Binomial vs. Hypergeometric Distributions - Extra Practice
Overview
Welcome! This module tests your intuition regarding two fundamental counting distributions: the Binomial distribution and the Hypergeometric distribution.
Practice Questions
Question 1: Binomial versus Hypergeometric
Make up a probability question involving draws from a basket contains 15 gold tokens and 35 silver tokens so that the answer involves the binomial formula and give the answer. Next change the probability question so that the answer involves the hypergeometric formula and give the answer.
**Correct Answer:
Binomial: Find the probability that 8 draws with replacement from the basket result in 3 gold tolkens. Answ: \(\binom{8}{3}(.3)^3(.7)^5\)
Hypergeometric: Find the probability that 8 draws without replacement from the basket result in 3 gold tolkens. Answ: \(\frac{\binom{15}{3}\binom{35}{5}}{\binom{50}{8}}\)
Question 2: Computing the Problem in R
Imagine you want to calculate the probability of getting at most 3 gold tokens when sampling 10 tokens without replacement from the basket (15 gold, 35 silver). What is the cleanest built-in syntax command to execute this in R?
Correct Answer: phyper(3, m = 15, n = 35, k = 10)
To find the probability of a cumulative threshold (at most 3), we use a cumulative distribution function (CDF) starting with p, not a probability mass function (d). Because this is sampling without replacement, we use the hypergeometric function family (*hyper):
q = 3(the number of successes we want to calculate up to)m = 15(number of success tokens in the basket)n = 35(number of failure tokens in the basket)k = 10(total number of trials/draws)