13 Calculating Expected Offspring

Problem

For a random variable XX taking integer values between 1 and nn, the expected value of XX is E(X)=∑nk=1k×Pr(X=k)E(X)=∑k=1nk×Pr(X=k). The expected value offers us a way of taking the long-term average of a random variable over a large number of trials.

As a motivating example, let XX be the number on a six-sided die. Over a large number of rolls, we should expect to obtain an average of 3.5 on the die (even though it's not possible to roll a 3.5). The formula for expected value confirms that E(X)=∑6k=1k×Pr(X=k)=3.5E(X)=∑k=16k×Pr(X=k)=3.5.

More generally, a random variable for which every one of a number of equally spaced outcomes has the same probability is called a uniform random variable (in the die example, this "equal spacing" is equal to 1). We can generalize our die example to find that if XX is a uniform random variable with minimum possible value aa and maximum possible value bb, then E(X)=a+b2E(X)=a+b2. You may also wish to verify that for the dice example, if YY is the random variable associated with the outcome of a second die roll, then E(X+Y)=7E(X+Y)=7.

Given: Six nonnegative integers, each of which does not exceed 20,000. The integers correspond to the number of couples in a population possessing each genotype pairing for a given factor. In order, the six given integers represent the number of couples having the following genotypes:

1. AA-AA
2. AA-Aa
3. AA-aa
4. Aa-Aa
5. Aa-aa
6. aa-aa

Return: The expected number of offspring displaying the dominant phenotype in the next generation, under the assumption that every couple has exactly two offspring.

Sample Dataset

1 0 0 1 0 1

Sample Output

3.5

# coding='utf-8'
# method1
def fun(a, b, c, d, e, f):
    x1 = 1 * a
    x2 = 1 * b
    x3 = 1 * c
    x4 = 0.75 * d
    x5 = 0.5 * e
    x6 = 0 * f

    return sum([x1, x2, x3, x4, x5, x6]) * 2


print fun(16634, 19016, 18660, 17721, 19835, 16233)


# method2

input = '16298 16360 18376 16233 18250 19449'
nums = [int(i) for i in input.split(' ')]
es = [0.75*nums[3],0.5*nums[4]]
for i in xrange(3):
    es.append(nums[i])
print sum(es)*2