13 Calculating Expected Offspring

Problem

For a random variable $X$ taking integer values between 1 and $n$ , the expected value of $X$ is $E (X) = \sum_{k = 1}^{n} k \times P r (X = k)$ . 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 $X$ 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 $E (X) = \sum_{k = 1}^{6} k \times P r (X = k) = 3.5$ .

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 $X$ is a uniform random variable with minimum possible value $a$ and maximum possible value $b$ , then $E (X) = \frac{a + b}{2}$ . You may also wish to verify that for the dice example, if $Y$ is the random variable associated with the outcome of a second die roll, then $E (X + Y) = 7$ .

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:

AA-AA
AA-Aa
AA-aa
Aa-Aa
Aa-aa
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