Mean of the Binomial Distribution
A pollster asks $1{,}000$ randomly selected voters how they intend to vote. If $45\%$ of the population supports candidate A, how many "yes" responses do we expect on average? You don't need to compute a 1000-term sum — there is a single elegant formula: $E(X) = np$. This lesson derives that result from first principles using indicator (Bernoulli) variables, and shows you how to apply it.
Recall: a Bernoulli variable $Y$ takes value $1$ with probability $p$ and $0$ with probability $1-p$. Before any algebra — write down what $E(Y)$ equals, using the definition $E(Y) = \sum y\cdot P(Y=y)$. Then explain in words why this is the average value of $Y$.
Every binomial expectation question rewards two habits: decompose $X$ into indicator variables $X = Y_1 + Y_2 + \dots + Y_n$, then use linearity of expectation: $E(X) = E(Y_1)+E(Y_2)+\dots+E(Y_n) = np$.
The decompose-and-sum strategy: (1) write $X$ as a sum of $n$ identical Bernoulli variables, (2) note each has mean $p$, (3) apply linearity to get $E(X) = n \cdot p$.
Bernoulli: $E(Y_i) = p$ · Linearity: $E(\sum Y_i) = \sum E(Y_i) = np$
Key facts
- For Bernoulli $Y$: $E(Y) = p$
- For $X \sim B(n, p)$: $E(X) = np$
- Linearity of expectation: $E(\sum Y_i) = \sum E(Y_i)$
Concepts
- Why a binomial $X$ can be written as a sum of $n$ Bernoulli indicators
- Why linearity gives $E(X) = np$ instantly — without a brute-force sum
- Why $E(X)$ need not be an integer even when $X$ is integer-valued
Skills
- State and apply $E(X) = np$ in real-world contexts
- Derive the result by decomposing $X$ into Bernoulli variables
- Solve inverse problems: given $E(X)$ and one of $n,p$, find the other
The cleanest derivation uses indicator variables. Let $X \sim B(n, p)$ count successes in $n$ trials. Define:
$$Y_i = \begin{cases} 1 & \text{if trial } i \text{ is a success} \\ 0 & \text{otherwise} \end{cases}$$
Then $X = Y_1 + Y_2 + \dots + Y_n$. The expected value of each indicator is:
$$E(Y_i) = 0\cdot(1-p) + 1\cdot p = p.$$
Applying linearity of expectation:
$$E(X) = E\!\left(\sum_{i=1}^{n} Y_i\right) = \sum_{i=1}^{n} E(Y_i) = \sum_{i=1}^{n} p = np.$$
Worked through the hook: Fair die rolled $30$ times. Let $X$ = number of sixes. Each roll is a Bernoulli trial with $p = \tfrac{1}{6}$, so $X \sim B(30, \tfrac{1}{6})$. Then:
$$E(X) = np = 30 \cdot \tfrac{1}{6} = 5.$$
So on average we expect $5$ sixes — matches intuition (one in six rolls is a six, so $30/6 = 5$).
$E(X)=np$ for $X\sim B(n,p)$. Derivation: write $X=Y_1+\cdots+Y_n$ where $E(Y_i)=p$; apply linearity to get $E(X)=np$.
Pause — copy the indicator-variable derivation of $E(X)=np$: define $Y_i$, note $E(Y_i)=p$, apply linearity of expectation into your book.
Quick check: A multiple-choice test has $40$ questions, each with $5$ options. A student guesses every answer. What is the expected number of correct guesses?
We just saw that $E(X)=np$ is derived via indicator variables $Y_i$: $X=\sum Y_i$, each $E(Y_i)=p$, so by linearity $E(X)=\sum p=np$. That raises a question: given $E(X)=np$, how do you solve the three inverse problem types where either $n$ or $p$ is unknown? This card answers it → forward: $E(X)=np$; inverse for $p$: $p=E(X)/n$; inverse for $n$: $n=E(X)/p$.
The formula $E(X) = np$ supports three question types:
- Forward: given $n$ and $p$, compute $E(X) = np$. Trivial.
- Inverse for $p$: given $n$ and $E(X)$, find $p = E(X)/n$.
- Inverse for $n$: given $p$ and $E(X)$, find $n = E(X)/p$.
Example. A pollster surveys $200$ voters and expects $84$ to support candidate A. What is the support rate $p$?
$$E(X) = np \;\Rightarrow\; 84 = 200p \;\Rightarrow\; p = 0.42.$$
So $42\%$ of voters support candidate A — an inverse problem solved in one line.
The formula $E(X) = np$ supports three question types:
Pause — copy the three $E(X)=np$ problem types: forward ($E=np$), solve for $p$ ($p=E/n$), solve for $n$ ($n=E/p$), with a worked example of each into your book.
Did you get this? True or false: for $X \sim B(7, 0.4)$, the expected value $E(X) = 2.8$ is impossible because $X$ only takes integer values.
Worked examples · 3 in a row, reveal as you go
A biased coin lands heads with probability $0.35$. It is tossed $80$ times. Find the expected number of heads.
A quality-control inspector samples $50$ items from a production line and expects, on average, $3$ to be defective. Estimate the defect rate $p$.
Let $X \sim B(n, p)$. Prove that $E(X) = np$ by decomposing $X$ into Bernoulli indicators and using linearity of expectation.
Fill the gap: For $X \sim B(120, 0.25)$, the expected value is $E(X) = np = 120 \times 0.25 = $.
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: the derivation of $E(X) = np$ via indicator variables requires the trials to be independent.
Activities · practice with the ideas
A fair coin is tossed $50$ times. Find the expected number of heads. Use $E(X) = np$ and state $n$ and $p$.
A fair die is rolled $60$ times. Find the expected number of times an even number appears.
In a sample of $200$ randomly selected voters, the expected number supporting candidate B is $76$. Find the support rate $p$.
Derive $E(X) = np$ from first principles for $X \sim B(n,p)$ by writing $X$ as a sum of Bernoulli indicators. Show every step.
A telemarketer's call has a $12\%$ chance of resulting in a sale. How many calls $n$ should be planned so the expected number of sales is $30$?
Odd one out: Three of these statements about $X \sim B(20, 0.3)$ are correct. Which one is NOT?
Earlier you guessed the expected number of sixes when a fair die is rolled $30$ times.
The answer is $E(X) = np = 30 \cdot \tfrac{1}{6} = 5$, and you can prove it cleanly by decomposing $X$ into $30$ Bernoulli indicators (one per roll) and applying linearity. The brute-force computation $\sum_{k=0}^{30} k\binom{30}{k}(1/6)^k(5/6)^{30-k}$ gives the same answer but requires far more work.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. A fair die is rolled $48$ times. Find the expected number of times a five or six appears. (2 marks)
Q2. A factory tests $250$ light bulbs, and on average $15$ are found to be faulty. (a) Find the failure rate $p$. (b) If a new sample of $400$ is tested, find the expected number of faulty bulbs. (3 marks)
Q3. Let $X \sim B(n, p)$. Prove that $E(X) = np$ by writing $X$ as a sum of Bernoulli indicators $Y_1, Y_2, \dots, Y_n$ and applying linearity of expectation. Justify each step. (3 marks)
Comprehensive answers (click to reveal)
Activity answers:
1. $X \sim B(50, 0.5)$. $E(X) = 50 \cdot 0.5 = 25$ heads.
2. $P(\text{even}) = 3/6 = 1/2$; $X \sim B(60, 1/2)$. $E(X) = 60 \cdot 1/2 = 30$.
3. $76 = 200p \Rightarrow p = 76/200 = 0.38$ (i.e. $38\%$ support).
4. Let $Y_i = 1$ if trial $i$ succeeds, else $0$. Then $X = \sum_{i=1}^n Y_i$. $E(Y_i) = 0\cdot(1-p)+1\cdot p = p$. By linearity, $E(X) = \sum_{i=1}^n E(Y_i) = \sum_{i=1}^n p = np$. $\blacksquare$
5. $E(X) = np$: $30 = 0.12n \Rightarrow n = 250$ calls.
Q1 (2 marks): $P(5 \text{ or } 6) = 2/6 = 1/3$ [1]. $X \sim B(48, 1/3)$; $E(X) = 48 \cdot 1/3 = 16$ [1].
Q2 (3 marks): (a) $E(X) = np$: $15 = 250p \Rightarrow p = 0.06$ [1]. (b) For $X \sim B(400, 0.06)$: $E(X) = 400 \cdot 0.06 = 24$ [1]. Final answer with units stated [1].
Q3 (3 marks): Define $Y_i$: $1$ if trial $i$ is a success, $0$ otherwise [1]. $E(Y_i) = 0\cdot(1-p)+1\cdot p = p$, and $X = Y_1+\dots+Y_n$ [1]. By linearity, $E(X) = E(Y_1)+\dots+E(Y_n) = np$ [1].
Five timed questions on the binomial mean — forward applications, inverse problems for $p$ or $n$, and derivation by indicators. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering binomial-mean questions. Lighter alternative to the boss.
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