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
genot(1/2 Aa + 1/2 aa)
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:
genot(1/2 Aa + 1/2 aa)
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
genot= (1/2)
= 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.
