Binomial vs. Hypergeometric Distributions - Extra Practice

Question 1: The Core Distinguishing Assumption

Suppose a basket contains 15 gold tokens and 35 silver tokens. You are instructed to select exactly 8 tokens from the basket, one by one. Under which of the following scenarios is the total count of selected gold tokens modeled perfectly by a Binomial distribution?

You draw 8 tokens simultaneously in one quick handful.
You draw a token, record its color, put it back into the basket, shake the basket thoroughly, and repeat this process until you have 8 observations.
You draw a token, give it to a classmate sitting next to you, and repeat this process until you have 8 observations.
None of the above setups can be modeled by a Binomial distribution because the overall population size is small (50 total tokens).

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Correct Answer: Option 2 (Sampling with replacement)

For a random variable to follow a Binomial distribution, it must satisfy the BINS criteria:

Binary: Each trial results in a success (gold) or failure (silver).
Independent: The outcome of one trial must not affect the next.
Number of trials is fixed (n = 8).
Success probability remains constant ((p = = 0.30)).

Placing the token back into the basket (sampling with replacement) ensures that the probability of drawing a gold token remains exactly 0.30 for every single draw. If you keep the tokens out (without replacement), the trials become dependent, which shifts the problem into a Hypergeometric framework.

Question 2: Probability Algebra Formulation

Let’s change the scenario. You now draw 5 tokens from the same basket (15 gold, 35 silver) without replacement. Which mathematical expression evaluates to the exact probability of drawing exactly 2 gold tokens?

\(\binom{5}{2} (0.30)^2 (0.70)^3\)
\(\frac{\binom{15}{2} + \binom{35}{3}}{\binom{50}{5}}\)
\(\frac{\binom{15}{2} \binom{35}{3}}{\binom{50}{5}}\)
\(\binom{15}{2} \left(\frac{15}{50}\right)^2 \left(\frac{35}{50}\right)^3\)

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Correct Answer: Option 3 (Hypergeometric mass formula)

Because we are sampling without replacement, we must track combinations using the Hypergeometric distribution PMF:

\[\frac{\binom{\text{Successes Available}}{\text{Successes Wanted}} \times \binom{\text{Failures Available}}{\text{Failures Wanted}}}{\binom{\text{Total Population}}{\text{Total Sample Size}}}\]

Mapping our variables directly:

Successes available: 15 gold tokens. We want to choose exactly (2 ).
Failures available: 35 silver tokens. Since total sample size is 5, we must choose exactly 5 - 2 = 3 silver tokens \(\implies \binom{35}{3}\).
Total sample pool: 50 tokens. We draw a total of \(5 \implies \binom{50}{5}\).

Multiplying the success/failure paths over the total sample space gives us: \(\frac{\binom{15}{2}\binom{35}{3}}{\binom{50}{5}}\).

Question 3: 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?

pbinom(3, size = 10, prob = 0.30)
dhyper(3, m = 15, n = 35, k = 10)
phyper(3, m = 15, n = 35, k = 10)

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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)

Binomial vs. Hypergeometric Distributions - Extra Practice

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