brainwagon

The Amateur Extra test is 50 questions, multiple choice, with 4 answers per question. A passing grade is 35 or more. A few minutes of programming this morning, even before I had any coffee yielded that the exact probability of passing was:

      4677523340461106447
------------------------------
158456325028528675187087900672

or about 1 in 33.9 billion.

This wasn’t that interesting of a question, but to solve it, I hacked up a quick but limited implementation of rational arithmetic in Python. I was wondering if there was a better way to implement this in Python so overloading would “just work”. I didn’t know how, and the problem was simple enough, so I didn’t try. Here’s my solution.

#!/usr/bin/env python

def gcd(a, b):
    if (a < b):
        a, b = b, a
    while b != 0:
        a, b = b, a%b
    return a 

class Rational:
    def __init__(self, a, b):
        self.a = a 
        self.b = b 
    def __str__(self):
        return "[%d / %d]" % (self.a, self.b)
    def pow(self, p):
        return Rational(pow(self.a, p), pow(self.b, p))
    def mult(self, r):
        tmpa = self.a * r.a ;
        tmpb = self.b * r.b ;
        d = gcd(tmpa, tmpb)
        return Rational(tmpa//d, tmpb//d)
    def imult(self, i):
        tmpa = self.a * i ;
        tmpb = self.b ;
        d = gcd(tmpa, tmpb)
        return Rational(tmpa//d, tmpb//d)
    def add(self, r):
        tmpa = self.a * r.b + r.a * self.b
        tmpb = self.b * r.b 
        d = gcd(tmpa, tmpb)
        return Rational(tmpa//d, tmpb//d)

p = Rational(1, 4)
q = Rational(3, 4)

def fact(n):
        c = 1 
        while n > 1:
                c = c * n 
                n = n - 1 
        return c 

def comb(a, b):
        return fact(a)/(fact(b)*fact(a-b))

total = Rational(0, 1)

for t in range(35, 51):
        x = p.pow(t).mult(q.pow(50-t)).imult(comb(50, t))
        total = total.add(x)

print "Exact probability is",  total
print "Only about 1 in", total.b // total.a, "will pass"

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