# 3. Classroom tasks and solutions / Аудиторні задачі та розв'язання

> **Source of tasks.** The problems below are taken from the practical-class
> assignment for this topic (combinatorics and probability); only the wording has
> been translated to English. Every task is followed by a complete solution. Theory
> and notation are in [2method.md](2method.md): the product rule, $P(n,k)$
> (arrangements), $C(n,k)=\binom{n}{k}$ (combinations), $\binom{n+k-1}{k}$ (multisets),
> and the classical probability $P(A)=|A|/|\Omega|$.

## Combinatorics

### Задача 1 — car plates

**A car plate consists of four letters of an alphabet (12 letters are used) and four
digits. How many different plates are there?**

Each of the four letter positions is filled independently from $12$ letters, and each
of the four digit positions from $10$ digits ($0$–$9$); repetition is allowed. By the
product rule,

$$N = 12^{4}\cdot 10^{4} = 20\,736\cdot 10\,000 = \boxed{207\,360\,000}.$$

### Задача 2 — captain and deputy

**A football team has 11 players; a captain and a deputy captain must be chosen. In
how many ways?**

The two roles are **distinct**, so order matters, and one person cannot hold both, so
there is no repetition — an **arrangement** of $2$ out of $11$:

$$P(11,2) = 11\cdot 10 = \boxed{110}.$$

### Задача 3 — a football-pools card

**A card must predict the results of thirteen matches. Each result is a win for one of
the two teams or a draw (three outcomes; the score does not matter). How many ways can
one card be filled in?**

Each of the $13$ matches has $3$ possible outcomes, chosen independently:

$$3^{13} = \boxed{1\,594\,323}.$$

### Задача 4 — balls into boxes

**In how many ways can 4 black, 4 white, and 4 blue balls be placed into 6 different
boxes?**

Balls of the same colour are identical, so placing the $4$ balls of one colour into
$6$ distinct boxes is a **multiset** (stars-and-bars) count — $4$ stars and $5$ bars:

$$\binom{6+4-1}{4}=\binom{9}{4}=126.$$

The three colours are placed independently, so by the product rule

$$\left[\binom{9}{4}\right]^{3}=126^{3}=\boxed{2\,000\,376}.$$

### Задача 5 — a detachment from a company

**A company has 3 officers, 6 sergeants, and 60 privates. In how many ways can a
detachment of one officer, two sergeants, and twenty privates be formed?**

Order within each rank does not matter, so each rank is a **combination**, and the
three independent choices multiply:

$$C(3,1)\,C(6,2)\,C(60,20)=3\cdot 15\cdot\binom{60}{20}.$$

With $\binom{60}{20}=4\,191\,844\,505\,805\,495$,

$$45\cdot 4\,191\,844\,505\,805\,495=\boxed{188\,633\,002\,761\,247\,275}.$$

## Probability

### Задача 6 — one ball from each of two boxes

**Box 1 holds 2 white and 10 red balls; Box 2 holds 8 white and 4 red. One ball is
drawn from each box. What is the probability that both are white?**

![Two boxes: Box 1 with 2 white and 10 red balls, Box 2 with 8 white and 4 red balls](img/urns_t6.png)

The two draws are **independent** (different boxes), so the probabilities multiply:

$$P = \frac{2}{12}\cdot\frac{8}{12}=\frac16\cdot\frac23=\boxed{\tfrac{1}{9}\approx 0.111}.$$

### Задача 7 — two balls from one box

**A box holds 10 white and 14 red balls (24 in all). Two balls are drawn. What is the
probability that both are white?**

![One box with 10 white and 14 red balls](img/urn_t7.png)

All $\binom{24}{2}$ two-ball subsets are equally likely, and $\binom{10}{2}$ of them are
all-white:

$$P = \frac{C(10,2)}{C(24,2)}=\frac{45}{276}=\boxed{\tfrac{15}{92}\approx 0.163}.$$

Equivalently, drawing one after another, $\dfrac{10}{24}\cdot\dfrac{9}{23}=\dfrac{15}{92}$.

### Задача 8 — grade of a part

**A part is chosen at random. It is known that 4% of parts are defective, and that 75%
of the non-defective parts are of the first grade. Find the probability that the part
is of the first grade.**

![Probability tree: a part is good (0.96) or defective (0.04); a good part is first grade (0.75) or lower grade (0.25)](img/tree_t8.png)

A first-grade part must first be **non-defective** — probability $1-0.04=0.96$ — and
then **first grade among the good ones** — probability $0.75$. Multiply along the
branch (the conditional/multiplication rule):

$$P(\text{1st grade}) = P(\text{good})\cdot P(\text{1st}\mid\text{good}) = 0.96\cdot 0.75 = \boxed{0.72}.$$

### Задача 9 — winning at least once

**Of 100 lottery tickets, 5 are winning. Two tickets are bought. What is the
probability of winning by at least one ticket?**

Use the **complement**: "at least one win" is the opposite of "no win". Both tickets
missing means both come from the $95$ losing tickets:

$$P(\text{no win}) = \frac{C(95,2)}{C(100,2)}=\frac{4465}{4950}=\frac{893}{990}.$$

$$P(\text{at least one win}) = 1-\frac{893}{990}=\boxed{\tfrac{97}{990}\approx 0.098}.$$

As a check, counting the winning cases directly gives the same value:
$\dfrac{C(5,1)\,C(95,1)+C(5,2)}{C(100,2)}=\dfrac{475+10}{4950}=\dfrac{485}{4950}=\dfrac{97}{990}$.