Algorithmic / Probabilistic Exercise

I am trying to solve an exercise from the rosalind project, but apparently make a mistake. The full text is available here , but my shorter abstract interpretation and attempt is as follows. Please help me find what I'm doing wrong:

We have a group of 3 elements: AA, AA, AA. We start at 1 in AAand do kiterations of generating new elements. At each iteration, each element in the group:

  • AAcan produce: AA(25%), AA(50%), AA(25%)
  • AAcan produce: AA(50%), AA(50%)
  • AAcan produce: AA(50%), AA(50%)

As a result of the iteration, we calculate the expected number of elements for each group, assuming that we generate 2 new elements from each in the previous iteration. Thus, we get:

  • 0th inertia: AA: 0 AA: 1 AA: 0
  • 1st inertia: AA.5, AA1, AA.5
  • Iter 2nd: AA: 1 AA: 2 AA: 1
  • etc .. - the proportions remain at 1: 2: 1 between the groups.

The sum of the expected values ​​/ population at each iteration is equal 2^iteration, and the probability that the element in the group is AAalways equal to 50%.

, , : N , AA, , . ( : N AaBb, AaBb, AaBb,.... )

, , AA, 50%, ( 2^iteration) scipy (k = 2, N = 1) AA:

In [75]: bin = scipy.stats.binom(4, .5)
In [76]: sum(b.pmf(x) for x in range(1, 4+1))
Out[76]: 0.93750000000000022

, , AaBb:

In [77]: sum(b.pmf(x) for x in range(1, 4+1))**2
Out[77]: 0.87890625000000044

: 0.684

? ( , , , , , ).

+5
1

, , , .

:

Aa-- --Bb. , , AaBb, Aa-- --Bb .

, , : aaBb, AABb, Aabb, AaBB Aa--, --Bb, AaBb.

+1

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