Clear Notion
Login

Concepts

Read through concept notes with worked explanations and diagrams, chapter by chapter.

Concept 07 / 08Statistics and Probability

The Binomial Distribution

Bernoulli trials and the binomial probability nCk p^k q^(n-k), mean np and variance npq, the most probable number of successes, 'at least' and 'at most' calculations, and recognising when the binomial model applies.

Figure 1
A Bernoulli trial has two outcomes, success (probability pp) and failure (probability q=1−pq = 1 - p). Repeat it nn times, with the trials independent and pp the same each time. The number of successes XX then has the binomial distribution B(n,p)B(n, p): P(X=k)=(nk)pkqn−kP(X = k) = \binom{n}{k}p^{k}q^{n-k}, for k=0,1,…,nk = 0, 1, \ldots, n. Reason: any particular sequence with kk successes and n−kn - k failures has probability pkqn−kp^{k}q^{n-k} by independence, and there are (nk)\binom{n}{k} such sequences (choose which trials are successes). The probabilities are the terms of (q+p)n(q + p)^{n}, which sum to 11; hence the name. Conditions to check. Fixed number of trials; independence; constant pp; only two outcomes per trial. Drawing without replacement from a small population violates independence and constant pp (that situation is hypergeometric), while drawing with replacement satisfies them. Mean and variance: E(X)=npE(X) = np and Var(X)=npq\text{Var}(X) = npq. The quickest proof writes XX as a sum of nn independent indicator variables, each with mean pp and variance pqpq. So Var(X)≤n4\text{Var}(X) \le \frac{n}{4}, with the maximum at p=12p = \frac{1}{2}. In questions giving the mean and variance, Varmean=q\frac{\text{Var}}{\text{mean}} = q, which determines pp and then nn. If the stated variance exceeds the mean, no binomial distribution fits. Shape and mode. For p=12p = \frac{1}{2} the distribution is symmetric about n2\frac{n}{2}; otherwise it is skewed towards the smaller end, as in the figure. The most probable value (mode) comes from the ratio P(X=k)P(X=k−1)=(n−k+1)pkq\frac{P(X = k)}{P(X = k - 1)} = \frac{(n - k + 1)p}{kq}, which exceeds 11 iff k<(n+1)pk < (n + 1)p. So the mode is ⌊(n+1)p⌋\lfloor(n + 1)p\rfloor; if (n+1)p(n + 1)p is an integer, both (n+1)p(n + 1)p and (n+1)p−1(n + 1)p - 1 are modes. This is the same ratio method used for the greatest term of a binomial expansion. Cumulative questions. P(X≥1)=1−qnP(X \ge 1) = 1 - q^{n}. 'At least kk' =∑j≥k= \sum_{j \ge k}, often computed as 1−∑j<k1 - \sum_{j < k}. 'At most kk' =∑j≤k= \sum_{j \le k}. A common type asks for the least nn such that P(X≥1)≥P(X \ge 1) \ge some value: solve 1−qn≥c1 - q^{n} \ge c, i.e. qn≤1−cq^{n} \le 1 - c, using logarithms. Key Takeaways: - P(X=k)=(nk)pkqn−kP(X = k) = \binom{n}{k}p^{k}q^{n-k} for independent trials with constant pp. - Mean npnp; variance npqnpq; variancemean=q<1\frac{\text{variance}}{\text{mean}} = q < 1. - Mode =⌊(n+1)p⌋= \lfloor(n + 1)p\rfloor (two modes if (n+1)p(n + 1)p is an integer). - P(at least one)=1−qnP(\text{at least one}) = 1 - q^{n}. - Symmetric only for p=12p = \frac{1}{2}. - Without replacement from a small population: not binomial. - Pitfall: forgetting the factor (nk)\binom{n}{k}, which counts the orders. Brain-Teaser 1: The mean and variance of a binomial distribution are 44 and 43\frac{4}{3}. Find P(X≥1)P(X \ge 1). Solution: q=4/34=13q = \frac{4/3}{4} = \frac{1}{3}, so p=23p = \frac{2}{3} and n=4p=6n = \frac{4}{p} = 6. Then P(X≥1)=1−(13)6=1−1729=728729P(X \ge 1) = 1 - \left(\frac{1}{3}\right)^{6} = 1 - \frac{1}{729} = \frac{728}{729}. The ratio of variance to mean identifies qq immediately. Brain-Teaser 2: How many times must a fair coin be tossed so that the probability of getting at least one head is at least 0.990.99? Solution: 1−(12)n≥0.991 - \left(\frac{1}{2}\right)^{n} \ge 0.99 requires 2n≥1002^{n} \ge 100, so n≥7n \ge 7 (since 26=642^{6} = 64 and 27=1282^{7} = 128). The least number is 77. Converting 'at least one' to the complement gives a clean exponential inequality. Brain-Teaser 3: A die is thrown 1010 times. What is the most likely number of sixes, and what is the probability of getting exactly that many? Solution: n=10n = 10, p=16p = \frac{1}{6}, (n+1)p=116≈1.83(n + 1)p = \frac{11}{6} \approx 1.83, so the mode is 11. P(X=1)=10⋅16⋅(56)9≈0.323P(X = 1) = 10 \cdot \frac{1}{6} \cdot \left(\frac{5}{6}\right)^{9} \approx 0.323. Compare P(X=2)=45⋅136⋅(56)8≈0.291P(X = 2) = 45 \cdot \frac{1}{36} \cdot \left(\frac{5}{6}\right)^{8} \approx 0.291 and P(X=0)=(56)10≈0.162P(X = 0) = \left(\frac{5}{6}\right)^{10} \approx 0.162, confirming that 11 is the most likely value although the mean is 106≈1.67\frac{10}{6} \approx 1.67.

Practice this concept

Free · no account required · instant solutions

DPP: The Binomial Distribution

15 questionsSee the solution the moment you answer

Practice