect the following distribution
of the number of Aa progeny:
#
of Aa
progeny |
8 |
7 |
6 |
5 |
4 |
3 |
2 |
1 |
0 |
| probability |
1/256 |
8/256 |
28/256 |
56/256 |
70/256 |
56/256 |
28/256 |
8/256 |
1/256 |
We can check this table using the binomial expansion:
(1/2
Aa + 1/2 aa)8
P (8 Aa progeny) = 1/2 x 1/2 x 1/2 x
1/2 x 1/2 x 1/2 x 1/2 x 1/2
= (1/2)8=
1/256
This situation is more complicated when we want the
probability of 7 Aa and 1 aa progeny in a family, with
order ignored.
