B. Probability Distributions · Discrete Probability Distributions

Comparing Discrete Distributions: Binomial and Hypergeometric

B1.6 B1.4 B1.5About 70 minutes

Learning goals

I can compare a binomial and a hypergeometric distribution for the same situation, with and without replacement.
I can describe how a binomial distribution changes as the number of trials increases.
I can explain when a hypergeometric distribution is close to a binomial one.
I can use technology (a spreadsheet) to build and graph distributions for comparison.

Words to know

with replacement
Each item is put back before the next draw, so the trials are independent and the probability stays the same (binomial).

without replacement
Items are not put back, so the trials are dependent (hypergeometric).

sampling fraction
n/N, the fraction of the population that is sampled. When it is small (a common guideline is under about 5%), sampling without replacement behaves almost like sampling with replacement.

Same question, two models

You have now met three discrete distributions: uniform, binomial and hypergeometric. A data analyst is often asked: which model fits, and how much does the choice matter? In this lesson you put distributions side by side and compare their shape, centre and spread.

We start with an example from the curriculum: drawing face cards (jacks, queens and kings) from a standard deck. There are 12 face cards in 52.

Worked example

Example 1: Face cards with and without replacement

Draw 3 cards from a standard deck. Let X be the number of face cards. Find the distribution (a) if each card is replaced and the deck reshuffled before the next draw, (b) if the cards are not replaced.

(a) With replacement: independent trials, p = 12/52 = 3/13, so X ~ B(3, 3/13).
P(0) = (10/13)³ = 1000/2197 ≈ 0.455; P(1) = 3(3/13)(10/13)² = 900/2197 ≈ 0.410; P(2) = 3(3/13)²(10/13) = 270/2197 ≈ 0.123; P(3) = (3/13)³ = 27/2197 ≈ 0.012.
(b) Without replacement: hypergeometric with N = 52, r = 12, n = 3. C(52, 3) = 22 100.
P(0) = C(40, 3)/22 100 = 9880/22 100 ≈ 0.447; P(1) = 12 × 780/22 100 = 9360/22 100 ≈ 0.424; P(2) = 66 × 40/22 100 = 2640/22 100 ≈ 0.119; P(3) = 220/22 100 ≈ 0.010.
Both have mean 3 × 12/52 ≈ 0.692.

The two distributions are very close. Without replacement, the extreme values 0 and 3 are slightly less likely and the middle value 1 is slightly more likely.

Face cards in 3 draws. The pairs of bars are nearly equal because 3 cards is a small part of a 52-card deck. Without replacement, the distribution is a little more concentrated around the mean.

Why without replacement is less spread out

With replacement, you could draw the same face card three times. Without replacement, every face card you draw makes the next one a little less likely, and every non-face card makes the next face card a little more likely. This pulls results toward the middle, so the hypergeometric distribution has the same mean but a smaller spread.

The difference is big when the sample is a large part of the population, and tiny when it is a small part.

Sampling 5 from 10 without replacement is very different from with replacement.
Sampling 5 from 500 is almost the same as with replacement.

A common guideline: if the sample is less than about 5% of the population (n/N < 0.05), the binomial distribution with p = r/N is a good approximation to the hypergeometric. Polling companies rely on this when they sample about a thousand people from millions.

x (successes in a sample of 5)	0	1	2	3	4	5
Hypergeometric, N = 10, r = 4	0.0238	0.2381	0.4762	0.2381	0.0238	0
Hypergeometric, N = 50, r = 20	0.0673	0.2587	0.3641	0.2341	0.0686	0.0073
Hypergeometric, N = 500, r = 200	0.0767	0.2592	0.3473	0.2308	0.0760	0.0099
Binomial, B(5, 0.4)	0.0778	0.2592	0.3456	0.2304	0.0768	0.0102

In every row, 40% of the population are successes and the sample has 5 items. As the population N grows, the hypergeometric row approaches the binomial row.

Worked example

Example 2: Reading the comparison table

Use the table. (a) For N = 10, why is P(X = 5) = 0? (b) For which N is the hypergeometric distribution within 0.002 of the binomial for every x?

(a) With N = 10 and r = 4, there are only 4 successes, so you cannot get 5 in the sample.
(b) For N = 500 the largest difference is at x = 2: 0.3473 − 0.3456 = 0.0017. For N = 50 the difference at x = 2 is 0.3641 − 0.3456 = 0.0185, too big.

(a) Only 4 successes exist. (b) N = 500, where the sample is only 1% of the population.

Binomial distributions as n grows

Keep p fixed and increase the number of trials n. Three things happen:

The centre np moves to the right.
The spread σ = √(npq) grows, but more slowly than n (it grows like √n).
The shape becomes more symmetric and bell-shaped, even when p is not 0.5.

The third fact is the reason the normal distribution can approximate binomial probabilities, a key idea in Continuous Distributions and the Normal Distribution.

Binomial distributions with p = 0.3 and n = 5, 10, 20 and 40, each drawn across the same width. The small-n histogram is clearly skewed right; by n = 40 it looks like a symmetric bell.

Worked example

Example 3: Using a spreadsheet to compare

Describe how to compare B(10, 0.3) and B(40, 0.3) in a spreadsheet, and state their means and standard deviations.

Column A: 0 to 40. Column B: =BINOM.DIST(A1, 10, 0.3, FALSE) (for A1 > 10 this returns an error or 0; leave those blank). Column C: =BINOM.DIST(A1, 40, 0.3, FALSE).
Insert a column chart for each column.
B(10, 0.3): mean 3, σ = √2.1 ≈ 1.45. B(40, 0.3): mean 12, σ = √8.4 ≈ 2.90.
n grew 4 times; the mean grew 4 times, but σ only doubled (√4 = 2).

Means 3 and 12; σ ≈ 1.45 and 2.90. The larger-n graph is wider, more symmetric and has a lower peak.

Worked example

Example 4: Choosing and comparing models

A school has 1200 students; 30% walk to school. A survey asks 10 randomly chosen students (without replacement). Should you use the binomial or the hypergeometric model for the number who walk? Does it matter much here?

Strictly, choosing 10 different students is without replacement, so the exact model is hypergeometric with N = 1200, r = 360, n = 10.
The sampling fraction is 10/1200 ≈ 0.8%, well under 5%.
So B(10, 0.3) will give almost the same probabilities and is easier to compute.
Both models have mean 10 × 0.3 = 3 students.

Hypergeometric is exact, but the binomial B(10, 0.3) is an excellent approximation because the sample is less than 1% of the school.

Just for fun

Pollsters can survey about a thousand people to learn about a country of millions because, once the population is huge, sampling without replacement behaves almost exactly like sampling with replacement. The population could double overnight and the poll would barely notice.

Hands-on activity

With or without? A card experiment

You need: a standard deck of cards, pencil, paper and (optional) a spreadsheet.

Without replacement: shuffle, draw 3 cards, record the number of face cards. Return them and reshuffle. Do this 30 times.
With replacement: draw one card, record whether it is a face card, put it back and shuffle; do this 3 times for one trial. Do 30 trials.
Make a table of relative frequencies for X = 0, 1, 2, 3 for each method, and compare them with the theoretical values in Example 1.
Now make the population small: use only the 12 face cards and 12 number cards (24 cards). Repeat steps 1 and 2 with 6 cards drawn each time. Predict first: will the difference between "with" and "without" be bigger or smaller than with the full deck?

What to look for: with the full deck the two methods give very similar results (30 trials is too few to see the small difference reliably). With the 24-card deck and samples of 6 (25% of the population), without replacement gives noticeably fewer results at 0 and 6 and more near 3.

Your turn

Two cards are drawn from a standard deck. Find P(both aces) with replacement and without replacement.

Do the binomial B(n, r/N) and the hypergeometric distribution have the same mean? Which has the larger spread?

A sample of 20 is taken from a population of 2000. Is the binomial a good approximation to the hypergeometric? Explain.

A sample of 8 is taken from a population of 40. Is the binomial a good approximation? Explain.

B(16, 0.5) and B(64, 0.5): find each mean and σ.

As n increases with p = 0.1 fixed, how does the shape of B(n, 0.1) change?

From the comparison table, what is P(X = 2) for the hypergeometric with N = 10 and for the binomial? Why is the hypergeometric value so much larger?

More practice: warm-up

Which model fits sampling with replacement?

Which model fits sampling without replacement?

How many face cards are in a standard deck?

For B(n, p), what happens to the mean when n doubles?

For B(n, p), what happens to σ when n is multiplied by 4?

More practice: on your own

A bag has 6 red and 4 blue marbles. Draw 3. Find P(X = 3 red) with and without replacement.

For the bag in the previous question, find P(X = 2) with and without replacement. Which is larger?

A population has N = 1000 items, 100 of them defective. A sample of 10 is taken without replacement. Estimate P(no defectives) with the binomial model.

Explain why the shape of B(n, p) gets more symmetric as n grows, even when p = 0.2.

A pollster samples 1000 Ontarians from about 16 million. Should the pollster worry that sampling is without replacement?

Challenge

The hypergeometric variance is n(r/N)(1 − r/N)(N − n)/(N − 1). Compute σ for the face-card example (N = 52, r = 12, n = 3) and compare with the binomial σ.

For N = 10, r = 4, n = 5, the table gives P(X = 2) = 0.4762. Verify with combinations.

Show that if you sample the entire population (n = N) without replacement, X always equals r. What does the formula (N − n)/(N − 1) say about the spread?

Check your understanding

Quiz

Compared with sampling with replacement, sampling without replacement (same n and same population) gives...

When is the binomial a good approximation to the hypergeometric?

Draw 3 cards with replacement. What is P(no face cards)?

As n increases with p fixed, a binomial histogram becomes...

B(25, 0.5) has σ = 2.5. What is σ for B(100, 0.5)?

A sample of 5 is drawn from 10 items (4 successes). Which statement is true?

For parents and teachers

Big idea (B1.6): comparing distributions side by side shows when the choice of model matters. With and without replacement give the same mean, but without replacement has less spread; the difference is tiny when the sample is a small part of a large population. Binomial distributions become more bell-shaped as the number of trials grows.

How to help: do the card experiment together, once with replacement and once without, and talk about why polls of about a thousand people can describe a whole country.

Common difficulties:

Thinking without-replacement changes the mean (it does not; it changes the spread).
Applying the binomial to small populations where the sample is a large fraction.
Expecting σ to double when n doubles (it grows by √2).