Venn Diagram, Permutation, Combination & Probability
Venn Diagram
Key Ideas
- Rectangle represents the universal set.
- Circles depict subsets and their intersections.
- Inclusion-exclusion principle counts combined set elements.
Venn Diagram is a pictorial description of set relationships where a rectangle represents the universal set and enclosed circles depict subsets.
Two-Set Venn Diagrams
- Universal Set: rectangle holding all elements.
- Subsets: circles representing individual sets.
- Intersection: overlapping region belonging to both.
- Union: combined region of all set elements.
Two-Set Union Principle
Example 1: In a class of 50 students, 35 like Algebra and 30 like Geometry. If every student likes at least one of the two subjects, how many students like both Algebra and Geometry?
Solution: Let \(x\) be the number of students who like both subjects.
Answer: 15 students
Example 2: A class of 60 students took examinations in Physics and Chemistry. If 45 passed Physics, 42 passed Chemistry, and 5 failed both subjects, how many students passed both subjects?
Solution: Total students who passed at least one subject is \(60 - 5 = 55\). Let \(x\) be the number of students who passed both subjects.
Answer: 32 students
Example 3: The probability of examinees passing Mathematics is \(\dfrac{4}{9}\) and for Science is \(\dfrac{7}{9}\). If no examinee fails both subjects and exactly 6 examinees pass both subjects, find the total number of examinees.
Solution: Let \(N\) be the total number of examinees.
Answer: 27 examinees
Three-Set Venn Diagrams
- Start with the center three-set intersection.
- Subtract center value from two-set intersections.
- Subtract overlapping values to find single-set regions.
- Sum all disjoint regions for total.
Three-Set Inclusion-Exclusion Principle
Example 4: A survey evaluated student preferences for Mathematics, Physics, and Chemistry. Results showed 65 like Mathematics, 55 like Physics, 50 like Chemistry, 35 like Math and Physics, 30 like Math and Chemistry, 25 like Physics and Chemistry, and 12 like all three subjects. How many students were surveyed?
Solution: Compute disjoint regions starting from the center intersection of all three sets:
Center (all three): 12
Math and Physics only: \(35 - 12 = 23\)
Math and Chemistry only: \(30 - 12 = 18\)
Physics and Chemistry only: \(25 - 12 = 13\)
Math only: \(65 - (23 + 18 + 12) = 12\)
Physics only: \(55 - (23 + 13 + 12) = 7\)
Chemistry only: \(50 - (18 + 13 + 12) = 7\)
Summing all regions:
Answer: 92 students
Example 5: In a survey of 800 people on brand preference, 100 prefer Brand A only, 180 prefer Brand B only, and 140 prefer Brand C only. Statistics indicate 340 prefer Brand A or B but not C, 400 prefer Brand B or C but not A, and 300 prefer Brand C or A but not B. If no person prefers all three brands, how many persons have no brand preference?
Solution: Determine pairwise intersections using single-brand counts:
Intersection A and B: \(340 - (100 + 180) = 60\)
Intersection B and C: \(400 - (180 + 140) = 80\)
Intersection C and A: \(300 - (140 + 100) = 60\)
Total with brand preference: \(100 + 180 + 140 + 60 + 80 + 60 + 0 = 620\)
Persons with no brand preference: \(800 - 620 = 180\)
Answer: 180 persons
Survey Inconsistency Verification
- Three-set overlap cannot exceed two-set overlap.
- Individual regions cannot yield negative values.
- Region total cannot exceed given sample size.
Intersection Subset Rule
Example 6: A market researcher claims that in a survey of consumers, 8 people use all three product brands A, B, and C, but only 5 people use both brands A and B. Is this survey reliable?
Solution: The set of consumers using all three brands \((A \cap B \cap C)\) is a subset of consumers using both brands A and B \((A \cap B)\). Therefore, \(n(A \cap B \cap C) \le n(A \cap B)\). Since \(8 > 5\), the claim is logically impossible.
Answer: No, the survey is unreliable
Permutation and Combination
Key Ideas
- Counting principles determine total arrangements or selections.
- Permutations account for specific order of elements.
- Combinations select objects without regard to order.
Permutation and Combination provide systematic mathematical principles for counting the number of possible outcomes, arrangements, or selections from a set of objects.
Fundamental Principle of Counting
- Sequential Operations: Multiply number of ways for each step.
- Independent Choices: Total outcomes equal the product of choices.
- Multi-Stage Decision: Applies when events occur successively.
Fundamental Counting Principle
Example 1: An engineering store sells 8 different models of calculators, each available in 2 screen colors, 3 casing materials, and 2 battery types. How many distinct calculator variants are available?
Solution: Apply the fundamental counting principle across all independent choices:
Answer: 96 variants
Example 2: There are 4 main highways connecting City A to City B, and 3 highways connecting City B to City C. In how many ways can a driver travel from A to C and return to A without using any highway more than once?
Solution: Calculate options for each leg of the round trip without repeating roads:
A to B: 4 ways; B to C: 3 ways; C to B (return): 2 ways; B to A (return): 3 ways.
Answer: 72 ways
Permutations
- Order Matters: Arrangement sequence changes the outcome.
- Linear Arrangement: Lining up distinct items in a row.
- Factorial Rule: Arranging all \(n\) distinct items equals \(n!\).
- Partial Selection: Arranging \(r\) items chosen from \(n\) distinct items.
Permutation Formula
Example 3: How many distinct 4-letter code words without repeating any letter can be formed from the 6 letters in the word MASTER?
Solution: Calculate permutations of \(n = 6\) taken \(r = 4\) at a time:
Answer: 360 code words
Example 4: In how many ways can a board of 12 directors elect a President, Vice President, Secretary, and Treasurer if no member can hold more than one office?
Solution: Electing 4 distinct officer positions from 12 candidates where position order matters:
Answer: 11,880 ways
Permutations with Restrictions and Grouping
- Unit Grouping: Treat items that must stay together as 1 unit.
- Internal Arrangement: Multiply by internal permutations of the grouped unit.
- Alternating Pattern: Arrange separate groups in alternating slots.
- Position Constraint: Fill restricted positions first before open slots.
Grouped Permutation Rule
Example 5: Four distinct mathematics books, 3 distinct physics books, and 2 distinct chemistry books are arranged on a shelf. In how many ways can they be arranged if books of the same subject must stay together?
Solution: Treat the 3 subject categories as 3 units. Arrange units in \(3!\) ways, then arrange books within each category:
Answer: 1,728 ways
Example 6: In how many ways can 4 men and 4 women be seated in a row of 8 chairs such that men and women alternate?
Solution: Alternating pattern can start with a man or a woman (2 starting arrangements). For each pattern, arrange men in \(4!\) ways and women in \(4!\) ways:
Answer: 1,152 ways
Example 7: How many 4-letter words starting and ending with a vowel without letter repetition can be formed from the word CANDIDATE?
Solution: Distinct vowels available are A, I, E (3 vowels). Distinct consonants available are C, N, D, T (4 consonants). Fill 1st position with vowel (3 choices), 4th position with remaining vowel (2 choices). Fill 2nd and 3rd positions with remaining available letters (5 remaining letters taken 2 at a time):
Answer: 120 words
Distinguishable Permutations
- Identical Elements: Objects of the same type are indistinguishable.
- Division by Factorials: Divide total factorial by factorials of identical counts.
- Partitioning Sets: Distributing identical items into distinct groups.
Distinguishable Permutation Formula
Example 8: How many distinct permutations can be made from the letters of the word PHILIPPINES?
Solution: Total letters \(n = 11\). Counts of repeated letters: P appears 3 times, I appears 3 times, E appears 1 time, H, L, N, S appear 1 time each.
Answer: 1,108,800 permutations
Example 9: In how many ways can 2 identical balls, 4 identical dolls, and 6 identical toy cars be distributed among 12 children if each child receives exactly one toy?
Solution: Distributable arrangements of \(n = 12\) toys with repeated identical items of 2, 4, and 6:
Answer: 13,860 ways
Circular Permutations
- No Fixed Beginning: Rotating seats does not create a new seating.
- Fixed Anchor: Fix one object position to establish relative order.
- Cyclic Formula: Permutation of \(n\) items in a circle is \((n - 1)!\).
Circular Permutation Formula
Example 10: How many different ways can 6 delegates be seated around a circular conference table?
Solution: Apply circular permutation formula for \(n = 6\):
Answer: 120 ways
Example 11: How many ways can 5 men and 5 women form a circle with men and women sitting in alternating seats?
Solution: Fix 1 man seat in the circle. Arrange remaining 4 men in \((5-1)! = 4!\) ways. The 5 women then occupy the 5 alternating seats between men in \(5!\) ways:
Answer: 2,880 ways
Combinations
- Order Irrelevant: Selection sequence does not affect the outcome.
- Unordered Subsets: Choosing \(r\) objects from \(n\) distinct items.
- Combinatorial Relation: Combination equals permutation divided by \(r!\).
Combination Formula
Example 12: In how many ways can a committee of 5 members be selected from an organization of 20 members?
Solution: Calculate combination of \(n = 20\) taken \(r = 5\) at a time:
Answer: 15,504 ways
Example 13: A sports league has 10 teams. If each team plays every other team exactly once, what is the total number of games played?
Solution: Each game requires selecting 2 distinct teams from 10 teams regardless of order:
Answer: 45 games
Example 14: How many distinct line segments can be formed by connecting 15 distinct points on a plane, no three of which are collinear?
Solution: Each line segment is formed by choosing 2 points from 15 points:
Answer: 105 line segments
Compound Combinations
- Multi-Group Selection: Selecting subsets independently from separate sets.
- Product Rule: Multiply combinations from each category together.
- Joint Outcomes: Combines independent selections into total choices.
Compound Combination Principle
Example 15: A company needs to hire 5 civil engineers and 3 electrical engineers. In how many ways can the choices be made from 8 qualified civil engineers and 5 qualified electrical engineers?
Solution: Calculate combinations for each engineering field and multiply:
Answer: 560 ways
Combinations Taken Any Number at a Time
- Sum of Combinations: Taking 1, 2, 3, up to \(n\) items.
- Binary Power Formula: Total combinations equals \(2^n - 1\).
- Exclusion of Empty Set: Subtract 1 to exclude choosing zero items.
Total Combinations Formula
Example 16: In how many ways can a person invite one or more of 6 friends to a dinner party?
Solution: Apply total combination formula for \(n = 6\) distinct friends:
Answer: 63 ways
Probability
Key Ideas
- Probability quantifies the likelihood of an event occurring.
- Values range from 0 (impossible) to 1 (certain).
- Combined events use addition, multiplication, or binomial principles.
Probability measures the likelihood that a specific event will happen, expressed as the ratio of favorable outcomes to total possible outcomes in a sample space.
Single Event Probability
- Favorable Outcomes: number of ways an event succeeds.
- Sample Space: total number of possible equal outcomes.
- Complementary Event: probability that an event fails to occur.
- Probability Sum Rule: sum of success and failure probabilities equals 1.
Single Event Probability Formula
Example 1: A single die is rolled once. What is the probability of obtaining a number greater than 4?
Solution: The sample space is \(\{1, 2, 3, 4, 5, 6\}\), so \(T = 6\). Numbers greater than 4 are 5 and 6, so \(S = 2\).
Answer: 1/3
Example 2: In a single draw from a standard pack of 52 playing cards, what is the probability of drawing an Ace?
Solution: Total cards \(T = 52\). Number of Aces \(S = 4\).
Answer: 1/13
Complementary Probability Formula
Example 3: An urn contains 5 red balls, 4 green balls, and 6 blue balls. If one ball is drawn at random, what is the probability of not drawing a red ball?
Solution: Total balls \(T = 5 + 4 + 6 = 15\). Number of non-red balls \(S = 4 + 6 = 10\).
Answer: 2/3
Mutually Exclusive and Combined Events
- Mutually Exclusive Events: events that cannot occur at the same time.
- Addition Rule: sum individual probabilities for mutually exclusive events.
- Non-Mutually Exclusive Events: overlapping events require subtracting intersection.
Addition Rule for Mutually Exclusive Events
Example 4: Two fair dice are rolled simultaneously. What is the probability that the sum of the numbers shown is either 10 or 12?
Solution: Total outcomes \(T = 6 \times 6 = 36\). Outcomes for sum of 10: \((4,6), (5,5), (6,4)\) (3 ways). Outcomes for sum of 12: \((6,6)\) (1 way). Since the sums are mutually exclusive:
Answer: 1/9
General Addition Rule for Overlapping Events
Example 5: A card is drawn at random from a standard deck of 52 playing cards. What is the probability of drawing either a King or a Spade?
Solution: Number of Kings \(n(K) = 4\), number of Spades \(n(S) = 13\), and King of Spades \(n(K \cap S) = 1\).
Answer: 4/13
Independent and Dependent Events
- Independent Events: occurrence of one event does not affect another.
- Dependent Events: occurrence of one event changes the probability of another.
- Multiplication Rule: multiply probabilities for joint occurrence of events.
- Sampling With Replacement: preserves constant probabilities across draws.
- Sampling Without Replacement: alters available total and favorable outcomes.
Multiplication Rule for Independent Events
Example 6: A bag contains 4 white balls and 5 black balls. Two balls are drawn one at a time with replacement. What is the probability that both balls drawn are white?
Solution: Total balls \(T = 9\). Probability of white on 1st draw \(P_1 = \dfrac{4}{9}\). Since the ball is replaced, probability of white on 2nd draw \(P_2 = \dfrac{4}{9}\).
Answer: 16/81
Multiplication Rule for Dependent Events
Example 7: A box contains 3 white balls and 5 black balls. If two balls are drawn in succession without replacement, what is the probability that both balls are black?
Solution: Probability of black on 1st draw \(P_1 = \dfrac{5}{8}\). After drawing one black ball, 4 black balls and 7 total balls remain. Probability of black on 2nd draw \(P_2 = \dfrac{4}{7}\).
Answer: 5/14
Example 8: The reliability of the first stage of a two-stage rocket is 0.96, and the overall reliability of both stages functioning correctly is 0.912. What is the reliability of the second stage?
Solution: Let \(P_1 = 0.96\) and total \(P_{\text{total}} = 0.912\). Since stages operate in series:
Answer: 0.95
Repeated Trials and Binomial Probability
- Fixed Trials: experiment consists of \(n\) independent identical trials.
- Two Outcomes: each trial results in either success (\(p\)) or failure (\(q\)).
- Constant Probability: value of \(p\) remains constant throughout trials.
Binomial Probability Formula
Example 9: A fair coin is tossed 5 times. What is the probability of getting exactly 3 heads and 2 tails?
Solution: Number of trials \(n = 5\), successful trials \(r = 3\), \(p = \dfrac{1}{2}\), and \(q = \dfrac{1}{2}\).
Answer: 5/16
Example 10: In a 10-question multiple-choice test, each question has 5 choices with only 1 correct answer. What is the probability of getting exactly 6 questions correct by pure guesswork?
Solution: \(n = 10\), \(r = 6\), \(p = \dfrac{1}{5} = 0.20\), and \(q = \dfrac{4}{5} = 0.80\).
Answer: 0.55%
Example 11: An examinee has an \(80\%\) chance of passing each subject in a licensure board exam. What is the probability that the examinee passes at least 2 out of 3 subjects?
Solution: \(n = 3\), \(p = 0.8\), \(q = 0.2\). “At least 2” means exactly 2 or exactly 3 passing subjects.
Answer: 89.6%
The “At Least One” Rule
- Complementary Strategy: calculate probability of total failure first.
- Total Failure: event fails in all \(n\) independent trials.
- Subtraction Principle: subtract total failure probability from 1.
At Least One Success Formula
Example 12: The probability of obtaining credit in an examination is \(\dfrac{1}{3}\). If 3 students take the examination, what is the probability that at least one student obtains credit?
Solution: Failure probability for each student \(q = 1 - \dfrac{1}{3} = \dfrac{2}{3}\). Total failure for all 3 students \(Q = \left(\dfrac{2}{3}\right)^3 = \dfrac{8}{27}\).
Answer: 19/27
Example 13: A missile has an \(80\%\) probability of missing its target on any single shot. How many missiles must be fired to ensure at least an \(80\%\) chance of hitting the target?
Solution: Probability of hit \(p = 0.20\) or miss \(q = 0.80\). Set \(P(\text{at least 1 hit}) \ge 0.80\):
Rounding up to the nearest whole number:
Answer: 8 missiles
Expected Value
- Weighted Average: average expected outcome over many repetitions.
- Payoff Product: multiply each outcome value by its probability.
- Fair Game: expected net winning value equals zero.
Expected Value Formula
Example 14: In a game using a fair six-sided die, a player wins 20 pesos if a 1 or 6 turns up, and loses 10 pesos if any other face turns up. What is the expected winning per roll?
Solution: Probability of winning \(P(\text{win}) = \dfrac{2}{6} = \dfrac{1}{3}\) with payoff \(+20\). Probability of losing \(P(\text{lose}) = \dfrac{4}{6} = \dfrac{2}{3}\) with payoff \(-10\).
Answer: 0 pesos (Fair game)
Formula Summary
| Formula Name | Formula |
|---|---|
| Two-Set Union Principle | \(n(A \cup B) = n(A) + n(B) - n(A \cap B)\) |
| Three-Set Inclusion-Exclusion Principle | \(n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(C \cap A) + n(A \cap B \cap C)\) |
| Intersection Subset Rule | \(n(A \cap B \cap C) \le n(A \cap B)\) |
| Fundamental Counting Principle | \(N = m \cdot n \cdot p\) |
| Permutation Formula | \({}_n P_r = \dfrac{n!}{(n - r)!}\) |
| Grouped Permutation Rule | \(N = N_{\text{groups}}! \times n_1! \times n_2!\) |
| Distinguishable Permutation Formula | \(P = \dfrac{n!}{q! \cdot r! \cdot s!}\) |
| Circular Permutation Formula | \(P_{\text{circular}} = (n - 1)!\) |
| Combination Formula | \({}_n C_r = \dfrac{n!}{(n - r)! \cdot r!}\) |
| Compound Combination Principle | \(N = {}_{n_1} C_{r_1} \times {}_{n_2} C_{r_2}\) |
| Total Combinations Formula | \(C_{\text{total}} = 2^n - 1\) |
| Single Event Probability Formula | \(P(E) = \dfrac{S}{T} = \dfrac{h}{h + f}\) |
| Complementary Probability Formula | \(P(\text{not } E) = 1 - P(E)\) |
| Addition Rule for Mutually Exclusive Events | \(P(A \cup B) = P(A) + P(B)\) |
| General Addition Rule for Overlapping Events | \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\) |
| Multiplication Rule for Independent Events | \(P(A \cap B) = P(A) \cdot P(B)\) |
| Multiplication Rule for Dependent Events | \(P(A \cap B) = P(A) \cdot P(B|A)\) |
| Binomial Probability Formula | \(P(r) = {}_n C_r \cdot p^r \cdot q^{n-r}\) |
| At Least One Success Formula | \(P(\text{at least 1}) = 1 - q^n\) |
| Expected Value Formula | \(E(X) = \sum x_i \cdot P(x_i)\) |