The binomial theorem and probability theory are two fundamental branches of mathematics that are widely used in science, statistics, and engineering. On the YÖS exam, these topics require an understanding of combinatorics, polynomial expansions, and the fundamentals of probability theory.
Definition: The factorial of a natural number $n$ (denoted $n!$) is the product of all natural numbers from 1 to $n$.
$$n! = 1 \cdot 2 \cdot 3 \cdot ... \cdot n$$
Special cases:- $0! = 1$ (by definition)- $1! = 1$
Examples:- $2! = 1 \cdot 2 = 2$- $3! = 1 \cdot 2 \cdot 3 = 6$- $4! = 1 \cdot 2 \cdot 3 \cdot 4 = 24$- $5! = 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 = 120$- $6! = 720$- $7! = 5040$- $10! = 3,628,800$
Recurrence relation:$$n! = n \cdot (n-1)!$$
Example 1: Using this property:$$7! = 7 \cdot 6! = 7 \cdot 720 = 5040$$
Definition: A permutation of $n$ elements is any ordering of these elements. The number of all permutations of $n$ elements is denoted $P(n)$ or $P_n$.
$$P(n) = n!$$
Intuition:- For the first position we have $n$ choices- For the second position — $(n-1)$ choices- For the third position — $(n-2)$ choices- And so on...
By the product rule: $n \cdot (n-1) \cdot (n-2) \cdot ... \cdot 1 = n!$
Example 2: How many ways are there to arrange 5 students in a row?
Solution:$$P(5) = 5! = 120$$
Example 3: There are 3 books on a shelf. How many ways are there to arrange them?
Solution:$$P(3) = 3! = 6$$
Arrangements: ABC, ACB, BAC, BCA, CAB, CBA
Definition: An arrangement of $n$ elements taken $r$ at a time (where $0 \leq r \leq n$) is an ordered selection of $r$ elements from $n$ elements without repetition. The number of arrangements is denoted $A(n, r)$, $A_n^r$ or $P(n, r)$.
$$A(n, r) = \frac{n!}{(n-r)!}$$
Intuition:- For the first position we choose from $n$ elements: $n$ choices- For the second position we choose from the $(n-1)$ remaining: $(n-1)$ choices- For the third position we choose from the $(n-2)$ remaining: $(n-2)$ choices- And so on, $r$ times
By the product rule: $n \cdot (n-1) \cdot (n-2) \cdot ... \cdot (n-r+1) = \frac{n!}{(n-r)!}$
Example 4: How many ways are there to choose and arrange 3 students out of 5 students?
Solution:$$A(5, 3) = \frac{5!}{(5-3)!} = \frac{5!}{2!} = \frac{120}{2} = 60$$
Or: $5 \cdot 4 \cdot 3 = 60$
Example 5: 8 athletes compete in a championship. How many ways are there to determine the first, second, and third place?
Solution:$$A(8, 3) = \frac{8!}{5!} = 8 \cdot 7 \cdot 6 = 336$$
Definition: A combination of $n$ elements taken $r$ at a time (where $0 \leq r \leq n$) is an unordered selection of $r$ elements from $n$ elements without repetition. The number of combinations is denoted $C(n, r)$, $C_n^r$ or $\binom{n}{r}$.
$$\binom{n}{r} = \frac{n!}{r!(n-r)!}$$
Difference from arrangements: In combinations, the order of the selected elements does not matter.
Relation to arrangements:$$A(n, r) = \binom{n}{r} \cdot r!$$
because each combination of $r$ elements can be arranged in $r!$ ways.
Example 6: How many ways are there to choose 3 students out of 5?
Solution:$$\binom{5}{3} = \frac{5!}{3! \cdot 2!} = \frac{120}{6 \cdot 2} = \frac{120}{12} = 10$$
This is less than the number of arrangements ($60 > 10$), because the order does not matter.
Example 7: There are 6 red and 4 blue balls in a box. How many ways are there to choose 2 red and 2 blue balls?
Solution:- Ways to choose 2 red out of 6: $\binom{6}{2} = \frac{6!}{2! \cdot 4!} = 15$- Ways to choose 2 blue out of 4: $\binom{4}{2} = \frac{4!}{2! \cdot 2!} = 6$- Total ways: $15 \cdot 6 = 90$
Property 1: Boundary cases$$\binom{n}{0} = \binom{n}{n} = 1$$
Property 2: Complementarity$$\binom{n}{r} = \binom{n}{n-r}$$
Proof:$$\binom{n}{n-r} = \frac{n!}{(n-r)!(n-(n-r))!} = \frac{n!}{(n-r)! \cdot r!} = \binom{n}{r}$$
Example 8: $\binom{10}{7} = \binom{10}{3} = \frac{10!}{3! \cdot 7!} = 120$
Property 3: Recurrence relation (Pascal's)$$\binom{n}{r} = \binom{n-1}{r-1} + \binom{n-1}{r}$$
This property is used to build Pascal's triangle.
Definition: Pascal's triangle is a numerical triangle in which each number equals the sum of the two numbers above it.
Structure:
``` 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1```
Relation to combinations: The $n$-th row contains the coefficients $\binom{n}{0}, \binom{n}{1}, \binom{n}{2}, ..., \binom{n}{n}$.
Example:- Row 0: $1$- Row 1: $1, 1$- Row 2: $1, 2, 1$- Row 3: $1, 3, 3, 1$- Row 4: $1, 4, 6, 4, 1$
Application: The coefficients of the expansion $(a+b)^n$ are found in the $n$-th row of Pascal's triangle.
Theorem (Binomial theorem): For any numbers $a$ and $b$ and any natural number $n$, the following formula holds:
$$(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$$
Expanded form:$$(a+b)^n = \binom{n}{0} a^n + \binom{n}{1} a^{n-1}b + \binom{n}{2} a^{n-2}b^2 + ... + \binom{n}{n-1} ab^{n-1} + \binom{n}{n} b^n$$
Example 9: Expand $(a+b)^3$.
Solution:$$(a+b)^3 = \binom{3}{0}a^3 + \binom{3}{1}a^2b + \binom{3}{2}ab^2 + \binom{3}{3}b^3$$$$= 1 \cdot a^3 + 3 \cdot a^2b + 3 \cdot ab^2 + 1 \cdot b^3$$$$= a^3 + 3a^2b + 3ab^2 + b^3$$
Example 10: Expand $(2x - 3)^4$.
Solution: We use $a = 2x$, $b = -3$, $n = 4$:
$$(2x-3)^4 = \sum_{k=0}^{4} \binom{4}{k} (2x)^{4-k} (-3)^k$$
$$= \binom{4}{0}(2x)^4 + \binom{4}{1}(2x)^3(-3) + \binom{4}{2}(2x)^2(-3)^2 + \binom{4}{3}(2x)(-3)^3 + \binom{4}{4}(-3)^4$$
$$= 16x^4 - 4 \cdot 8x^3 \cdot 3 + 6 \cdot 4x^2 \cdot 9 - 4 \cdot 2x \cdot 27 + 81$$
$$= 16x^4 - 96x^3 + 216x^2 - 216x + 81$$
Definition: The general (or $(k+1)$-th) term of the expansion of $(a+b)^n$ has the form:
$$T_{k+1} = \binom{n}{k} a^{n-k} b^k$$
(taking the first term $T_1$ at $k = 0$)
Example 11: In the expansion of $(x+2)^5$, find the third term.
Solution:The third term is $T_3$ (at $k = 2$):
$$T_3 = \binom{5}{2} x^{5-2} \cdot 2^2 = 10 \cdot x^3 \cdot 4 = 40x^3$$
Example 12: In the expansion of $(3a - b)^6$, find the term containing $a^3b^3$.
Solution:
General term: $T_{k+1} = \binom{6}{k} (3a)^{6-k} (-b)^k = \binom{6}{k} 3^{6-k} (-1)^k a^{6-k} b^k$
We need: $6 - k = 3$ and $k = 3$
$$T_4 = \binom{6}{3} 3^3 (-1)^3 a^3 b^3 = 20 \cdot 27 \cdot (-1) \cdot a^3 b^3 = -540a^3b^3$$
Theorem: The sum of all coefficients in the expansion of $(a+b)^n$ equals $2^n$.
Proof: Substitute $a = 1$ and $b = 1$:
$$(1+1)^n = \sum_{k=0}^{n} \binom{n}{k} \cdot 1^{n-k} \cdot 1^k = \sum_{k=0}^{n} \binom{n}{k}$$
$$2^n = \binom{n}{0} + \binom{n}{1} + \binom{n}{2} + ... + \binom{n}{n}$$
Example 13: Find the sum of coefficients in the expansion of $(x+y)^7$.
Solution:$$\text{Sum} = 2^7 = 128$$
Theorem: The sum of coefficients with alternating signs equals $0$ (if $n > 0$).
Proof: Substitute $a = 1$ and $b = -1$:
$$(1-1)^n = \sum_{k=0}^{n} \binom{n}{k} \cdot 1^{n-k} \cdot (-1)^k = \sum_{k=0}^{n} \binom{n}{k} (-1)^k$$
$$0 = \binom{n}{0} - \binom{n}{1} + \binom{n}{2} - \binom{n}{3} + ...$$
$$(a+b)^2 = a^2 + 2ab + b^2$$
$$(a-b)^2 = a^2 - 2ab + b^2$$
$$(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$$
$$(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$$
$$a^2 - b^2 = (a-b)(a+b)$$
Definition: The probability of an event is a measure of the possibility of the event occurring.
Basic terms:- Random experiment (or experiment): An action whose outcome is not known in advance- Elementary event: An indivisible outcome of an experiment- Sample space: The set of all possible elementary events, denoted $S$ or $\Omega$- Event: A subset of the sample space- Certain event: An event that will always occur (probability 1)- Impossible event: An event that cannot occur (probability 0)
Definition: If an experiment has $n$ equally likely outcomes, and $m$ of them are favorable to event $A$, then the probability of event $A$ equals:
$$P(A) = \frac{|A|}{|S|} = \frac{m}{n}$$
where $|A|$ is the number of favorable outcomes, and $|S|$ is the number of all possible outcomes.
Properties of probability:1. $0 \leq P(A) \leq 1$ for any event $A$2. $P(S) = 1$ (probability of a certain event)3. $P(\emptyset) = 0$ (probability of an impossible event)
Example 14: The probability of rolling an even number on a die.
Solution:- Total outcomes: $n = 6$ (the numbers 1, 2, 3, 4, 5, 6)- Favorable outcomes (even numbers): $m = 3$ (the numbers 2, 4, 6)- Probability: $P(\text{even}) = \frac{3}{6} = \frac{1}{2}$
Example 15: An urn contains 5 red and 3 blue balls. Probability of drawing a red ball.
Solution:- Total balls: $n = 5 + 3 = 8$- Red balls: $m = 5$- Probability: $P(\text{red}) = \frac{5}{8}$
Definition: The complementary event to event $A$ (denoted $\overline{A}$ or $A'$) is the event that event $A$ did not occur.
Formula:$$P(\overline{A}) = 1 - P(A)$$
Proof: $A$ and $\overline{A}$ are mutually exclusive events, and $A \cup \overline{A} = S$, so:$$P(A) + P(\overline{A}) = P(S) = 1$$$$P(\overline{A}) = 1 - P(A)$$
Example 16: The probability that NO 6 is rolled on a die.
Solution:$$P(\text{not 6}) = 1 - P(\text{6}) = 1 - \frac{1}{6} = \frac{5}{6}$$
Example 17: The probability that we do NOT win in a lottery, if the probability of winning equals 0.2.
Solution:$$P(\text{not winning}) = 1 - 0.2 = 0.8 = 80\%$$
Definition: Events $A$ and $B$ are called mutually exclusive if they cannot occur simultaneously, that is, $A \cap B = \emptyset$.
Formula for mutually exclusive events:$$P(A \cup B) = P(A) + P(B)$$
Definition: Events $A$ and $B$ are called compatible if they can occur simultaneously.
Formula for compatible events:$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$
Proof: The region $A \cup B$ consists of three parts:- Elements only in $A$: $P(A) - P(A \cap B)$- Elements only in $B$: $P(B) - P(A \cap B)$- Elements in both: $P(A \cap B)$
Sum: $(P(A) - P(A \cap B)) + (P(B) - P(A \cap B)) + P(A \cap B) = P(A) + P(B) - P(A \cap B)$
Example 18: The probability of rolling either a 3 or a 5 on a die.
Solution:The events are mutually exclusive (they cannot occur simultaneously).$$P(3 \text{ or } 5) = P(3) + P(5) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$$
Example 19: The probability that a randomly chosen number from 1 to 10 is divisible by 2 OR by 3.
Solution:- Divisible by 2: 2, 4, 6, 8, 10 (5 numbers)- Divisible by 3: 3, 6, 9 (3 numbers)- Divisible by 2 AND by 3: 6 (1 number)
The events are compatible (the number 6 is divisible by both numbers).$$P(2 \text{ or } 3) = \frac{5}{10} + \frac{3}{10} - \frac{1}{10} = \frac{7}{10}$$
Definition: Events $A$ and $B$ are called independent if the occurrence of one does not affect the probability of the other occurring.
Formula for independent events:$$P(A \cap B) = P(A) \cdot P(B)$$
Definition: Events $A$ and $B$ are called dependent if the occurrence of one affects the probability of the other occurring.
Conditional probability: The probability of event $B$ given that event $A$ has occurred is denoted $P(B|A)$.
Formula for dependent events:$$P(A \cap B) = P(A) \cdot P(B|A)$$
or equivalently:$$P(A \cap B) = P(B) \cdot P(A|B)$$
Example 20: The probability that a 6 is rolled both times in two rolls of a die.
Solution:The events are independent (the result of the first roll does not affect the second).$$P(\text{6 and 6}) = P(\text{6}) \cdot P(\text{6}) = \frac{1}{6} \cdot \frac{1}{6} = \frac{1}{36}$$
Example 21: An urn contains 5 red and 3 blue balls. We draw two balls without replacement. Probability that both are red.
Solution:The events are dependent (after the first draw, the number of balls changes).- $P(\text{first red}) = \frac{5}{8}$- $P(\text{second red | first red}) = \frac{4}{7}$ (7 balls remain, of which 4 are red)
$$P(\text{both red}) = \frac{5}{8} \cdot \frac{4}{7} = \frac{20}{56} = \frac{5}{14}$$
In the expansion of the binomial $(2x - \frac{1}{x})^9$, find:a) The number of termsb) The term not containing $x$c) The coefficient of $x^5$
General term of the expansion:$$T_{k+1} = \binom{9}{k} (2x)^{9-k} \left(-\frac{1}{x}\right)^k = \binom{9}{k} 2^{9-k} (-1)^k x^{9-k} \cdot x^{-k} = \binom{9}{k} 2^{9-k} (-1)^k x^{9-2k}$$
a) Number of terms: The number of terms in the expansion of $(a+b)^n$ is always $n + 1 = 9 + 1 = 10$
b) Term without $x$: We need $9 - 2k = 0$, which gives $k = 4.5$. Since $k$ must be an integer, there is no term without $x$.
c) Coefficient of $x^5$: We need $9 - 2k = 5$, which gives $k = 2$.
$$T_3 = \binom{9}{2} 2^7 (-1)^2 = 36 \cdot 128 \cdot 1 = 4608$$
Answer: a) 10 terms; b) there is no term without $x$; c) coefficient 4608
A box contains 4 white and 6 black balls. We draw one ball. Find the probability that it is:a) Whiteb) Not whitec) Red
Total balls: $n = 4 + 6 = 10$
a) White: There are 4 white balls.$$P(\text{white}) = \frac{4}{10} = \frac{2}{5} = 0.4$$
b) Not white: We use the complementary event formula.$$P(\text{not white}) = 1 - P(\text{white}) = 1 - \frac{2}{5} = \frac{3}{5} = 0.6$$
Or: the number of red and black balls is $6 + 0 = 6$.$$P(\text{not white}) = \frac{6}{10} = \frac{3}{5}$$
c) Red: The number of red balls is 0.$$P(\text{red}) = \frac{0}{10} = 0$$
In a class of 30 students: 18 study English, 15 study French, 10 study both languages. Find the probability that a randomly chosen student:a) Studies English OR Frenchb) Studies English AND Frenchc) Studies neither English nor French
a) English OR French:
We use the formula for compatible events:$$P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{18}{30} + \frac{15}{30} - \frac{10}{30} = \frac{23}{30}$$
b) English AND French:
$$P(A \cap B) = \frac{10}{30} = \frac{1}{3}$$
c) Neither English nor French:
Number of students studying at least one language: $23$ (from part a)
Number of students not studying any language: $30 - 23 = 7$
$$P(\text{neither}) = \frac{7}{30}$$
In two independent attempts, a shooter hits the target with probability 0.7. Find the probability of:a) Both hitsb) One hit (either the first or the second)c) At least one hitd) No hits
Denote: $P(A) = P(\text{hit}) = 0.7$, $P(\overline{A}) = P(\text{miss}) = 0.3$
a) Both hits:$$P(\text{both}) = 0.7 \cdot 0.7 = 0.49$$
b) One hit:This means (first hit AND second miss) OR (first miss AND second hit).
The events are mutually exclusive:$$P(\text{one}) = (0.7 \cdot 0.3) + (0.3 \cdot 0.7) = 0.21 + 0.21 = 0.42$$
c) At least one hit:
Method 1 (via complementary event):$$P(\text{at least one}) = 1 - P(\text{none}) = 1 - (0.3 \cdot 0.3) = 1 - 0.09 = 0.91$$
Method 2 (by addition):$$P(\text{at least one}) = 0.49 + 0.42 = 0.91$$
d) No hits:$$P(\text{none}) = 0.3 \cdot 0.3 = 0.09$$
Answer: a) 0.49; b) 0.42; c) 0.91; d) 0.09
In a lottery there are 100 tickets, of which 5 are winning. 3 tickets are bought. Find the probability:a) That all three are winningb) That exactly two are winningc) That exactly one is winningd) That none are winning
Total ways to choose 3 tickets out of 100: $\binom{100}{3}$
a) All three winning:
Ways to choose 3 winning out of 5: $\binom{5}{3}$
$$P(\text{all three}) = \frac{\binom{5}{3}}{\binom{100}{3}} = \frac{10}{161700} \approx 0.0000618$$
b) Exactly two winning:
Ways: choose 2 out of 5 winning AND 1 out of 95 non-winning.
$$P(\text{exactly two}) = \frac{\binom{5}{2} \cdot \binom{95}{1}}{\binom{100}{3}} = \frac{10 \cdot 95}{161700} = \frac{950}{161700} \approx 0.00588$$
c) Exactly one winning:
Ways: choose 1 out of 5 winning AND 2 out of 95 non-winning.
$$P(\text{exactly one}) = \frac{\binom{5}{1} \cdot \binom{95}{2}}{\binom{100}{3}} = \frac{5 \cdot 4465}{161700} = \frac{22325}{161700} \approx 0.1381$$
d) None winning:
Ways: choose 0 out of 5 winning AND 3 out of 95 non-winning.
$$P(\text{none}) = \frac{\binom{5}{0} \cdot \binom{95}{3}}{\binom{100}{3}} = \frac{1 \cdot 141860}{161700} \approx 0.8772$$
Verification: $0.0000618 + 0.00588 + 0.1381 + 0.8772 \approx 1.021 \approx 1$ (rounding)
### Mistake 1: Confusing arrangements and combinationsWrong: Using an arrangement when a combination is needed, and vice versa.Correct: Arrangement — if the order matters, combination — if it does not.
### Mistake 2: Forgetting the definition of the binomialWrong: $(a+b)^2 = a^2 + b^2$Correct: $(a+b)^2 = a^2 + 2ab + b^2$
### Mistake 3: Incorrect application of the probability formulaWrong: Forgetting that the outcomes must be equally likely.Correct: Make sure all outcomes have equal probability.
### Mistake 4: Confusing addition and multiplication of probabilitiesWrong: Using multiplication for mutually exclusive events.Correct: Addition for mutually exclusive events (union), multiplication for independent events (intersection).
### Mistake 5: Forgetting the complementary event formulaWrong: Trying to find $P(A)$ directly when it is easier to find $P(\overline{A})$.Correct: Use $P(A) = 1 - P(\overline{A})$ when necessary.
### Mistake 6: Incorrect application of the addition formula for compatible eventsWrong: $P(A \cup B) = P(A) + P(B)$ for compatible eventsCorrect: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$
### Mistake 7: Incorrectly identifying dependent eventsWrong: Considering events independent when they are dependent (for example, drawing balls without replacement).Correct: Check whether the occurrence of one event affects the probability of another.
### Tip 1: Memorize the formulas for combinations$$\binom{n}{0} = 1, \binom{n}{1} = n, \binom{n}{2} = \frac{n(n-1)}{2}$$
### Tip 2: Use Pascal's triangle for small values of $n$For $(a+b)^n$ with $n \leq 7$, it is often faster to use Pascal's triangle than to compute combinations.
### Tip 3: Transform expressions before expansionBefore expanding, simplify the expression:$(2x - \frac{1}{x})^9$ is better rewritten as $\frac{(2x^2 - 1)^9}{x^9}$
### Tip 4: Create a table for probability problemsCompile a table of all known probabilities and use it to compute the unknown ones.
### Tip 5: Check the sum of probabilitiesThe sum of probabilities of all mutually exclusive events must equal 1.
### Tip 6: Use the complementary event formulaWhen the probability of an event is difficult to compute, find the probability of the complementary event.
### Tip 7: Distinguish "AND" from "OR"- "AND" (intersection) → multiplication of probabilities for independent events- "OR" (union) → addition of probabilities for mutually exclusive events
### Tip 8: Read the problem statement carefullyDetermine whether the selection is made with or without replacement. This affects the independence of events.
### Tip 9: Practice with concrete examplesWork with real examples (a die, a coin, cards, balls in an urn) for better understanding.
### Tip 10: Check boundary casesWhen expanding a binomial, check the first and last terms:- First: $\binom{n}{0} a^n$- Last: $\binom{n}{n} b^n$
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Conclusion
The binomial theorem and probability theory are powerful tools that require a solid understanding of combinatorics and logic. Regular practice with a variety of problems will help you confidently solve any problem on the YÖS exam.
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