-- IEV.hs

  - Problem 
  - 
  - For a random variable X taking integer values between 1 and n , the expected value of X is E(X)=∑nk=1k×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 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)=∑6k=1k×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 X is a uniform random variable with minimum possible value a and maximum possible value b , then E(X)=a+b2 . 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 positive 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 
  -         
  -         
  - 
  - 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 

 - Problem
 -
 - For a random variable X taking integer values between 1 and n, the expected value of X is E(X)=∑nk=1k×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 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)=∑6k=1k×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 X is a uniform random variable with minimum possible value a and maximum possible value b, then E(X)=a+b2. 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 positive 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
-}
import System.Environment (getArgs)

main = do
    args <- getArgs
    let inp = map read args
    print $ expDominant inp

expDominant [dd,dm,dr,mm,mr,rr] =
    (dd + dm + dr + 0.75*mm + 0.5*mr) * 2

 IEV.hs

  - Problem 
  - 
  - For a random variable X taking integer values between 1 and n , the expected value of X is E(X)=∑nk=1k×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 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)=∑6k=1k×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 X is a uniform random variable with minimum possible value a and maximum possible value b , then E(X)=a+b2 . 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 positive 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 
  -         
  -         
  - 
  - 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 

 - Problem
 -
 - For a random variable X taking integer values between 1 and n, the expected value of X is E(X)=∑nk=1k×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 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)=∑6k=1k×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 X is a uniform random variable with minimum possible value a and maximum possible value b, then E(X)=a+b2. 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 positive 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
-}
import System.Environment (getArgs)

main = do
    args <- getArgs
    let inp = map read args
    print $ expDominant inp

expDominant [dd,dm,dr,mm,mr,rr] =
    (dd + dm + dr + 0.75*mm + 0.5*mr) * 2
