Binomial Distribution — Formula, Conditions & Example

A binomial distribution tells you the probability of getting exactly $k$ successes in $n$ trials. Use it only when each trial has two outcomes for the event you care about, the trials are independent, and the success probability stays the same each time.

If one of those conditions fails, the arithmetic can still look correct while the model itself is wrong.

What the binomial distribution means

Suppose you repeat the same kind of trial $n$ times. On each trial, you label one outcome as a success and the other as a failure.

If the probability of success is $p$ on every trial, then the random variable $X$, the number of successes, can follow a binomial distribution.

You will often see this written as

$$X \sim \text{Bin}(n,p)$$

This notation means:

• $n$ is the number of trials
• $p$ is the probability of success on each trial
• $X$ counts how many successes occur

This is a counting model. It does not ask which trial succeeded. It asks how many successes happened in total.

Binomial distribution formula

For exactly $k$ successes, the probability is

$$P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}$$

Each part has a job:

• $\binom{n}{k}$ counts how many ways the $k$ successes can be arranged among $n$ trials
• $p^k$ gives the probability of those $k$ successes
• $(1-p)^{n-k}$ gives the probability of the remaining failures

The formula works for $k=0,1,2,\dots,n$.

When you can use the binomial formula

Use a binomial model only when all of these conditions are true:

Fixed number of trials

You know in advance how many trials there are. For example, flipping a coin $8$ times fits this condition.

Two outcomes per trial

For the event you are tracking, each trial must be classified as success or failure. A die roll can still fit if you define success as something like "rolling a $6$."

Independent trials

One trial should not change the probability on the next. Sampling with replacement can fit this condition. Sampling without replacement from a small group usually does not.

Constant success probability

The value of $p$ must stay the same from trial to trial. If the chance changes each time, a simple binomial model is not appropriate.

Worked example: exactly 3 heads in 5 flips

Suppose a biased coin lands heads with probability $0.6$. You flip it $5$ times. What is the probability of getting exactly $3$ heads?

Let heads be the success event. Then

$$n=5,\quad p=0.6,\quad k=3$$

Use the formula:

$$P(X=3)=\binom{5}{3}(0.6)^3(0.4)^2$$

Now compute each part:

$$\binom{5}{3}=10,\quad (0.6)^3=0.216,\quad (0.4)^2=0.16$$

So

$$P(X=3)=10(0.216)(0.16)=0.3456$$

The probability of getting exactly $3$ heads is $0.3456$, or $34.56\%$.

Why is the binomial model valid here? The experiment has a fixed $n$, two outcomes per flip, independent trials, and the same probability $p=0.6$ on each flip.

A fast shortcut for "at least one"

For questions like "at least one success," the complement is often faster than adding many terms.

For example, if $X \sim \text{Bin}(5,0.6)$, then

$$P(X \ge 1)=1-P(X=0)=1-(0.4)^5$$

This works because "at least one success" is the complement of "zero successes."

Common mistakes in binomial distribution problems

Ignoring the conditions

A common mistake is using the binomial formula when trials are not independent. A classic example is drawing items without replacement from a small set and still pretending that $p$ never changes.

Misreading what "success" means

In a binomial problem, success does not have to mean something good. It only means the outcome you chose to count.

Mixing up "exactly," "at least," and "at most"

These phrases lead to different calculations even in the same experiment. "Exactly $3$" means one term, "at least $3$" means several terms, and "at most $3$" means a different sum.

When the binomial distribution is used

The binomial distribution appears when you count repeated yes-or-no style outcomes such as defective vs. not defective, pass vs. fail, click vs. no click, or heads vs. tails.

It is useful in quality control, survey sampling under the right assumptions, reliability questions, and basic probability models in statistics.

Try a similar problem

Try your own version with $8$ flips of a coin where $p=0.4$. First find $P(X=2)$, then find $P(X \ge 1)$ using the complement. If you want another case, compare what changes when the trials are no longer independent.