---
title: "Expected Successes & Failures — AP Stats Definition"
description: "Expected successes (np) and expected failures (n(1-p)) must both be at least 10 for p̂'s sampling distribution to be approximately normal. Core Unit 5 check."
canonical: "https://fiveable.me/ap-stats/key-terms/expected-successes-and-expected-failures"
type: "key-term"
subject: "AP Statistics"
unit: "Unit 3"
---

# Expected Successes & Failures — AP Stats Definition

## Definition

In AP Statistics, expected successes (np) and expected failures (n(1-p)) are the counts you check to verify the Large Counts Condition; if both are at least 10, the sampling distribution of the sample proportion p̂ is approximately normal (Topic 5.5).

## What It Is

Expected successes and expected failures are the two counts np and n(1-p), where n is your [sample size](/ap-stats/key-terms/sample-size "fv-autolink") and p is the [population proportion](/ap-stats/unit-3/sampling-distributions-for-sample-proportions/study-guide/Ezxev8MPpv3mFKjV4Gq3 "fv-autolink"). They answer a simple question. If you took a sample of size n, how many successes would you expect on average, and how many failures? If p = 0.3 and n = 50, you expect 50(0.3) = 15 successes and 50(0.7) = 35 failures.

Why do these numbers matter? The [sampling distribution](/ap-stats/unit-2/central-limit-theorem/study-guide/DPmpebCrsJBYfpSgOKn3 "fv-autolink") of p̂ is only *approximately* normal when the sample is big enough relative to how lopsided p is. The CED's rule (EK 5.5.B) is that both np ≥ 10 and n(1-p) ≥ 10. Think of it this way. A proportion near 0 or 1 makes the distribution of p̂ skewed, because p̂ can't go below 0 or above 1. Requiring at least 10 expected successes AND at least 10 expected failures guarantees the distribution has enough room on both sides to look like a normal curve instead of getting squished against a boundary.

## Why It Matters

This lives in Topic 5.5 (Sampling Distributions for Sample Proportions) in [Unit 5](/ap-stats/unit-5 "fv-autolink") and directly supports learning objective [AP Stats](/ap-stats "fv-autolink") 5.5.B, deciding whether the sampling distribution of p̂ can be described as approximately normal. It's the gatekeeper step. You can always compute the mean μp̂ = p and standard deviation σp̂ = √(p(1-p)/n) under 5.5.A, but you're only allowed to use normalcdf or a z-score to find a probability about p̂ if the expected counts check passes. Skipping or botching this check is one of the most common ways to lose points, and the habit you build here carries straight into every proportion inference procedure in Units 6 and beyond.

## Connections

### [Large Counts Condition (Unit 5)](/ap-stats/key-terms/large-counts-condition)

Expected successes and failures ARE the [Large Counts Condition](/ap-stats/key-terms/large-counts-condition "fv-autolink"). The condition is the rule (both counts ≥ 10), and np and n(1-p) are the actual numbers you compute and show to prove it holds. On the exam, naming the condition without showing the two counts usually isn't enough.

### Binomial mean np (Unit 4)

Expected successes is just the binomial mean from [Unit 4](/ap-stats/unit-4 "fv-autolink") wearing a new hat. The number of successes in n independent trials is binomial with mean np, so 'expected successes' literally means the average count of successes you'd see across many samples.

### Ten percent condition and sampling without replacement (Unit 5)

These are partner checks, but they police different things. [Large counts](/ap-stats/key-terms/large-counts "fv-autolink") (np and n(1-p)) earns you the normal shape; the 10% condition earns you the standard deviation formula when sampling without replacement. You need both, and mixing up which condition justifies what is a classic point-loser.

### Inference for proportions (Unit 6)

The same check resurfaces every time you build a confidence interval or run a significance test for a proportion. The twist is which proportion you plug in. In Unit 5 you use the population p, but in Unit 6 you'll use p̂ for intervals or the null value p₀ for tests, because the true p is unknown.

## On the AP Exam

This shows up two ways. In multiple choice, a stem might give you n and p and ask whether the sampling distribution of p̂ is approximately normal, or ask for the smallest n that makes it so. The trap answers usually check only one of the two counts, so always verify both np ≥ 10 and n(1-p) ≥ 10. In free response, this is a 'show your work' condition check inside a probability calculation about p̂, and it appeared on the 2025 exam in FRQ Q4. To earn credit, compute both numbers with the actual values (like 50(0.3) = 15 ≥ 10 and 50(0.7) = 35 ≥ 10), compare each to 10, and state the conclusion that the sampling distribution of p̂ is approximately normal. Writing 'large counts is met' with no numbers typically won't score.

## expected successes and expected failures vs Observed successes and failures (np̂ and n(1-p̂))

Expected counts use the population proportion p, because in Unit 5 sampling distribution problems you're told the true p. Observed counts use the sample proportion p̂, which is what you switch to in Unit 6 confidence intervals when p is unknown. Same ≥ 10 logic, different proportion plugged in. Use whatever proportion the procedure actually relies on.

## Key Takeaways

- Expected successes is np and expected failures is n(1-p), and both must be at least 10 for the sampling distribution of p̂ to be approximately normal.
- You must check BOTH counts; a small p can pass the failures check while failing the successes check, and vice versa.
- This check only justifies the normal shape; the 10% condition is a separate check that justifies the standard deviation formula when sampling without replacement.
- On FRQs, show the actual computed numbers compared to 10, not just the phrase 'large counts is met,' or you risk losing the point.
- The same idea returns in Unit 6 inference, but there you plug in p̂ (for intervals) or p₀ (for tests) instead of the true p.
- The cutoff of 10 exists because a p near 0 or 1 squishes p̂ against a boundary and skews its distribution; enough expected counts on both sides keeps it bell-shaped.

## FAQs

### What are expected successes and expected failures in AP Stats?

They're the counts np and n(1-p), the average number of successes and failures you'd expect in a sample of size n from a population with proportion p. If both are at least 10, the sampling distribution of the sample proportion p̂ is approximately normal (Topic 5.5).

### Why do np and n(1-p) both have to be at least 10?

Because p̂ is trapped between 0 and 1, a proportion close to either boundary makes its sampling distribution skewed. Requiring at least 10 expected successes and 10 expected failures guarantees enough room on both sides for the distribution to be roughly bell-shaped.

### Is checking just np ≥ 10 enough to say the distribution is approximately normal?

No. You have to verify both np ≥ 10 and n(1-p) ≥ 10. For example, with n = 200 and p = 0.97, np = 194 passes but n(1-p) = 6 fails, so the normal approximation is not justified.

### What's the difference between expected counts (np) and observed counts (np̂)?

Expected counts use the true population proportion p and show up in Unit 5 sampling distribution problems where p is given. Observed counts use the sample proportion p̂ and show up in Unit 6 confidence intervals, where p is unknown. Same ≥ 10 threshold, different proportion.

### Is the expected successes check the same as the 10% condition?

No. Expected successes and failures (the Large Counts Condition) justify the normal shape of p̂'s sampling distribution. The 10% condition (sample size under 10% of the population) justifies using the standard deviation formula √(p(1-p)/n) when sampling without replacement. They're separate checks for separate things.

## Related Study Guides

- [3.2 Sampling Distributions for Sample Proportions](/ap-stats/unit-3/sampling-distributions-for-sample-proportions/study-guide/Ezxev8MPpv3mFKjV4Gq3)

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