Binomial and Poisson distributions flashcards

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Binomial distribution B(n, p)
The number of successes in a fixed number n of independent trials, each with the same probability p of success.
Conditions for a binomial model
Fixed number of trials; two outcomes per trial; trials independent; constant probability of success. State them in context.
Poisson distribution Po(λ)
The number of events in a fixed interval of time or space, when events occur singly, independently, at random and at a constant mean rate λ.
Conditions for a Poisson model
Events occur singly, independently of each other, at random, at a constant mean rate. State them in context.
Possible values of X for B(n, p) and for Po(λ)
Binomial: 0, 1, …, n. Poisson: 0, 1, 2, … with no upper limit.
Cumulative probability
P(X ≤ x): the probability of x or fewer. The binomial and Poisson tables give this, to 4 decimal places.
Binomial probability function
P(X = r) = ⁿCᵣ pʳ(1 − p)ⁿ⁻ʳ, r = 0, 1, …, n. (Yes)
Poisson probability function
P(X = r) = e^(−λ) λʳ / r!, r = 0, 1, 2, … (Yes)
Mean and variance of B(n, p)
E(X) = np; Var(X) = np(1 − p). (Yes)
Mean and variance of Po(λ)
E(X) = Var(X) = λ; standard deviation √λ. (Yes)
Additive property of the Poisson distribution
X ~ Po(λ), Y ~ Po(μ) independent ⇒ X + Y ~ Po(λ + μ). (No — learn it)
Poisson approximation to the binomial
B(n, p) ≈ Po(np) when n is large and p is small. (No — learn it)
P(X = r) and P(X ≥ r) from cumulative values
P(X = r) = P(X ≤ r) − P(X ≤ r − 1); P(X ≥ r) = 1 − P(X ≤ r − 1). (No — derive it)
Shape of B(n, p) for p < 0.5, p = 0.5, p > 0.5
Skewed right (positive skew), symmetrical, skewed left; the peak is near np.
Shape of Po(λ) as λ increases
Small λ: strongly skewed right with the peak at 0 or 1. Larger λ: peak near λ and closer to symmetrical.
Bar chart of B(150, 0.02) beside Po(3)
The pairs of bars are almost identical (largest gap about 0.002): n large and p small, so Po(np) fits.
Number-line picture of P(2 ≤ X < 7)
Shade 0–6 for P(X ≤ 6), remove 0–1 for P(X ≤ 1): what is left is 2, 3, 4, 5, 6.
Data bar chart with fitted Poisson expected frequencies
If sample mean ≈ sample variance, the expected frequencies N × P(X = x) track the observed bars closely.
Why scale λ with the interval?
Events occur at a constant mean rate, so twice the time (or area) means twice as many events on average.
Using tables when p > 0.5
Count the other outcome: if X ~ B(n, p) then n − X ~ B(n, 1 − p). Large X becomes small n − X.
How mean and variance help choose a model
Poisson has mean = variance; binomial has variance < mean. Data with mean ≈ variance support a Poisson model.
Why Po(np) approximates B(n, p) when p is small
The binomial variance np(1 − p) is then almost np, the same as its mean, just like a Poisson.
Why independence matters when adding Poisson variables
The additive property needs independent counts; X − Y is never Poisson (it can be negative).
X ~ B(8, 0.35). Find P(X = 3).
⁸C₃(0.35)³(0.65)⁵ = 0.2786.
X ~ B(20, 0.15). Find P(2 ≤ X < 7).
P(X ≤ 6) − P(X ≤ 1) = 0.9781 − 0.1756 = 0.8025.
Calls arrive at 3 per 10 minutes. P(more than 8 calls in 30 minutes)?
Po(9): 1 − P(X ≤ 8) = 1 − 0.4557 = 0.5443.
Independent Po(2.5) and Po(1.5) per hour. P(total ≥ 6 in an hour)?
Total ~ Po(4): 1 − P(T ≤ 5) = 1 − 0.7851 = 0.2149.
X ~ B(n, p) with mean 6 and variance 4.2. Find n and p.
1 − p = 4.2/6 = 0.7, so p = 0.3 and n = 6/0.3 = 20.
X ~ B(150, 0.02). Approximate P(X ≤ 2).
Po(3): P(X ≤ 2) ≈ 0.4232 (exact 0.4209).
p = 0.3 per trial. Least n so that P(at least one success) > 0.95?
1 − 0.7ⁿ > 0.95 ⇒ n > ln 0.05/ln 0.7 = 8.40 ⇒ n = 9.
Calls at 3 per 10 minutes. P(no calls in 2 minutes)?
Po(0.6): e^(−0.6) = 0.5488.
P(X ≥ 6) from the tables
1 − P(X ≤ 5), not 1 − P(X ≤ 6).
P(X < 4) for a discrete X
Equals P(X ≤ 3). For B(10, 0.4): 0.3823, whereas P(X ≤ 4) = 0.6331.
Starting a binomial or Poisson question
Define X in words and state its distribution with parameters, e.g. X ~ Po(4.5) for 3 months.
'State an assumption' for a binomial model
Give it in context: 'whether one seed germinates does not affect another', not just 'trials are independent'.
Approximating B(60, 0.4) by Po(24)
Not valid: p is not small (mean 24, variance 14.4). Large n alone is not enough.
'Exactly 1 of each' with two independent Poissons
Multiply P(X = 1) × P(Y = 1); do not use P(X + Y = 2).

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