Binomial distribution — PMF, mean, variance: Exercises

Intuition How to use this page

Work each problem before opening its solution. The levels climb a ladder: L1 Recognition (is it even binomial?) → L2 Application (plug into the PMF/mean/variance) → L3 Analysis (combine events, use complements) → L4 Synthesis (mix binomial with other ideas) → L5 Mastery (prove, generalise, connect).

Everything you need lives in the parent note the topic page. We reuse three tools constantly, so let's name them once in plain words:

PMF — "probability mass function", the machine $P (X = k) = (k n) p^{k} (1 - p)^{n - k}$ that spits out the chance of exactly $k$ successes.
$(k n)$ — read " $n$ choose $k$ ": the count of ways to pick which $k$ of the $n$ trials are the successes. It equals $\frac{n !}{k ! ( n - k )!}$ .
complement — instead of computing a hard event directly, compute "1 minus the easy opposite".

Below, $q$ always means $1 - p$ (the failure probability), and $X \sim Bin (n, p)$ means " $X$ is a binomial count with $n$ trials and success chance $p$ ".

Level 1 — Recognition

Recall Solution

Check the BINS conditions: Binary, Independent, fixed Number, Same $p$ .

(a) Binomial. Each flip is head/tail (Binary), flips don't influence each other (Independent), exactly $12$ of them (Number fixed), and $p = 0.5$ every time (Same $p$ ). So $X \sim Bin (12, 0.5)$ .
(b) NOT binomial. Removing a card changes the deck, so $p$ (red) shifts each draw and draws are dependent. This is the without-replacement cousin → Hypergeometric distribution.
(c) NOT binomial. The number of trials is not fixed in advance — you stop when the six appears. That's a geometric waiting-time question, not a fixed- $n$ count.
(d) Binomial (approximately). Binary Yes/No, essentially constant $p$ and independence because the population is huge relative to $30$ , and $n = 30$ is fixed. So $X \sim Bin (30, p)$ .

Recall Solution

$n = 10$ (fixed number of shots), $p = 0.8$ (same each shot, independent, made/missed is binary). So $X \sim Bin (10, 0.8)$ , and $X$ can be any integer $k = 0, 1, 2, \dots, 10$ .

Level 2 — Application

Recall Solution

WHAT: exactly $k = 2$ successes in $n = 6$ trials with $p = 0.4$ , $q = 0.6$ . HOW: plug into the PMF. $P (X = 2) = (2 6) (0.4)^{2} (0.6)^{4} .$ $(2 6) = \frac{6 !}{2 ! 4 !} = 15$ (ways to place the 2 successes). Then $(0.4)^{2} = 0.16$ and $(0.6)^{4} = 0.1296$ . $P (X = 2) = 15 \times 0.16 \times 0.1296 = 0.31104.$

Recall Solution

$E [X] = n p = 200 \times 0.05 = 10$ defects on average. $Var (X) = n p (1 - p) = 200 \times 0.05 \times 0.95 = 9.5$ . $σ = 9.5 \approx 3.082$ . Reading it: expect about $10$ defects, usually within $\pm 3$ of that. A batch of $25$ defects sits $(25 - 10) /3.08 \approx 4.9$ standard deviations high — a genuine alarm.

Recall Solution

" $X \leq 1$ " means $X = 0$ or $X = 1$ ; these are separate outcomes, so add their probabilities. $P (X = 0) = (0 4) (0.5)^{0} (0.5)^{4} = \frac{1}{16}, P (X = 1) = (1 4) (0.5)^{1} (0.5)^{3} = 4 \cdot \frac{1}{16} = \frac{4}{16} .$ $P (X \leq 1) = \frac{1}{16} + \frac{4}{16} = \frac{5}{16} = 0.3125.$

Level 3 — Analysis

Recall Solution

Computing $P (X \geq 1) = P (1) + P (2) + \dots + P (50)$ is $50$ terms. Use the complement instead: $P (X \geq 1) = 1 - P (X = 0) = 1 - (0 50) (0.02)^{0} (0.98)^{50} = 1 - (0.98)^{50} .$ $(0.98)^{50} \approx 0.36417$ , so $P (X \geq 1) \approx 1 - 0.36417 = 0.63583$ . Why complement wins: " $X = 0$ " is a single term $(1 - p)^{n}$ — one line replaces fifty.

Recall Solution

The range ${2, 3, 4}$ is smaller to remove than to add — but here it's clean either way. Directly: $P (X = 2) = (2 5) (0.6)^{2} (0.4)^{3} = 10 (0.36) (0.064) = 0.2304,$ $P (X = 3) = (3 5) (0.6)^{3} (0.4)^{2} = 10 (0.216) (0.16) = 0.3456,$ $P (X = 4) = (4 5) (0.6)^{4} (0.4)^{1} = 5 (0.1296) (0.4) = 0.2592.$ Sum: $0.2304 + 0.3456 + 0.2592 = 0.8352$ . Check by complement: the outside is ${0, 1, 5}$ : $P (0) = 0. 4^{5} = 0.01024$ , $P (1) = 5 (0.6) (0. 4^{4}) = 0.0768$ , $P (5) = 0. 6^{5} = 0.07776$ ; total $0.16480$ , and $1 - 0.16480 = 0.8352$ ✓.

Recall Solution

Conditional probability: $P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}$ . Here $A = {X = 3}$ , $B = {X \geq 1}$ , and $A \subset B$ so $A \cap B = A$ . $P (X \geq 1) = 1 - P (X = 0) = 1 - (0.5)^{3} = 1 - \frac{1}{8} = \frac{7}{8}, P (X = 3) = (0.5)^{3} = \frac{1}{8} .$ $P (X = 3 ∣ X \geq 1) = \frac{1/8}{7/8} = \frac{1}{7} \approx 0.142857.$

Level 4 — Synthesis

Recall Solution

Failure of the whole system = all servers down. Each is down with probability $q = 0.1$ . $P (all down) = 0. 1^{n} \leq 0.0001 = 1 0^{- 4} .$ Take base-10 logs: $n lo g_{10} (0.1) \leq lo g_{10} (1 0^{- 4})$ , i.e. $- n \leq - 4$ , so $n \geq 4$ . The smallest integer is $n = 4$ . Check: $0. 1^{4} = 0.0001$ , giving $P (at least one up) = 1 - 0.0001 = 0.9999$ ✓.

Recall Solution

Let $X \sim Bin (8, 0.25)$ = number of wins. Net profit $= 5 X - 1 (8 - X) = 6 X - 8$ . By Linearity of expectation, $E [profit] = 6 E [X] - 8 = 6 (n p) - 8 = 6 (8 \cdot 0.25) - 8 = 6 (2) - 8 = 4$ . Expected net profit $=\$ 4 $. * * I n t er p r e t a t i o n : * * o na v er a g ey o ug ain, d r i v e n e n t i r e l y b y$ E[X]=np=2$ wins.

Recall Solution

$λ = n p = 500 \times 0.004 = 2$ . Exact: $P (X = 2) = (2 500) (0.004)^{2} (0.996)^{498}$ . $(2 500) = 124750$ , $(0.004)^{2} = 1.6 \times 1 0^{- 5}$ , $(0.996)^{498} \approx 0.13629$ . $P (X = 2) \approx 124750 \times 1.6 \times 1 0^{- 5} \times 0.13629 \approx 0.27201$ . Poisson: $P = \frac{λ ^{2} e ^{- λ}}{2 !} = \frac{4 e ^{- 2}}{2} = 2 e^{- 2} \approx 0.27067$ . They agree to about $3$ decimals — expected, since $n$ is large and $p$ tiny with $n p$ moderate. This is exactly the regime where binomial $\to$ Poisson.

Level 5 — Mastery

Recall Solution

Write both PMF values and divide — the factorials telescope: $\frac{P ( k )}{P ( k - 1 )} = \frac{( k n ) p ^{k} q ^{n - k}}{( k - 1 n ) p ^{k - 1} q ^{n - k + 1}} = \frac{( k n )}{( k - 1 n )} \cdot \frac{p}{q} .$ Now $\frac{( k n )}{( k - 1 n )} = \frac{n ! / ( k ! ( n - k )!)}{n ! / (( k - 1 )! ( n - k + 1 )!)} = \frac{( k - 1 )! ( n - k + 1 )!}{k ! ( n - k )!} = \frac{n - k + 1}{k}$ . So the ratio is $\frac{n - k + 1}{k} \cdot \frac{p}{q}$ . ✓ Deduce the mode: the PMF rises ( $P (k) > P (k - 1)$ ) exactly while this ratio $> 1$ : $(n - k + 1) p > k (1 - p) ⟺ n p + p - k p > k - k p ⟺ k < (n + 1) p .$ So probabilities increase up to $k = ⌊(n + 1) p ⌋$ and decrease after — the mode is $⌊(n + 1) p ⌋$ (and if $(n + 1) p$ is an integer, both it and $(n + 1) p - 1$ tie). See the staircase figure.

Figure — Binomial distribution — PMF, mean, variance

Recall Solution

Fix $n > 0$ . Maximise $f (p) = p (1 - p) = p - p^{2}$ . Complete the square: $p - p^{2} = - (p^{2} - p) = - (p - \frac{1}{2})^{2} + \frac{1}{4} .$ Since $(p - \frac{1}{2})^{2} \geq 0$ and is subtracted, $f (p) \leq \frac{1}{4}$ , with equality iff $p = \frac{1}{2}$ . Therefore $Var = n p (1 - p) \leq \frac{n}{4}$ , maximised at $p = \frac{1}{2}$ , giving $Var_{m a x} = \frac{n}{4}$ ; and $Var = 0$ at $p = 0$ or $p = 1$ (no randomness). See the parabola figure.

Recall Solution

Conceptual proof (cleanest). $X$ counts successes in $m$ independent trials each with success chance $p$ ; $Y$ counts successes in another $n$ such trials. Pool all $m + n$ trials: they are independent, binary, fixed in number, and share the same $p$ — that is exactly the BINS setup for $Bin (m + n, p)$ . The total count $X + Y$ is that binomial. ∎ Consistency checks (these need $p$ equal, and independence for variance): $E [X + Y] = m p + n p = (m + n) p ✓, Var (X + Y) = m p (1 - p) + n p (1 - p) = (m + n) p (1 - p) ✓ .$ Caveat: if the two $p$ 's differed, pooling would violate Same $p$ and $X + Y$ would not be binomial — the sum theorem hinges on identical $p$ .

Connections

Bernoulli distribution — each single trial in every problem above.
Linearity of expectation — powers L4.2's payoff shortcut.
Variance and covariance — behind the "variance adds" step in L5.3.
Poisson distribution — the large- $n$ , small- $p$ limit checked in L4.3.
Normal approximation to binomial — the next tool once $n$ is large.
Hypergeometric distribution — the without-replacement case that killed L1.1(b).
Binomial Theorem — the identity guaranteeing the PMF sums to 1.