Why 10 successes and 10 failures? Large Counts rule

Q: Where does the number 10 actually come from?

It is a convention chosen for convenience, not a result anyone proved. Other textbooks use 5 or 15. Ten is round, and it holds the leftover skewness under about 0.32 while keeping the count at least 3.16 standard deviations clear of 0 and of n. A bar of 5 would give 0.45 and 2.24 instead. On the AP exam, use 10.

Q: Why is a large sample size not enough on its own?

Because a large n paired with a tiny p is still badly skewed. Take n = 200 and p = 0.02: that is 200 observations but only 4 expected successes, a standard deviation of sqrt(200(0.02)(0.98)) = 1.98, and a floor at 0 sitting just 2.02 standard deviations below the center. The condition counts outcomes, not observations, which is exactly why it catches this.

Q: Do I use p-hat or the null value in the check?

Use whatever the procedure assumes. A confidence interval assumes nothing, so use p-hat, which means counting the successes and failures you actually observed. A significance test assumes H₀: p = p₀, so use p₀. A two-proportion test assumes the proportions are equal, so use the pooled p-hat_c on all four counts.

Q: Is Large Counts the same as the 10% condition?

No, and mixing them up is common because both mention 10. Large Counts asks whether the sampling distribution is close enough to normal. The 10% condition, n ≤ 0.10N, asks whether sampling without replacement leaves the observations close enough to independent for the standard error formula to hold. You check both, separately.

By Jude Wallis · Published July 28, 2026

Because the sample proportion is a binomial count in disguise, and a binomial is noticeably skewed unless both outcomes are common. Ten expected successes and ten expected failures keep the count about three standard deviations clear of both 0 and the sample size, so a normal curve is safe to use.

AP Statistics: Unit 3 (topics 3.2 Sampling Distributions for Sample Proportions, 3.3 Constructing a Confidence Interval for a Population Proportion, 3.5 Setting Up a Test for a Population Proportion). The Large Counts condition first appears in Unit 3 topic 3.2 of the Fall 2026 AP Statistics course and is then required in every proportion interval (3.3) and proportion test (3.5) you build.

The short answer

The Large Counts condition says you may treat the sampling distribution of the sample proportion $\hat{p}$ as approximately normal only when the expected number of successes and the expected number of failures are both at least 10:

np \geq 10 \quad \text{and} \quad n(1 - p) \geq 10

The condition exists because $\hat{p}$ is not normal. It is a binomial count rescaled, and a binomial is symmetric only when $p = 0.5$ . As $p$ slides toward 0 or 1 the distribution piles up against a wall and grows a long tail on the other side. A normal curve laid over that shape gives wrong probabilities, wrong p-values, and intervals that miss more often than advertised.

Two details students skip. A large $n$ is not something you can check on its own, because how large it has to be depends on $p$ , which is why the condition counts outcomes instead of counting observations. And 10 is a convention, not a theorem: some textbooks use 5, some use 15. AP Statistics standardized on 10, so use 10.

The sample proportion is a binomial count in disguise

Take a random sample of size $n$ and let $X$ be the number of successes. If the trials are independent with a fixed success probability $p$ , then $X$ has a binomial distribution, and

\hat{p} = \frac{X}{n}

Dividing by $n$ relabels the horizontal axis. It does not change the shape. So every claim about the shape of the distribution of $\hat{p}$ is really a claim about the shape of a binomial.

That shape has a known skewness:

\text{skewness} = \frac{1 - 2p}{\sqrt{np(1 - p)}}

Read the two pieces. The numerator $1 - 2p$ is zero exactly when $p = 0.5$ , which is the only perfectly symmetric case, and its size grows toward 1 as $p$ heads for 0 or for 1. The denominator grows with $n$ , so a bigger sample pulls the skewness back down.

So there are two ways to be nearly symmetric: have $p$ near the middle, or have $n$ large enough for the $p$ you have. The Large Counts condition handles both at once because it is written in expected counts, and an expected count already multiplies $n$ by $p$ .

More about the shape itself sits in what a sampling distribution is and in the binomial distribution.

Why 10, and why both sides

Setting the bar at 10 buys two specific things.

Room from the walls. A count of successes cannot fall below 0 or rise above $n$ . The count has mean $np$ and standard deviation $\sqrt{np(1-p)}$ , so the distance from the center down to 0, measured in standard deviations, is

\frac{np}{\sqrt{np(1-p)}} = \sqrt{\frac{np}{1-p}} \geq \sqrt{np}

With $np \geq 10$ , the floor at 0 sits at least $\sqrt{10} \approx 3.16$ standard deviations below the center. Run the same algebra on failures and $n(1-p) \geq 10$ puts the ceiling at $n$ at least 3.16 standard deviations above the center. A normal curve keeps almost nothing past three standard deviations, so once you have that much headroom on both sides the model stops assigning real probability to impossible counts. This is the direct answer to "why both successes and failures": each count guards one wall, and one check cannot guard both.

Bounded leftover skew. Write $a = np$ and $b = n(1-p)$ for the two expected counts. Since $n = a + b$ , the skewness above can be rewritten as

\text{skewness} = \frac{b - a}{\sqrt{ab(a+b)}}

That expression is always smaller in size than $1/\sqrt{a}$ or $1/\sqrt{b}$ , whichever count is the smaller of the two. So once both counts clear 10, the skewness is under $1/\sqrt{10} \approx 0.32$ no matter how lopsided $p$ is. Drop the bar from 10 to 5 and that bound loosens to $1/\sqrt{5} \approx 0.45$ , while the headroom shrinks from 3.16 standard deviations to $\sqrt{5} \approx 2.24$ . That tradeoff is the whole disagreement between textbooks that use 5 and textbooks that use 10.

Because the requirement is $n \geq 10 / p$ and $n \geq 10 / (1-p)$ at the same time, rarer outcomes need much bigger samples:

True $p$	Smallest $n$ that passes
0.50	20
0.30	34
0.20	50
0.10	100
0.05	200
0.01	1000

A 1% rate needs roughly 1000 observations before a z procedure is defensible. That is the practical cost of the condition, and it is why studies of rare events need very large samples before these tools apply.

Which proportion goes in the check

This is the part that costs points on free response, and the detail most condition checklists leave out. Use the value that the procedure itself assumes about $p$ .

Situation	Check these	Why
Sampling distribution of $\hat{p}$ , with $p$ given	$np$ and $n(1-p)$	You were handed the true value, so use it.
Confidence interval for $p$	$n\hat{p}$ and $n(1-\hat{p})$	Nothing is assumed about $p$ , so the data is all you have.
Significance test for $p$	$np_0$ and $n(1-p_0)$	Everything in a test is computed as if $H_0: p = p_0$ were true, including the standard error.
Two-proportion z test	$n_1\hat{p}_c$ , $n_1(1-\hat{p}_c)$ , $n_2\hat{p}_c$ , $n_2(1-\hat{p}_c)$	$H_0$ says the two proportions are equal, so pool them.
Two-proportion confidence interval	Each group's own $n\hat{p}$ and $n(1-\hat{p})$	No equality is assumed, so no pooling.

The interval row has a shortcut worth knowing. Since $\hat{p}$ is the observed successes over $n$ , the quantity $n\hat{p}$ is literally the number of successes you counted, and $n(1-\hat{p})$ is the number of failures. Checking Large Counts for an interval is nothing more than reading your own data: did I see at least 10 of each? The arithmetic is only there so a grader can see it.

For the two-sample versions, one-proportion vs two-proportion z test walks through the pooled proportion in full.

What to do when the condition fails

First, say so. Write the failing count with its number, in context: " $n\hat{p} = 3$ , which is below 10, so the Large Counts condition is not met and a z-interval may not be valid." Naming the failed condition with its number is what a rubric is looking for. Computing silently is not, and it is the more common error, because a calculator will happily hand back an interval for any numbers you feed it.

Second, understand what actually goes wrong. Two symptoms show up:

The interval can run below 0 or above 1. A proportion cannot do either, and seeing it happen means the normal model is spending probability on values that cannot occur.
The capture rate degrades. A nominally 95% interval built from a badly skewed sampling distribution can capture the true $p$ noticeably less than 95% of the time, and the shortfall is worst for exactly the small- $p$ cases the condition is designed to catch.

Third, the only real fix inside the AP toolkit is more data. You need $n \geq 10 / p$ and $n \geq 10 / (1-p)$ , so plan the sample size around the rate you expect rather than discovering the problem afterward. The sample size calculator handles the margin-of-error side of that planning.

Do not invent a substitute procedure on the exam. Statisticians do have exact binomial methods for small counts, but they are not in the AP course and not on the formula sheet, so naming one is not what the question is asking for. State the failure, proceed only if the prompt tells you to, and flag that the result may not be trustworthy.

Mistakes to avoid

**Checking $n \geq 30$ .** That guideline belongs to the sample mean, not to proportions. A proportion problem never uses 30.
Checking only the successes. Both counts, every time. With $n = 500$ and $\hat{p} = 0.99$ you have 495 successes and only 5 failures, so the condition fails on the side you never looked at.
Multiplying by a percentage instead of a proportion. $n p_0$ with $p_0 = 60$ gives a meaningless number. Convert to 0.60 first.
**Using $\hat{p}$ in a test that names $p_0$ .** The null value is available and the test assumes it, so the check uses $np_0$ .
Writing symbols with no numbers. " $np_0 \geq 10$ " alone is not a check. " $np_0 = 200(0.60) = 120 \geq 10$ " is.
Confusing it with the 10% condition. Large Counts is about shape. The 10% condition, $n \leq 0.10N$ for a population of size $N$ , is about whether sampling without replacement keeps the observations close enough to independent. Both contain a 10 and they check completely different things.
Confusing it with the chi-square condition. Chi-square procedures want every expected count at least 5, a different rule for a different distribution. See chi-square tests explained.

A case that fails, and what it does to the interval

An inspector takes a random sample of 50 items from a large shipment and finds 3 defective. Check the Large Counts condition for a 95% confidence interval for the proportion defective, then compute the interval anyway to see what the failure does.

Find the sample proportion. $\hat{p} = 3/50 = 0.06$ , so the failure proportion is $1 - \hat{p} = 0.94$ .
A confidence interval assumes no value for $p$ , so the check uses $\hat{p}$ . Successes: $n\hat{p} = 50(0.06) = 3$ . Failures: $n(1 - \hat{p}) = 50(0.94) = 47$ .
Compare to 10. The 47 failures clear the bar easily; the 3 successes do not. Large Counts is not met, so a z-interval may not be valid. Write that sentence down.
Push through to see the damage. $SE = \sqrt{\frac{0.06(0.94)}{50}} = \sqrt{\frac{0.0564}{50}} = \sqrt{0.001128} = 0.0336$ .
Margin of error at 95%: $1.96(0.0336) = 0.0658$ .
Interval: $0.06 \pm 0.0658$ , giving $(-0.006, 0.126)$ .
The lower bound is negative, and no proportion can be negative. That is the normal model handing probability to values that cannot happen.
Confirm it in the counts. Estimating the centre and spread of the defective count from the data gives $n\hat{p} = 3$ and $\sqrt{n\hat{p}(1 - \hat{p})} = \sqrt{50(0.06)(0.94)} = \sqrt{2.82} = 1.68$ , so 0 sits only $3 / 1.68 = 1.79$ standard deviations below the center, not the 3.16 the condition would have guaranteed.

Successes $= 3 < 10$ , so the Large Counts condition fails. Forcing the interval gives $(-0.006, 0.126)$ , and the negative lower bound is the failure made visible. The correct exam response states the condition is not met and that the z-interval may not be trustworthy.

Same data, opposite verdicts: the null value vs p-hat

A company claims 25% of its customers use its mobile app. In a random sample of 40 customers, 8 use the app. Check the Large Counts condition for (a) a one-proportion z test of $H_0: p = 0.25$ and (b) a 95% confidence interval for $p$ .

Write down both proportions. The claim gives $p_0 = 0.25$ . The data gives $\hat{p} = 8/40 = 0.20$ .
(a) The test computes everything as if $H_0$ were true, so the check uses $p_0$ . Successes: $np_0 = 40(0.25) = 10$ . Failures: $n(1 - p_0) = 40(0.75) = 30$ .
Both are at least 10, and the condition reads 'at least', so 10 exactly passes. The test's Large Counts condition is met.
(b) The interval assumes nothing about $p$ , so the check uses $\hat{p}$ . Successes: $n\hat{p} = 40(0.20) = 8$ . Failures: $n(1 - \hat{p}) = 40(0.80) = 32$ .
The 8 successes fall short of 10, so the interval's Large Counts condition fails.
Nothing is contradictory here. The two procedures assume different things about $p$ , so they inspect different expected counts, and with $\hat{p}$ below $p_0$ the success count for the interval is the smaller of the two.
For completeness, the test statistic: $SE = \sqrt{\frac{0.25(0.75)}{40}} = \sqrt{0.0046875} = 0.0685$ , so $z = \frac{0.20 - 0.25}{0.0685} = -0.73$ .
And the interval you would have to flag: $SE = \sqrt{\frac{0.20(0.80)}{40}} = \sqrt{0.004} = 0.0632$ , so $0.20 \pm 1.96(0.0632)$ gives $(0.076, 0.324)$ . The numbers look fine, which is precisely why the condition has to be checked rather than eyeballed.

The test passes ( $np_0 = 10$ and $n(1-p_0) = 30$ ). The interval fails ( $n\hat{p} = 8 < 10$ ). Use $p_0$ whenever the null names a value and $\hat{p}$ when nothing is assumed, and expect the two checks to disagree when $\hat{p}$ is far from $p_0$ .

Frequently asked questions

Where does the number 10 actually come from?

It is a convention chosen for convenience, not a result anyone proved. Other textbooks use 5 or 15. Ten is round, and it holds the leftover skewness under about 0.32 while keeping the count at least 3.16 standard deviations clear of 0 and of $n$ . A bar of 5 would give 0.45 and 2.24 instead. On the AP exam, use 10.

Why is a large sample size not enough on its own?

Because a large $n$ paired with a tiny $p$ is still badly skewed. Take $n = 200$ and $p = 0.02$ : that is 200 observations but only 4 expected successes, a standard deviation of $\sqrt{200(0.02)(0.98)} = 1.98$ , and a floor at 0 sitting just 2.02 standard deviations below the center. The condition counts outcomes, not observations, which is exactly why it catches this.

Do I use p-hat or the null value in the check?

Use whatever the procedure assumes. A confidence interval assumes nothing, so use $\hat{p}$ , which means counting the successes and failures you actually observed. A significance test assumes $H_0: p = p_0$ , so use $p_0$ . A two-proportion test assumes the proportions are equal, so use the pooled $\hat{p}_c$ on all four counts.

Is Large Counts the same as the 10% condition?

No, and mixing them up is common because both mention 10. Large Counts asks whether the sampling distribution is close enough to normal. The 10% condition, $n \leq 0.10N$ , asks whether sampling without replacement leaves the observations close enough to independent for the standard error formula to hold. You check both, separately.