The Binomial Probability Formula

From a Binomial Setting to an Exact Probability

In Defining $n$, $p$, and $X$ in Context, you identified the fixed number of trials, the probability of success on one trial, and the random variable that counts successes. Now use those ingredients to find the probability of one exact count. For instance, if $X$ counts successes in 10 trials, how can you calculate $P(X=4)$?

Exactly four successes means that four trials have the success outcome and the other six have the failure outcome. Those successes could occur in many different positions among the 10 trials. The binomial formula accounts for all those possible arrangements while using the probability of any one arrangement.

Formula: If $X$ counts successes in a binomial setting with $n$ trials and probability of success $p$ on each trial, then the probability of exactly $k$ successes is $$ P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}. $$ Here, $\binom{n}{k}$ is the number of ways to place $k$ successes among the $n$ trials. The factor $p^k$ accounts for the successes, and $(1-p)^{n-k}$ accounts for the failures.

The formula relies on the binomial conditions introduced in Recognizing a Binomial Setting and The BINS Checklist for Binomial Conditions: each trial has two outcomes, the trials are independent, the number of trials is fixed, and the probability of success stays the same. Independence lets us multiply the probabilities for one particular sequence of outcomes.

Why the Formula Has These Parts

Consider one particular sequence of 10 trials with exactly four successes. For example, the first four trials might be successes and the remaining six failures. If a success has probability $p$, then a failure has probability $1-p$. By independence, the probability of this particular sequence is

p\cdot p\cdot p\cdot p\cdot(1-p)\cdot(1-p)\cdot(1-p)\cdot(1-p)\cdot(1-p)\cdot(1-p) =p^4(1-p)^6.

But the four successes do not have to be the first four. They could occupy any four of the ten trial positions. The coefficient $\binom{10}{4}$ counts how many different selections of four positions are possible. Each arrangement has the same probability, so multiply the probability of one arrangement by the number of arrangements.

For the specific event $X=4$ when $n=10$, the formula becomes

P(X=4)=\binom{10}{4}p^4(1-p)^6.

The exponent on $p$ is the number of successes, $4$. The exponent on $1-p$ is the number of failures, $10-4=6$. Together those exponents add to the total number of trials. The coefficient counts arrangements; it is not an additional probability for one trial.

For this calculation, the number of arrangements is

\binom{10}{4} =\frac{10\cdot9\cdot8\cdot7}{4\cdot3\cdot2\cdot1} =210.

This calculation gives the coefficient needed here. The next tutorial, Understanding Combinations in the Binomial Formula, examines combinations and this counting notation in more depth.

A Hand Calculation Routine

Before substituting numbers, translate the event into successes and failures. Then calculate the arrangement count and the two probability factors. Keeping these pieces separate makes it easier to catch an incorrect exponent or a missing factor.

Identify the binomial ingredients.
State what counts as success, and identify $n$, $p$, and the requested value $k$. Check that the binomial conditions are reasonable.

Count failures.
For exactly $k$ successes in $n$ trials, the number of failures is $n-k$.

Substitute into the formula.
Use $\binom{n}{k}p^k(1-p)^{n-k}$, matching each exponent to the corresponding count.

Calculate and interpret.
Evaluate the coefficient and powers, multiply, round as needed, and explain what the probability represents in context.

The formula gives the probability of an exact count. For example, $P(X=4)$ includes outcomes with four successes and no other counts. It does not include outcomes with five, six, or more successes. Questions about several possible counts require adding the relevant exact probabilities; the current focus is calculating one of them.

Worked Example: Exactly Four Damaged Packages

A shipping model treats the condition of each package as independent of the others and assigns each package a probability of 0.30 of arriving damaged. For a shipment of 10 packages, let $X$ be the number that arrive damaged. Find the probability that exactly four packages arrive damaged.

State. One trial is observing whether one package arrives damaged. Define a damaged package as a success. The requested event is $X=4$.

Plan. Use the four binomial conditions. There are two outcomes for each package, damaged or not damaged. The model states that package outcomes are independent. The shipment has a fixed 10 packages, and each package has the same damage probability, $p=0.30$. Thus $n=10$, $p=0.30$, and $k=4$, so the binomial formula applies.

Do. Exactly four packages are damaged, so $10-4=6$ are not damaged. The number of arrangements is $\binom{10}{4}=210$. Substitute and calculate:

\begin{aligned} P(X=4) &=\binom{10}{4}(0.30)^4(1-0.30)^6\\ &=210(0.30)^4(0.70)^6\\ &=210(0.0081)(0.117649)\\ &=210(0.0009529569)\\ &=0.200120949\\ &\approx 0.2001. \end{aligned}

Conclude. Under this model, the probability that exactly four of the 10 packages arrive damaged is about $0.2001$, or about 20.01%. In many comparable shipments of 10 packages, about 20.01% would have exactly four damaged packages, if the model’s assumptions hold.

The multiplication can also be checked in pieces: $0.0081\cdot0.117649=0.0009529569$, and $210\cdot0.0009529569=0.200120949$. Keeping extra digits until the last step avoids rounding error.

Worked Example: Exactly Four Successful Connections

A technician tests 10 independent wireless connections. A model assigns each connection a 0.20 probability of working on the first attempt, with the same probability for every connection. Let $W$ be the number that work on the first attempt. Find $P(W=4)$.

State. One trial is testing one connection. A connection that works on the first attempt is a success. We want exactly four successes out of 10.

Plan. The two outcomes are works or does not work on the first attempt. The model states that trials are independent and have the same success probability, and the number tested is fixed at 10. These are the binomial conditions, so use $n=10$, $p=0.20$, and $k=4$.

Do. There are six failures when there are exactly four successes. The arrangement count remains $\binom{10}{4}=210$; the probabilities change with $p$.

\begin{aligned} P(W=4) &=\binom{10}{4}(0.20)^4(0.80)^6\\ &=210(0.0016)(0.262144)\\ &=210(0.0004194304)\\ &=0.088080384\\ &\approx 0.0881. \end{aligned}

Conclude. The model gives a probability of about $0.0881$, or 8.81%, that exactly four of these 10 connections work on the first attempt.

The coefficient is unchanged because $n$ and $k$ are unchanged. The probability is different from the package example because the chance of success on each trial is different.

Worked Example: Four Successes with Equal Chances

A student makes 10 independent guesses on questions with two equally likely answer choices. Let $C$ be the number answered correctly. Find the probability of exactly four correct answers.

State. One trial is answering one question. A correct answer is a success, and the requested count is $C=4$.

Plan. Each answer is either correct or incorrect. The questions are modeled as independent, there are exactly 10 questions, and guessing gives the same success probability, $p=0.50$, each time. Thus the binomial model has $n=10$, $p=0.50$, and $k=4$.

Do. Four answers are correct and six are incorrect. Using $\binom{10}{4}=210$,

\begin{aligned} P(C=4) &=210(0.50)^4(0.50)^6\\ &=210(0.50)^{10}\\ &=210\left(\frac{1}{1024}\right)\\ &=\frac{210}{1024}\\ &=0.205078125\\ &\approx 0.2051. \end{aligned}

Conclude. Under the guessing model, the probability of exactly four correct answers is about $0.2051$, or 20.51%.

In this example, the success and failure probabilities are equal, so the product of the probability factors is $(0.50)^{10}$ for any particular arrangement of four correct answers. There are still 210 arrangements, which is why the probability for exactly four is larger than the probability of any one specified sequence.

Common Mistakes and AP Exam Tip

A complete response should show how the context determines $n$, $p$, and $k$, write the formula with the correct exponent for failures, and report a probability tied to the requested event. The following errors can change the answer substantially.

Using $p$ for failures. In the formula, $p$ is the probability of the outcome defined as success. The failure probability is $1-p$. If success is damaged, use the damage probability in $p$, even though damage is undesirable.
Using $k$ as both exponents. The success exponent is $k$, but the failure exponent is $n-k$. With four successes in 10 trials, the exponents must be 4 and 6.
Forgetting the arrangement count. The product $p^4(1-p)^6$ is the probability of one particular order of four successes and six failures. Multiply by $\binom{10}{4}$ to include every order.
Treating an exact probability as cumulative. $P(X=4)$ is not the probability of four or fewer successes, nor of at least four. The equality sign means exactly four.
Rounding too early. Keep the coefficient and unrounded power calculations until the final multiplication. Report the final value to a sensible number of decimal places and identify it as rounded.
Giving a number without context. A strong conclusion identifies the count and trial group. For example: “The model gives about a 20.01% chance that exactly four of the 10 packages arrive damaged.”

On an AP Statistics response, show enough work to make the model and calculation clear. A calculator can evaluate a binomial probability, but when asked to compute by hand, write the formula, substitute the values, show the success and failure powers, and include the arrangement count. Do not report only a decimal.

Key Takeaway

For exactly $k$ successes, first identify the success probability $p$ and the number of trials $n$. The success factor is $p^k$, the failure factor is $(1-p)^{n-k}$, and $\binom{n}{k}$ counts how many arrangements have that many successes. Multiply the three parts and interpret the result for the situation.

Key takeaway: For exactly four successes in 10 binomial trials, use $P(X=4)=\binom{10}{4}p^4(1-p)^6$. The coefficient counts arrangements, while the powers account for the four successes and six failures in each arrangement.

Check Your Understanding

For each question, identify the success probability and use the binomial formula. Show the coefficient and both probability factors before calculating.

A model gives each of 10 independent plants a 0.40 probability of flowering. If $F$ counts flowering plants, write the formula for $P(F=4)$. What does the probability describe?
A player makes each of 10 independent shots with probability 0.60. If $M$ counts made shots, what are the success exponent and failure exponent in $P(M=4)$?
For $n=10$ and $k=4$, what does $\binom{10}{4}$ count? Why is it multiplied by the probability of one particular arrangement?
In a binomial setting with $n=10$ and $p=0.25$, write the formula for exactly four successes. Identify the probability factor for failures.
Explain why $P(X=4)$ does not include outcomes with five successes, even though five is close to four.

The Binomial Probability Formula

What You'll Learn

From a Binomial Setting to an Exact Probability

Why the Formula Has These Parts

A Hand Calculation Routine

Worked Example: Exactly Four Damaged Packages

Worked Example: Exactly Four Damaged Packages

Worked Example: Exactly Four Successful Connections

Worked Example: Exactly Four Successful Connections

Worked Example: Four Successes with Equal Chances

Worked Example: Four Successes with Equal Chances

Common Mistakes and AP Exam Tip

Key Takeaway

Check Your Understanding

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