Binomial Probability Calculator

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What is the binomial distribution

Definition: The binomial distribution models the number of successes in a fixed number of independent trials. Each trial has only two possible outcomes (success or failure), and the probability of success remains constant across all trials.

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

Intro: This binomial probability calculator computes exact $P(X=k)$ for $X \sim \mathrm{Bin}(n,p)$ using the binomial probability mass function. Calculate individual probabilities or cumulative distributions with full working steps.

Binomial distribution formulas

Probability mass function: $P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}$
Expected value & variance: $\mu=np,\quad \sigma^2=np(1-p)$ ( $\sigma=\sqrt{np(1-p)}$ )
Cumulative probability: $P(X\le k)=\sum_{i=0}^{k}\binom{n}{i}p^i(1-p)^{n-i}$

How to use this calculator

Enter n, the total number of independent trials (positive integer).
Enter p, the probability of success on each trial (decimal between 0 and 1).
Enter k, the exact number of successes you want to find the probability for.
The calculator computes the binomial coefficient C(n,k), applies the PMF formula, and shows complete step-by-step working.

Worked example

n=10, p=0.3, k=4

Goal: For $X\sim\mathrm{Bin}(n,p)$ compute $P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}$.

Inputs: $n=10,\; p=0.3,\; k=4\Rightarrow 1-p=0.7$.

Coefficient: $\binom{10}{4}=\dfrac{10!}{4!6!}=210$.

Powers: $0.3^4=0.0081$, $0.7^6\approx0.117649$.

Multiply: $P=210\times0.0081\times0.117649\approx0.2000$.

Answer: $\boxed{P(X=4)\approx0.200}$.

A factory produces bolts with 5% defect rate. In batch of 50, find probability of exactly 3 defective bolts.

Model: Each bolt is a trial. Success = defective bolt. $n=50,\; p=0.05,\; k=3$.

Binomial applies because: fixed trials (50), two outcomes (defective/good), independent, constant probability.

Calculate $\binom{50}{3}$: $\dfrac{50!}{3!47!}=\dfrac{50\times49\times48}{6}=19{,}600$.

Success probability: $p^3=(0.05)^3=0.000125$.

Failure probability: $(1-p)^{47}=(0.95)^{47}\approx0.0973$.

Combine: $P(X=3)=19{,}600\times0.000125\times0.0973\approx0.238$.

Interpretation: $\boxed{23.8\%\text{ chance of exactly 3 defects}}$.

Weather forecast shows 40% chance of rain each day. Over 5 days, what's the probability of at most 2 rainy days?

Setup: Success = rainy day. $n=5,\; p=0.4,\; k\le2$.

Need cumulative: $P(X\le2)=P(X=0)+P(X=1)+P(X=2)$.

**For k=0**: $P(X=0)=\binom{5}{0}(0.4)^0(0.6)^5=(0.6)^5\approx0.0778$.

**For k=1**: $P(X=1)=\binom{5}{1}(0.4)^1(0.6)^4=5\times0.4\times0.1296\approx0.2592$.

**For k=2**: $P(X=2)=\binom{5}{2}(0.4)^2(0.6)^3=10\times0.16\times0.216\approx0.3456$.

Sum all three: $P(X\le2)=0.0778+0.2592+0.3456=0.6826$.

Answer: $\boxed{68.3\%\text{ probability of 2 or fewer rainy days}}$.

Interpretation: Over the 5-day period, there's about a 68% chance of having mostly dry days.

For $X\sim\text{Bin}(100,0.3)$, find the expected value and standard deviation.

Given: $n=100$ trials, $p=0.3$ success probability.

**Mean** (expected value): $\mu=E[X]=np=100\times0.3=30$.

**Variance**: $\sigma^2=\text{Var}(X)=np(1-p)=100\times0.3\times0.7=21$.

**Standard deviation**: $\sigma=\sqrt{\sigma^2}=\sqrt{21}\approx4.58$.

Interpretation: On average, expect 30 successes with typical variation of about ±4.58.

Answer: $\boxed{\mu=30,\;\sigma\approx4.58}$.

Basketball player makes 70% of free throws. In 10 attempts, find P(makes at least 8).

Setup: $n=10,\; p=0.7,\; k\ge8$ means find $P(X\ge8)$.

Use complement: $P(X\ge8)=1-P(X\le7)$ or directly compute $P(X=8)+P(X=9)+P(X=10)$.

**Direct approach** is simpler here with only 3 terms.

**k=8**: $P(X=8)=\binom{10}{8}(0.7)^8(0.3)^2=45\times0.0576\times0.09\approx0.2335$.

**k=9**: $P(X=9)=\binom{10}{9}(0.7)^9(0.3)^1=10\times0.0403\times0.3\approx0.1211$.

**k=10**: $P(X=10)=\binom{10}{10}(0.7)^{10}(0.3)^0=(0.7)^{10}\approx0.0282$.

Sum: $P(X\ge8)=0.2335+0.1211+0.0282=0.3828$.

Answer: $\boxed{38.3\%\text{ chance of making 8+ shots}}$.

A support team answers 65% of tickets within 2 hours. Over the next 18 tickets, what is the probability they meet the SLA on at least 14 tickets?

Model: Each ticket is a trial. Success = ticket answered within 2 hours. $n=18,\; p=0.65$.

We want: $P(X\ge 14)$ for $X\sim\mathrm{Bin}(18,0.65)$.

Use complement: $P(X\ge 14)=1-P(X\le 13)$.

Compute cumulative: $P(X\le 13)=\sum_{i=0}^{13}\binom{18}{i}(0.65)^i(0.35)^{18-i}$.

So $P(X\ge 14)=1-\sum_{i=0}^{13}\binom{18}{i}(0.65)^i(0.35)^{18-i}$.

A business picks up 72% of incoming calls. In 25 calls today, what is the probability they miss at most 3 calls?

Model: Each call is a trial. Let success = missed call. Then $n=25$ and $p=0.28$ (since miss rate = $1-0.72$).

We want: $P(X\le 3)$ for $X\sim\mathrm{Bin}(25,0.28)$.

Cumulative form: $P(X\le 3)=\sum_{i=0}^{3}\binom{25}{i}(0.28)^i(0.72)^{25-i}$.

This sum gives the probability of 0, 1, 2, or 3 missed calls out of 25.

Input validation and typical errors.

❌ **Using n=10.5**: Trials must be whole numbers. Use $n=10$ or $n=11$.

❌ **Setting p=75% instead of 0.75**: Probability must be decimal form (0 to 1).

❌ **Asking for k=15 when n=10**: Cannot have more successes than trials. Check $0\le k\le n$.

❌ **Negative values**: Both $n$ and $k$ must be non-negative integers; $p$ must be in [0,1].

✓ **Correct format**: $n=20$ (integer), $p=0.35$ (decimal), $k=7$ (integer $\le n$).

⚠️ **Large n**: For $n>100$, consider normal approximation: $X\sim N(np, np(1-p))$ when $np\ge5$ and $n(1-p)\ge5$.

FAQs

What makes an experiment binomial?

Four conditions: (1) Fixed number of trials, (2) Only two outcomes per trial, (3) Independent trials, (4) Constant probability of success. Common examples: coin flips, quality control testing, true/false quizzes.

How do I find cumulative probabilities P(X≤k) or P(X≥k)?

For P(X≤k), sum individual probabilities from 0 to k. For P(X≥k), either sum from k to n, or use complement: P(X≥k) = 1 - P(X≤k-1). Our calculator can compute both.

When should I use normal approximation instead?

Use normal approximation when n is large (typically n>30) AND both np≥5 and n(1-p)≥5. The binomial becomes approximately normal with mean np and variance np(1-p). Apply continuity correction for better accuracy.

What if my probability is given as a percentage?

Convert percentages to decimals: divide by 100. For example, 25% becomes p=0.25, and 3.5% becomes p=0.035.

Can I calculate 'at least' or 'at most' probabilities?

Yes. 'At least k' means P(X≥k) = sum from k to n. 'At most k' means P(X≤k) = sum from 0 to k. 'More than k' means P(X>k) = P(X≥k+1). 'Fewer than k' means P(X<k) = P(X≤k-1).

What's the relationship between binomial and Bernoulli distributions?

A Bernoulli distribution is a binomial with n=1. The binomial is the sum of n independent Bernoulli trials. If Y~Bernoulli(p), then X = Y₁+Y₂+...+Yₙ ~ Binomial(n,p).

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