Binomial Probability Formula: P(X=k) = C(n,k) × p^k × (1-p)^(n-k) Explained With Worked Examples

The binomial probability formula gives the probability of getting exactly k successes in n independent trials, where each trial has the same probability of success p: P(X = k) = C(n, k) × p^k × (1-p)^(n-k) Where: - n = total number of trials - k = number of successes - p = probability of success on each trial - (1-p) = probability of failure on each trial - C(n, k) = binomial coefficient = n choose k = n! / (k! × (n-k)!) The formula has three pieces multiplied together: the binomial coefficient (counting how many ways to arrange k successes among n trials), p raised to the k power (probability of those k successes happening in their specific positions), and (1-p) raised to the (n-k) power (probability of the (n-k) failures happening in their positions). For a binomial experiment to be valid, four conditions must hold: (1) Fixed number of trials n. (2) Each trial has only two possible outcomes (success or failure — hence binomial). (3) The probability of success p is the same for every trial. (4) The trials are independent of each other. Classic examples: flipping a coin n times and counting heads (p = 0.5), rolling a die n times and counting how many times you get a 6 (p = 1/6), surveying n people and counting how many support a candidate (p = the true proportion in the population), inspecting n manufactured parts and counting defects (p = the defect rate), etc. The binomial distribution is one of the most important discrete probability distributions in statistics because it models the count of successes in any independent-trial scenario. Understanding it cold is essential for hypothesis testing, confidence intervals for proportions, and many introductory statistics applications. Snap a photo of any binomial probability problem and StatsIQ identifies the n, k, and p, applies the formula, and shows the calculation step by step.

**Example 1**: A multiple-choice test has 10 questions, each with 4 options. A student guesses every answer randomly. What is the probability they get exactly 3 questions correct? Identify n, k, p: n = 10 (questions), k = 3 (correct answers), p = 1/4 = 0.25 (probability of correct guess on each question). Apply the formula: P(X = 3) = C(10, 3) × 0.25^3 × 0.75^7 C(10, 3) = 10! / (3! × 7!) = (10 × 9 × 8) / (3 × 2 × 1) = 120 0.25^3 = 0.015625 0.75^7 = 0.1335 P(X = 3) = 120 × 0.015625 × 0.1335 = 0.2503 The probability of getting exactly 3 right by guessing is approximately 0.250, or 25%. **Example 2**: A factory produces light bulbs with a 5% defect rate. In a sample of 20 bulbs, what is the probability of finding exactly 2 defective bulbs? n = 20, k = 2, p = 0.05. P(X = 2) = C(20, 2) × 0.05^2 × 0.95^18 C(20, 2) = 20! / (2! × 18!) = (20 × 19) / 2 = 190 0.05^2 = 0.0025 0.95^18 = 0.3972 P(X = 2) = 190 × 0.0025 × 0.3972 = 0.1887 The probability of finding exactly 2 defective bulbs in 20 is about 0.189, or 18.9%. **Example 3**: A baseball player has a batting average of 0.300. In 5 at-bats, what is the probability they get exactly 2 hits? n = 5, k = 2, p = 0.300. P(X = 2) = C(5, 2) × 0.3^2 × 0.7^3 C(5, 2) = 10 0.3^2 = 0.09 0.7^3 = 0.343 P(X = 2) = 10 × 0.09 × 0.343 = 0.3087 The probability of exactly 2 hits in 5 at-bats is about 0.309, or 30.9%. Notice that these are EXACT probabilities — the chance of getting a specific number, not a range. For range questions (at least, at most, between), you sum multiple individual probabilities. We will cover that next. StatsIQ handles binomial calculations automatically — snap a photo of any problem and it identifies n, k, and p, computes the binomial coefficient, applies the formula, and shows the full work.

Most real probability questions ask about a RANGE of outcomes, not a single exact number. These require summing the binomial probabilities across the range. **At least k successes**: P(X ≥ k) = P(X = k) + P(X = k+1) + ... + P(X = n). Example: a basketball player makes 80% of their free throws. In 10 attempts, what is the probability they make AT LEAST 8? P(X ≥ 8) = P(X = 8) + P(X = 9) + P(X = 10) P(X = 8) = C(10, 8) × 0.8^8 × 0.2^2 = 45 × 0.1678 × 0.04 = 0.3020 P(X = 9) = C(10, 9) × 0.8^9 × 0.2^1 = 10 × 0.1342 × 0.2 = 0.2684 P(X = 10) = C(10, 10) × 0.8^10 × 0.2^0 = 1 × 0.1074 × 1 = 0.1074 P(X ≥ 8) = 0.3020 + 0.2684 + 0.1074 = 0.6778 The probability of making at least 8 of 10 free throws is about 67.8%. **At most k successes**: P(X ≤ k) = P(X = 0) + P(X = 1) + ... + P(X = k). **Between (k1 ≤ X ≤ k2)**: sum from k1 to k2. **Complement trick**: it is often easier to use complements. P(X ≥ 1) = 1 - P(X = 0). This is much faster than summing P(X=1) + P(X=2) + ... + P(X=n). Example: in 20 random phone calls, what is the probability of AT LEAST one wrong number, if the wrong-number rate is 5%? Direct approach: sum P(X=1) + P(X=2) + ... + P(X=20). Tedious. Complement approach: P(X ≥ 1) = 1 - P(X = 0) = 1 - C(20, 0) × 0.05^0 × 0.95^20 = 1 - 1 × 1 × 0.3585 = 1 - 0.3585 = 0.6415. The probability of at least one wrong number in 20 calls is about 64.2%. Much faster than computing each individual probability. Always look for opportunities to use the complement trick. It is the difference between a 30-second calculation and a 5-minute calculation on exam problems. StatsIQ automatically applies the complement trick when it would simplify the calculation and shows both approaches when relevant for understanding.

The binomial distribution has a known mean and variance that you can compute directly without listing all the probabilities. **Mean of a binomial distribution**: μ = np This says the expected number of successes in n trials with probability p is simply n times p. Intuitively obvious — if you flip a coin 100 times with p=0.5, you expect 50 heads on average. For 200 trials with p=0.3, you expect 60 successes. **Variance of a binomial distribution**: σ² = np(1-p) **Standard deviation**: σ = sqrt(np(1-p)) Example: 100 coin flips, p = 0.5. μ = 100 × 0.5 = 50 (expected number of heads) σ² = 100 × 0.5 × 0.5 = 25 σ = 5 So we expect about 50 heads with a standard deviation of 5. Most outcomes (about 68% by the empirical rule for normal-ish distributions) fall within 1 SD of the mean — between 45 and 55 heads. Almost all outcomes (95%) fall within 2 SDs — between 40 and 60 heads. **Normal approximation to the binomial**: When n is large enough, the binomial distribution is well-approximated by a normal distribution with the same mean (np) and standard deviation (sqrt(np(1-p))). The standard rule of thumb: the approximation is good when both np ≥ 10 AND n(1-p) ≥ 10. (Some textbooks use 5 instead of 10; 10 is more conservative.) Why this matters: the binomial formula becomes computationally tedious for large n. Computing C(1000, 500) directly is impractical. Instead, use the normal approximation: convert the binomial problem to a z-score and look up the probability in the standard normal table. Example: in 200 coin flips, what is the probability of getting at least 110 heads? n = 200, p = 0.5, k = 110. μ = 200 × 0.5 = 100 σ = sqrt(200 × 0.5 × 0.5) = sqrt(50) = 7.07 z = (110 - 100) / 7.07 = 1.41 (using continuity correction, use 109.5 instead of 110: z = (109.5 - 100) / 7.07 = 1.34) P(Z ≥ 1.34) = 1 - 0.9099 = 0.0901, or about 9%. The continuity correction (subtracting 0.5 from k for an "at least" question) accounts for the fact that the binomial is discrete and the normal is continuous. It produces a more accurate approximation, especially for moderate n. StatsIQ automatically chooses between the exact binomial formula and the normal approximation based on whether the conditions np ≥ 10 and n(1-p) ≥ 10 are met, and applies the continuity correction when using the approximation.