A brief note on Counting Successes with the Binomial Model . Read it before a test. Try each question at the end before you open its answer.
When a count is binomial Two outcomes on each trial, success or failure. Independent trials. Picks from a group are close enough when the sample is under 10 % 10\% 10% of the group. A fixed number n n n of trials. The same chance p p p on every trial. One order: p k ( 1 − p ) n − k p^k(1 - p)^{n-k} p k ( 1 − p ) n − k . Number of orders: ( n k ) \binom{n}{k} ( k n ) , from Pascal's triangle or math.comb(n, k). Probability: P ( X = k ) = ( n k ) p k ( 1 − p ) n − k P(X = k) = \binom{n}{k}p^k(1 - p)^{n-k} P ( X = k ) = ( k n ) p k ( 1 − p ) n − k . Centre, spread and shape Mean: μ = n p \mu = np μ = n p . Standard deviation: σ = n p ( 1 − p ) \sigma = \sqrt{np(1 - p)} σ = n p ( 1 − p ) . Usual range: n p ± 2 σ np \pm 2\sigma n p ± 2 σ . Proportion: mean p p p , standard deviation p ( 1 − p ) / n \sqrt{p(1 - p)/n} p ( 1 − p ) / n . Shape: small p p p skews right, p = 0.5 p = 0.5 p = 0.5 is symmetric, large p p p skews left. A large n n n gives a bell. A worked example Ten made-up light switches each fail with probability 0.05 0.05 0.05 , independently.
None fail. 0.95 10 ≈ 0.5987 0.95^{10} \approx 0.5987 0.9 5 10 ≈ 0.5987 . Exactly one fails. 10 × 0.05 × 0.95 9 ≈ 0.3151 10 \times 0.05 \times 0.95^{9} \approx 0.3151 10 × 0.05 × 0.9 5 9 ≈ 0.3151 . At least two fail. 1 − 0.5987 − 0.3151 ≈ 0.0861 1 - 0.5987 - 0.3151 \approx 0.0861 1 − 0.5987 − 0.3151 ≈ 0.0861 . Centre and spread. μ = 0.5 \mu = 0.5 μ = 0.5 and σ = 0.475 ≈ 0.6892 \sigma = \sqrt{0.475} \approx 0.6892 σ = 0.475 ≈ 0.6892 switches. Practice with new numbers
A program to try The program checks the worked example.
Run it. Which line gives the answer to step 3?
1. Change n, p = 10, 0.05 to n, p = 20, 0.05. Is at least two failures more likely now?
Mistakes to avoid Leaving out ( n k ) \binom{n}{k} ( k n ) . Using the binomial when n n n is not fixed or the chance changes. Taking the variance for the standard deviation. Calling every binomial symmetric. Check yourself How many orders give exactly 2 2 2 successes in 8 8 8 trials? Stuck? Show a hint Show the answerA count is B ( 9 , 0.5 ) B(9, 0.5) B ( 9 , 0.5 ) . Find P ( X = 0 ) P(X = 0) P ( X = 0 ) . Stuck? Show a hint Show the answerFind the mean and the standard deviation of B ( 36 , 0.5 ) B(36, 0.5) B ( 36 , 0.5 ) . Stuck? Show a hint Show the answerIs B ( 15 , 0.85 ) B(15, 0.85) B ( 15 , 0.85 ) skewed to the left or to the right? Stuck? Show a hint Show the answerSources OpenStax, Introductory Statistics , Chapter 4, section on the binomial distribution.