Binomial Distribution

10 min · 2 drills

Introduction

The binomial distribution describes the probability of getting exactly $k$ successes in $n$ independent trials, where each trial has the same probability of success $p$.

On the EPM exam, this topic appears regularly:

• Repeated coin/dice tosses
• Multiple attempts (skier, shooter)
• Quality control (batch of parts)

---

Bernoulli Formula

$$P(X==k)==C_n^k\cdot \cdot p^k\cdot \cdot (1--p{)}^{n-k}=$$

$$\left(\genfrac{}{}{0ex}{}{n}{k}\right)p^k{q}^{n-k}$$

where:

• $n$: number of trials
• $k$: number of successes
• $p$: probability of success in one trial
• $q=1-p$: probability of failure

---

Breaking Down the Formula

Why $C_n^k$?

The order of successes doesn't matter. We choose $k$ positions for successes out of $n$, which is a combination!

Why $p^k\cdot q^{n-k}$?

• $p^k$: probability of $k$ successes
• $q^{n-k}$: probability of $(n-k)$ failures
• Trials are independent → multiply

---

Examples

Example 1: Coin

Tossing a coin 5 times. Probability of exactly 3 heads?

$n=5$, $k=3$, $p=0.5$

$$P(X==3)==C_5^3\cdot \cdot 0.5^3\cdot \cdot 0.5^2==10\cdot \cdot 0.125\cdot \cdot 0.25==0.3125$$

Example 2: Shooter

A shooter hits the target with probability 0.8. Makes 6 shots. Probability of exactly 4 hits?

$n=6$, $k=4$, $p=0.8$, $q=0.2$

$$P(X==4)==C_6^4\cdot \cdot 0.8^4\cdot \cdot 0.2^2==15\cdot \cdot 0.4096\cdot \cdot 0.04==0.2458$$

---

Typical EPM Questions

"At least one" (mindestens einmal)

$$P(X\ge \ge 1)==1--P(X==0)==1--q^n$$

Example: Rolling a die 4 times. Probability of at least one six?

$$P(X\ge \ge 1)==1--{\left(\frac{5}{6}\right)}^4==1--0.482==0.518$$

"At most k" (höchstens k)

$$P(X\le \le k)==\sum _{i=0}^k{C}_n^i\cdot \cdot p^i\cdot \cdot q^{n-i}$$

"More than k" (mehr als k)

$$P(X>>k)==1--P(X\le \le k)$$

When $p$ is small, most probability is concentrated near $k=0$, which is why "at least one" is such a useful shortcut. Try it: set $n=10$, $p=0.2$ below and see how skewed the distribution is.

---

Finding Minimum n

Typical EPM problem: How many minimum attempts are needed so that the probability of at least one success is $\ge$ a given threshold?

Algorithm

1. Write: $P(X\ge 1)\ge$ threshold
2. Transform: $1-q^n\ge$ threshold
3. Solve: $q^n\le 1-$ threshold
4. Take logarithm: $n\ge \frac{\ln (1-\text{threshold})}{\ln q}$
5. Round UP!

Example

Probability of success = 0.3. How many attempts for $P(\text{at least 1})\ge 0.95$?

$$1--0.7^n\ge \ge 0.95$$ $$0.7^n\le \le 0.05$$ $$n\cdot \cdot \ln (0.7)\le \le \ln (0.05)$$ $$n\ge \ge \frac{\ln (0.05)}{\ln (0.7)}==\frac{-2.996}{-0.357}==8.4$$

Answer: minimum 9 attempts

---

Distribution Parameters

Expected Value

A random variable $X$ with finitely many values $x_1,x_2,\dots ,x_m$ and the matching probabilities $P(X=x_i)$ has the expected value

$$\boxed{E(X)=\sum _i{x}_i\cdot P(X=x_i)}$$

Each value is weighted by its probability, and everything is added up. This definition assumes no binomial distribution: the $x_i$ may be any numbers at all, amounts of money with a sign among them.

When $X$ counts the successes in $n$ Bernoulli trials, the sum collapses to

$$\boxed{E(X)=n\cdot p}$$

Average number of successes in $n$ trials.

Variance

$$\boxed{D(X)=n\cdot p\cdot q}$$

Standard Deviation

$$\sigma ==\sqrt{n\cdot p\cdot q}$$

---

Fair Game

A game counts as fair (German faires Spiel) when neither the player nor the other side comes out ahead in the long run. The expected gain is then zero:

$$\boxed{E=\sum _i{x}_i\cdot p_i=0}$$

Every amount you receive counts positive, every amount you pay counts negative. One of those amounts is the unknown, usually written $X$.

Procedure

1. Build a payout table: one row per event, with its probability and its amount including the sign
2. Check completeness: the events must exclude one another; if probability is left over, a row "receive nothing, pay nothing" with amount $0$ joins the table, and the probabilities add up to $1$
3. Write the ansatz $E=0$
4. Solve for $X$

Clearing the common denominator of the probabilities before solving leaves a linear equation in $X$.

Example: Finding the Fair Amount

Problem: An urn holds four red, three blue and three yellow balls. Three balls are drawn without replacement. If all three balls are yellow, you receive 29 euros. If the three balls have three different colours, you receive $X$ euros. In every other case you pay 1 euro. How large must $X$ be for the game to be fair?

Solution:

There are $C_{10}^3=\left(\genfrac{}{}{0ex}{}{10}{3}\right)=120$ equally likely draws in total.

All three yellow: only the three yellow balls qualify, which is a single draw.

$$p_1==\frac{1}{120}$$

Three different colours: one ball from each colour group.

$$p_2==\frac{4\cdot 3\cdot 3}{120}==\frac{36}{120}$$

All remaining draws form the residual event.

$$p_3==1--\frac{1}{120}--\frac{36}{120}==\frac{83}{120}$$

Payout table:

Ansatz:

$$E==29\cdot \cdot \frac{1}{120}++X\cdot \cdot \frac{36}{120}--1\cdot \cdot \frac{83}{120}==0|\cdot \cdot 120$$ $$29++36X--83==0$$ $$36X==54$$ $$\boxed{X=1.50\text{euros}}$$

Variant: the Unknown Sits on the Paying Side

When $X$ is an amount you pay, $X$ enters the same equation with a negative sign.

Problem: A fair die is thrown three times. If the numbers 1, 2 and 3 come up in that order, you receive 281 euros. If at least one 6 comes up, you receive 1 euro. In every other case you pay $X$ euros. How large must $X$ be for the game to be fair?

Solution:

$$p_1=={\left(\frac{1}{6}\right)}^3==\frac{1}{216},p_2==1--{\left(\frac{5}{6}\right)}^3==\frac{91}{216}$$

The first sequence contains no 6, so the two events exclude one another.

$$p_3==1--\frac{1}{216}--\frac{91}{216}==\frac{124}{216}$$ $$E==281\cdot \cdot \frac{1}{216}++1\cdot \cdot \frac{91}{216}--X\cdot \cdot \frac{124}{216}==0|\cdot \cdot 216$$ $$281++91==124X$$ $$372==124X$$ $$\boxed{X=3\text{euros}}$$

---

Binomial Distribution Graph

When $p=0.5$, the distribution is symmetric around $E(X)=np=5$.

---

EPM Problem: Full Solution

Problem: A skier completes a run with success probability 0.6. Makes 5 attempts.

a) Probability of exactly 3 successful attempts? b) Probability of at least 4 successes? c) Minimum attempts for $P(\ge 1)\ge 0.99$?

Solution:

a) $P(X=3)=C_5^3\cdot 0.6^3\cdot 0.4^2=10\cdot 0.216\cdot 0.16=0.3456$

b) $P(X\ge 4)=P(X=4)+P(X=5)$

$=C_5^4\cdot 0.6^4\cdot 0.4^1+C_5^5\cdot 0.6^5\cdot 0.4^0$ $=5\cdot 0.1296\cdot 0.4+1\cdot 0.07776\cdot 1$ $=0.2592+0.07776=0.337$

c) $1-0.4^n\ge 0.99$

$0.4^n\le 0.01$ $n\ge \frac{\ln (0.01)}{\ln (0.4)}=\frac{-4.605}{-0.916}=5.03$

Answer: minimum 6 attempts

---

Common Mistakes

Mistake 1: Forgetting $C_n^k$

$P=p^k{q}^{n-k}$ covers only ONE sequence! Must multiply by number of ways: $C_n^k$

Mistake 2: Confusing p and q

Remember that $p$ stands for success and $q=1-p$ for failure. Check!

Mistake 3: Rounding down in minimum n problems

$n\ge 5.03$ → answer is 6, not 5!

---

Check Your Understanding

Your turn1 of 2

A basketball player misses the basket on each free throw with probability $0.3$, independently of the other throws. She throws three times. Which expression gives the probability that only the third throw is a hit?

A$C_3^1\cdot 0.3^2\cdot 0.7$B$0.3^2\cdot 0.7$C$C_3^1\cdot 0.7^2\cdot 0.3$D$0.7^2\cdot 0.3$

Check

Your turn2 of 2

A test has four questions with two answer choices each, exactly one of which is correct. Someone guesses on every question. What is the probability of answering at least two questions correctly?

A$\frac{11}{16}$B$\frac{5}{16}$C$\frac{14}{16}$D$\frac{13}{16}$

Check

---

Formulas to Remember

Bernoulli Formula:

$$P(X==k)==C_n^k\cdot \cdot p^k\cdot \cdot q^{n-k}$$

At least one:

$$P(X\ge \ge 1)==1--q^n$$

Expected value (general, discrete):

$$E(X)==\sum _i{x}_i\cdot \cdot P(X==x_i)$$

Fair game:

$$E==\sum _i{x}_i\cdot \cdot p_i==0$$

Expected value (binomial):

$$E(X)==np$$

Variance:

$$D(X)==npq$$

Minimum n attempts:

$$n\ge \ge \frac{\ln (1-\text{threshold})}{\ln q}$$