Probability - Ninth Grade Math
Introduction to Probability
Probability: The measure of likelihood that an event will occur
Range: $0 \leq P(E) \leq 1$ or 0% to 100%
Experiment: An action or process that produces outcomes
Outcome: A possible result of an experiment
Sample Space (S): The set of all possible outcomes
Event (E): A specific outcome or set of outcomes
Range: $0 \leq P(E) \leq 1$ or 0% to 100%
Experiment: An action or process that produces outcomes
Outcome: A possible result of an experiment
Sample Space (S): The set of all possible outcomes
Event (E): A specific outcome or set of outcomes
Probability Scale:
• $P(E) = 0$: Impossible event (never happens)
• $P(E) = 0.5$: Equally likely (50-50 chance)
• $P(E) = 1$: Certain event (always happens)
• Closer to 1: More likely
• Closer to 0: Less likely
• $P(E) = 0$: Impossible event (never happens)
• $P(E) = 0.5$: Equally likely (50-50 chance)
• $P(E) = 1$: Certain event (always happens)
• Closer to 1: More likely
• Closer to 0: Less likely
1. Theoretical Probability
Theoretical Probability: Probability based on mathematical reasoning and analysis
Assumes: All outcomes are equally likely
Does NOT require: Actually performing the experiment
Assumes: All outcomes are equally likely
Does NOT require: Actually performing the experiment
Theoretical Probability Formula:
$$P(E) = \frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Possible Outcomes}}$$
Or:
$$P(E) = \frac{n(E)}{n(S)}$$
where:
• $n(E)$ = number of ways event E can occur
• $n(S)$ = total number of possible outcomes in sample space
$$P(E) = \frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Possible Outcomes}}$$
Or:
$$P(E) = \frac{n(E)}{n(S)}$$
where:
• $n(E)$ = number of ways event E can occur
• $n(S)$ = total number of possible outcomes in sample space
Example 1: What is the probability of rolling a 4 on a fair die?
Sample space: {1, 2, 3, 4, 5, 6} → 6 possible outcomes
Favorable outcomes: {4} → 1 outcome
$$P(\text{rolling 4}) = \frac{1}{6}$$
Answer: $\frac{1}{6}$ ≈ 0.167 or about 16.7%
Sample space: {1, 2, 3, 4, 5, 6} → 6 possible outcomes
Favorable outcomes: {4} → 1 outcome
$$P(\text{rolling 4}) = \frac{1}{6}$$
Answer: $\frac{1}{6}$ ≈ 0.167 or about 16.7%
Example 2: Probability of drawing a heart from a standard deck of cards
Total cards: 52
Hearts in deck: 13
$$P(\text{heart}) = \frac{13}{52} = \frac{1}{4} = 0.25$$
Answer: $\frac{1}{4}$ or 25%
Total cards: 52
Hearts in deck: 13
$$P(\text{heart}) = \frac{13}{52} = \frac{1}{4} = 0.25$$
Answer: $\frac{1}{4}$ or 25%
Example 3: Probability of flipping two heads in a row
Sample space: {HH, HT, TH, TT} → 4 outcomes
Favorable: {HH} → 1 outcome
$$P(\text{two heads}) = \frac{1}{4}$$
Answer: $\frac{1}{4}$ or 25%
Sample space: {HH, HT, TH, TT} → 4 outcomes
Favorable: {HH} → 1 outcome
$$P(\text{two heads}) = \frac{1}{4}$$
Answer: $\frac{1}{4}$ or 25%
2. Experimental Probability
Experimental Probability: Probability based on actual experiments and observations
Also called: Empirical probability
Based on: What actually happened in trials
Requires: Conducting actual experiments
Also called: Empirical probability
Based on: What actually happened in trials
Requires: Conducting actual experiments
Experimental Probability Formula:
$$P(E) = \frac{\text{Number of Times Event Occurred}}{\text{Total Number of Trials}}$$
Key Point: More trials → more accurate results (closer to theoretical)
$$P(E) = \frac{\text{Number of Times Event Occurred}}{\text{Total Number of Trials}}$$
Key Point: More trials → more accurate results (closer to theoretical)
Example 1: A coin is flipped 50 times. Heads appeared 28 times.
Experimental probability of heads:
$$P(\text{heads}) = \frac{28}{50} = \frac{14}{25} = 0.56$$
Answer: 0.56 or 56%
Compare to theoretical: $P(\text{heads}) = \frac{1}{2} = 0.50$ or 50%
Note: Experimental is close but not exactly the same
Experimental probability of heads:
$$P(\text{heads}) = \frac{28}{50} = \frac{14}{25} = 0.56$$
Answer: 0.56 or 56%
Compare to theoretical: $P(\text{heads}) = \frac{1}{2} = 0.50$ or 50%
Note: Experimental is close but not exactly the same
Example 2: A spinner is spun 100 times with these results:
Experimental probabilities:
$P(\text{Red}) = \frac{32}{100} = 0.32$ or 32%
$P(\text{Blue}) = \frac{28}{100} = 0.28$ or 28%
$P(\text{Green}) = \frac{24}{100} = 0.24$ or 24%
$P(\text{Yellow}) = \frac{16}{100} = 0.16$ or 16%
Color | Red | Blue | Green | Yellow |
---|---|---|---|---|
Times | 32 | 28 | 24 | 16 |
Experimental probabilities:
$P(\text{Red}) = \frac{32}{100} = 0.32$ or 32%
$P(\text{Blue}) = \frac{28}{100} = 0.28$ or 28%
$P(\text{Green}) = \frac{24}{100} = 0.24$ or 24%
$P(\text{Yellow}) = \frac{16}{100} = 0.16$ or 16%
Theoretical vs Experimental Probability
Key Differences:
Law of Large Numbers: As number of trials increases, experimental probability approaches theoretical probability
Aspect | Theoretical | Experimental |
---|---|---|
Based on | Mathematical reasoning | Actual trials/experiments |
Formula | Favorable/Total possible | Occurred/Total trials |
Requires | Knowledge of outcomes | Performing experiment |
Result | What should happen | What did happen |
Example | P(heads) = 1/2 | P(heads) = 12/20 in 20 flips |
Law of Large Numbers: As number of trials increases, experimental probability approaches theoretical probability
3-4. Probabilities Using Two-Way Frequency Tables
Two-Way Frequency Table: A table showing frequencies of two categorical variables
Also called: Contingency table
Rows: One variable
Columns: Second variable
Cells: Show frequency of each combination
Also called: Contingency table
Rows: One variable
Columns: Second variable
Cells: Show frequency of each combination
Example 1: Survey of 100 students about favorite subject
Q: What is P(student likes Math)?
$$P(\text{Math}) = \frac{38}{100} = 0.38 \text{ or } 38\%$$
Q: What is P(student is a Girl)?
$$P(\text{Girl}) = \frac{55}{100} = 0.55 \text{ or } 55\%$$
Q: What is P(Girl AND Math)?
$$P(\text{Girl and Math}) = \frac{18}{100} = 0.18 \text{ or } 18\%$$
Math | Science | English | Total | |
---|---|---|---|---|
Boys | 20 | 15 | 10 | 45 |
Girls | 18 | 22 | 15 | 55 |
Total | 38 | 37 | 25 | 100 |
Q: What is P(student likes Math)?
$$P(\text{Math}) = \frac{38}{100} = 0.38 \text{ or } 38\%$$
Q: What is P(student is a Girl)?
$$P(\text{Girl}) = \frac{55}{100} = 0.55 \text{ or } 55\%$$
Q: What is P(Girl AND Math)?
$$P(\text{Girl and Math}) = \frac{18}{100} = 0.18 \text{ or } 18\%$$
Conditional Probability with Two-Way Tables
Conditional Probability: Probability of event A given that event B has occurred
Notation: $P(A|B)$ read as "probability of A given B"
Key: The condition limits the sample space
Notation: $P(A|B)$ read as "probability of A given B"
Key: The condition limits the sample space
Conditional Probability Formula:
$$P(A|B) = \frac{P(A \text{ and } B)}{P(B)}$$
Or using frequencies:
$$P(A|B) = \frac{\text{Frequency of A and B}}{\text{Frequency of B}}$$
In words: Look only at the row/column of the condition, then find the desired outcome within that row/column
$$P(A|B) = \frac{P(A \text{ and } B)}{P(B)}$$
Or using frequencies:
$$P(A|B) = \frac{\text{Frequency of A and B}}{\text{Frequency of B}}$$
In words: Look only at the row/column of the condition, then find the desired outcome within that row/column
Example 2: Using the table from Example 1
Q: What is P(Math | Boy)? "Probability student likes Math GIVEN student is a Boy"
Method 1 - Direct from table:
Look only at Boy row: 20 like Math out of 45 total boys
$$P(\text{Math}|\text{Boy}) = \frac{20}{45} = \frac{4}{9} \approx 0.44$$
Method 2 - Using formula:
$$P(\text{Math}|\text{Boy}) = \frac{P(\text{Math and Boy})}{P(\text{Boy})} = \frac{20/100}{45/100} = \frac{20}{45} = \frac{4}{9}$$
Answer: $\frac{4}{9}$ ≈ 44.4%
Q: What is P(Math | Boy)? "Probability student likes Math GIVEN student is a Boy"
Method 1 - Direct from table:
Look only at Boy row: 20 like Math out of 45 total boys
$$P(\text{Math}|\text{Boy}) = \frac{20}{45} = \frac{4}{9} \approx 0.44$$
Method 2 - Using formula:
$$P(\text{Math}|\text{Boy}) = \frac{P(\text{Math and Boy})}{P(\text{Boy})} = \frac{20/100}{45/100} = \frac{20}{45} = \frac{4}{9}$$
Answer: $\frac{4}{9}$ ≈ 44.4%
Example 3: From same table
Q: P(Girl | Science)? "Probability student is a Girl GIVEN student likes Science"
Look only at Science column: 22 girls out of 37 total who like Science
$$P(\text{Girl}|\text{Science}) = \frac{22}{37} \approx 0.59$$
Answer: $\frac{22}{37}$ ≈ 59.5%
Q: P(Girl | Science)? "Probability student is a Girl GIVEN student likes Science"
Look only at Science column: 22 girls out of 37 total who like Science
$$P(\text{Girl}|\text{Science}) = \frac{22}{37} \approx 0.59$$
Answer: $\frac{22}{37}$ ≈ 59.5%
5. Outcomes of Compound Events
Simple Event: Single outcome
Compound Event: Combination of two or more simple events
Types: AND events, OR events
Compound Event: Combination of two or more simple events
Types: AND events, OR events
Ways to Find All Outcomes
Method 1: Tree Diagram
• Shows all possible outcomes branching from each choice
• Good for visualizing compound events
Method 2: List
• Write out all possible combinations systematically
Method 3: Table/Grid
• Create table with outcomes of each event
• Good for two events
• Shows all possible outcomes branching from each choice
• Good for visualizing compound events
Method 2: List
• Write out all possible combinations systematically
Method 3: Table/Grid
• Create table with outcomes of each event
• Good for two events
Example 1: Flipping a coin and rolling a die
Sample Space using list:
{H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}
Total outcomes: 2 × 6 = 12
Q: P(Heads and even number)?
Favorable outcomes: {H2, H4, H6} = 3 outcomes
$$P(\text{H and even}) = \frac{3}{12} = \frac{1}{4}$$
Sample Space using list:
{H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}
Total outcomes: 2 × 6 = 12
Q: P(Heads and even number)?
Favorable outcomes: {H2, H4, H6} = 3 outcomes
$$P(\text{H and even}) = \frac{3}{12} = \frac{1}{4}$$
Example 2: Choosing outfit: 3 shirts (R, B, G) and 2 pants (J, K)
Tree Diagram outcomes:
R-J, R-K, B-J, B-K, G-J, G-K
Total outcomes: 3 × 2 = 6
Tree Diagram outcomes:
R-J, R-K, B-J, B-K, G-J, G-K
Total outcomes: 3 × 2 = 6
6-7. Independent and Dependent Events
Independent Events: The outcome of one event does NOT affect the other
Dependent Events: The outcome of one event DOES affect the other
Dependent Events: The outcome of one event DOES affect the other
Identifying Independent vs Dependent
Independent Events Examples:
• Flipping coin twice (first flip doesn't affect second)
• Rolling two dice (one die doesn't affect the other)
• Spinning spinner twice
• Sampling WITH replacement
Dependent Events Examples:
• Drawing cards without replacement (first draw affects second)
• Choosing students without replacement
• Taking items from a bag without putting back
• Sampling WITHOUT replacement
• Flipping coin twice (first flip doesn't affect second)
• Rolling two dice (one die doesn't affect the other)
• Spinning spinner twice
• Sampling WITH replacement
Dependent Events Examples:
• Drawing cards without replacement (first draw affects second)
• Choosing students without replacement
• Taking items from a bag without putting back
• Sampling WITHOUT replacement
Example 1: Identify if independent or dependent
A: Flip coin, then roll die
→ Independent (coin doesn't affect die)
B: Draw card, keep it, draw another
→ Dependent (first draw changes what's left)
C: Choose marble, replace it, choose again
→ Independent (replacement restores original situation)
D: Choose 2 students from class for a team
→ Dependent (can't choose same person twice)
A: Flip coin, then roll die
→ Independent (coin doesn't affect die)
B: Draw card, keep it, draw another
→ Dependent (first draw changes what's left)
C: Choose marble, replace it, choose again
→ Independent (replacement restores original situation)
D: Choose 2 students from class for a team
→ Dependent (can't choose same person twice)
Probability Formulas
Independent Events:
$$P(A \text{ and } B) = P(A) \times P(B)$$
Multiply the individual probabilities
Dependent Events:
$$P(A \text{ and } B) = P(A) \times P(B|A)$$
where $P(B|A)$ = probability of B after A has occurred
Key: Second probability changes based on first outcome
$$P(A \text{ and } B) = P(A) \times P(B)$$
Multiply the individual probabilities
Dependent Events:
$$P(A \text{ and } B) = P(A) \times P(B|A)$$
where $P(B|A)$ = probability of B after A has occurred
Key: Second probability changes based on first outcome
Example 2: Independent events
Q: Flip coin twice. P(two heads)?
$P(\text{H on first}) = \frac{1}{2}$
$P(\text{H on second}) = \frac{1}{2}$ (independent!)
$$P(\text{HH}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$$
Answer: $\frac{1}{4}$ or 25%
Q: Flip coin twice. P(two heads)?
$P(\text{H on first}) = \frac{1}{2}$
$P(\text{H on second}) = \frac{1}{2}$ (independent!)
$$P(\text{HH}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$$
Answer: $\frac{1}{4}$ or 25%
Example 3: Dependent events
Bag has 5 red and 3 blue marbles. Draw 2 without replacement. P(both red)?
$P(\text{1st red}) = \frac{5}{8}$
After removing 1 red: 4 red, 3 blue, 7 total remain
$P(\text{2nd red}|\text{1st red}) = \frac{4}{7}$
$$P(\text{both red}) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}$$
Answer: $\frac{5}{14}$ ≈ 35.7%
Bag has 5 red and 3 blue marbles. Draw 2 without replacement. P(both red)?
$P(\text{1st red}) = \frac{5}{8}$
After removing 1 red: 4 red, 3 blue, 7 total remain
$P(\text{2nd red}|\text{1st red}) = \frac{4}{7}$
$$P(\text{both red}) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}$$
Answer: $\frac{5}{14}$ ≈ 35.7%
8. Counting Principle
Fundamental Counting Principle: If one event can occur in m ways and another in n ways, then both can occur in m × n ways
Key Word: AND means multiply
Key Word: AND means multiply
Counting Principle Formula:
For multiple events:
$$\text{Total Outcomes} = n_1 \times n_2 \times n_3 \times ... \times n_k$$
where $n_i$ = number of choices for each event
Steps:
1. Count choices for first event
2. Count choices for second event
3. Count choices for each remaining event
4. Multiply all together
For multiple events:
$$\text{Total Outcomes} = n_1 \times n_2 \times n_3 \times ... \times n_k$$
where $n_i$ = number of choices for each event
Steps:
1. Count choices for first event
2. Count choices for second event
3. Count choices for each remaining event
4. Multiply all together
Example 1: Restaurant menu
Choose: 1 appetizer (4 choices), 1 main (6 choices), 1 dessert (3 choices)
Total meals possible:
$$4 \times 6 \times 3 = 72$$
Answer: 72 different meal combinations
Choose: 1 appetizer (4 choices), 1 main (6 choices), 1 dessert (3 choices)
Total meals possible:
$$4 \times 6 \times 3 = 72$$
Answer: 72 different meal combinations
Example 2: License plate with 2 letters followed by 3 digits
Letters: 26 choices each → $26 \times 26 = 676$
Digits: 10 choices each → $10 \times 10 \times 10 = 1000$
Total plates:
$$26 \times 26 \times 10 \times 10 \times 10 = 676,000$$
Answer: 676,000 possible plates
Letters: 26 choices each → $26 \times 26 = 676$
Digits: 10 choices each → $10 \times 10 \times 10 = 1000$
Total plates:
$$26 \times 26 \times 10 \times 10 \times 10 = 676,000$$
Answer: 676,000 possible plates
9-10. Permutations and Combinations
Permutation: Arrangement where ORDER MATTERS
Combination: Selection where ORDER DOES NOT MATTER
Key Question: Does the order of selection matter?
Combination: Selection where ORDER DOES NOT MATTER
Key Question: Does the order of selection matter?
Permutations
Permutation Formula:
Notation: $_nP_r$ or $P(n, r)$
Read as: "n permute r" or "permutations of n things taken r at a time"
$$_nP_r = \frac{n!}{(n-r)!}$$
where:
• $n$ = total number of items
• $r$ = number of items being arranged
• $n!$ (factorial) = $n \times (n-1) \times (n-2) \times ... \times 2 \times 1$
Special Case: Permutation of all n items
$$_nP_n = n!$$
Notation: $_nP_r$ or $P(n, r)$
Read as: "n permute r" or "permutations of n things taken r at a time"
$$_nP_r = \frac{n!}{(n-r)!}$$
where:
• $n$ = total number of items
• $r$ = number of items being arranged
• $n!$ (factorial) = $n \times (n-1) \times (n-2) \times ... \times 2 \times 1$
Special Case: Permutation of all n items
$$_nP_n = n!$$
Factorial Values:
$0! = 1$ (by definition)
$1! = 1$
$2! = 2 \times 1 = 2$
$3! = 3 \times 2 \times 1 = 6$
$4! = 4 \times 3 \times 2 \times 1 = 24$
$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$
$0! = 1$ (by definition)
$1! = 1$
$2! = 2 \times 1 = 2$
$3! = 3 \times 2 \times 1 = 6$
$4! = 4 \times 3 \times 2 \times 1 = 24$
$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$
Example 1: How many ways can 5 students line up?
Order matters (different positions in line)
This is a permutation of all 5
$$_5P_5 = 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$$
Answer: 120 ways
Order matters (different positions in line)
This is a permutation of all 5
$$_5P_5 = 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$$
Answer: 120 ways
Example 2: 8 runners in a race. How many ways can 1st, 2nd, 3rd place finish?
Order matters (1st place ≠ 2nd place)
Choose 3 from 8 where order matters
$$_8P_3 = \frac{8!}{(8-3)!} = \frac{8!}{5!} = \frac{8 \times 7 \times 6 \times 5!}{5!} = 8 \times 7 \times 6 = 336$$
Answer: 336 ways
Order matters (1st place ≠ 2nd place)
Choose 3 from 8 where order matters
$$_8P_3 = \frac{8!}{(8-3)!} = \frac{8!}{5!} = \frac{8 \times 7 \times 6 \times 5!}{5!} = 8 \times 7 \times 6 = 336$$
Answer: 336 ways
Combinations
Combination Formula:
Notation: $_nC_r$ or $C(n, r)$ or $\binom{n}{r}$
Read as: "n choose r" or "combinations of n things taken r at a time"
$$_nC_r = \frac{n!}{r!(n-r)!}$$
where:
• $n$ = total number of items
• $r$ = number of items being selected
Relationship to Permutations:
$$_nC_r = \frac{_nP_r}{r!}$$
Combinations are always less than or equal to permutations
Notation: $_nC_r$ or $C(n, r)$ or $\binom{n}{r}$
Read as: "n choose r" or "combinations of n things taken r at a time"
$$_nC_r = \frac{n!}{r!(n-r)!}$$
where:
• $n$ = total number of items
• $r$ = number of items being selected
Relationship to Permutations:
$$_nC_r = \frac{_nP_r}{r!}$$
Combinations are always less than or equal to permutations
Example 3: Choose 3 students from 10 for a committee
Order does NOT matter (same 3 people = same committee)
This is a combination
$$_{10}C_3 = \frac{10!}{3!(10-3)!} = \frac{10!}{3! \times 7!}$$
$$= \frac{10 \times 9 \times 8 \times 7!}{3 \times 2 \times 1 \times 7!} = \frac{10 \times 9 \times 8}{6} = \frac{720}{6} = 120$$
Answer: 120 ways
Order does NOT matter (same 3 people = same committee)
This is a combination
$$_{10}C_3 = \frac{10!}{3!(10-3)!} = \frac{10!}{3! \times 7!}$$
$$= \frac{10 \times 9 \times 8 \times 7!}{3 \times 2 \times 1 \times 7!} = \frac{10 \times 9 \times 8}{6} = \frac{720}{6} = 120$$
Answer: 120 ways
Example 4: Choose 2 toppings from 8 for a pizza
Order doesn't matter (pepperoni + mushroom = mushroom + pepperoni)
$$_8C_2 = \frac{8!}{2!(8-2)!} = \frac{8!}{2! \times 6!} = \frac{8 \times 7}{2 \times 1} = \frac{56}{2} = 28$$
Answer: 28 combinations
Order doesn't matter (pepperoni + mushroom = mushroom + pepperoni)
$$_8C_2 = \frac{8!}{2!(8-2)!} = \frac{8!}{2! \times 6!} = \frac{8 \times 7}{2 \times 1} = \frac{56}{2} = 28$$
Answer: 28 combinations
Permutation vs Combination Decision Guide
Use PERMUTATION when:
• Order matters
• Different arrangements count as different outcomes
• Keywords: arrange, order, rank, schedule, first/second/third
• Examples: Race finishes, passwords, seating arrangements
Use COMBINATION when:
• Order doesn't matter
• Only the group/selection matters
• Keywords: choose, select, committee, group, team
• Examples: Committee selection, pizza toppings, lottery numbers
• Order matters
• Different arrangements count as different outcomes
• Keywords: arrange, order, rank, schedule, first/second/third
• Examples: Race finishes, passwords, seating arrangements
Use COMBINATION when:
• Order doesn't matter
• Only the group/selection matters
• Keywords: choose, select, committee, group, team
• Examples: Committee selection, pizza toppings, lottery numbers
Example 5: Decide permutation or combination
A: Select 4 books from 12 to read
→ Order doesn't matter (same 4 books)
→ COMBINATION: $_{12}C_4$
B: Arrange 4 books from 12 on a shelf
→ Order matters (different arrangements)
→ PERMUTATION: $_{12}P_4$
C: Choose 5 students from 20 for a team
→ Order doesn't matter (same team)
→ COMBINATION: $_{20}C_5$
D: Assign 5 students to 5 different roles
→ Order matters (different roles)
→ PERMUTATION: $_{5}P_5 = 5!$
A: Select 4 books from 12 to read
→ Order doesn't matter (same 4 books)
→ COMBINATION: $_{12}C_4$
B: Arrange 4 books from 12 on a shelf
→ Order matters (different arrangements)
→ PERMUTATION: $_{12}P_4$
C: Choose 5 students from 20 for a team
→ Order doesn't matter (same team)
→ COMBINATION: $_{20}C_5$
D: Assign 5 students to 5 different roles
→ Order matters (different roles)
→ PERMUTATION: $_{5}P_5 = 5!$
Probability Formulas Summary
Type | Formula | When to Use |
---|---|---|
Theoretical | $P(E) = \frac{\text{Favorable}}{\text{Total Possible}}$ | Mathematical reasoning, equally likely outcomes |
Experimental | $P(E) = \frac{\text{Occurred}}{\text{Total Trials}}$ | Based on actual experiment results |
Conditional | $P(A|B) = \frac{P(A \text{ and } B)}{P(B)}$ | Probability of A given B occurred |
Independent AND | $P(A \text{ and } B) = P(A) \times P(B)$ | Both events, one doesn't affect other |
Dependent AND | $P(A \text{ and } B) = P(A) \times P(B|A)$ | Both events, first affects second |
Counting Methods Summary
Method | Formula | When to Use | Example |
---|---|---|---|
Counting Principle | $n_1 \times n_2 \times ... \times n_k$ | Multiple choices in sequence | Outfit with 3 shirts and 2 pants: 3×2=6 |
Permutation | $_nP_r = \frac{n!}{(n-r)!}$ | Arrange r from n, order matters | Arrange 3 books from 8: $_{8}P_3 = 336$ |
Permutation (all) | $n!$ | Arrange all n items | Arrange 5 people: 5! = 120 |
Combination | $_nC_r = \frac{n!}{r!(n-r)!}$ | Select r from n, order doesn't matter | Choose 3 from 8: $_{8}C_3 = 56$ |
Independent vs Dependent Events
Aspect | Independent | Dependent |
---|---|---|
Definition | Outcome of one doesn't affect other | Outcome of one affects the other |
Formula | $P(A \text{ and } B) = P(A) \times P(B)$ | $P(A \text{ and } B) = P(A) \times P(B|A)$ |
Examples | Two coin flips, rolling two dice, with replacement | Without replacement, choosing without returning |
P(B) changes? | NO - stays the same | YES - changes after first event |
Permutation vs Combination
Aspect | Permutation | Combination |
---|---|---|
Order | MATTERS | DOESN'T MATTER |
Formula | $_nP_r = \frac{n!}{(n-r)!}$ | $_nC_r = \frac{n!}{r!(n-r)!}$ |
Keywords | Arrange, order, schedule, rank | Choose, select, group, committee |
Examples | Race places, passwords, seating | Teams, committees, toppings |
Result | Usually LARGER number | Usually SMALLER number |
Relationship | $_nC_r = \frac{_nP_r}{r!}$ |
Success Tips for Probability:
✓ Probability is always between 0 and 1 (or 0% and 100%)
✓ Theoretical: favorable/possible; Experimental: occurred/trials
✓ For conditional probability, limit to the "given" condition row/column
✓ Independent: outcome doesn't change; Dependent: outcome changes
✓ AND means multiply probabilities
✓ Counting Principle: multiply number of choices
✓ Order matters → Permutation; Order doesn't matter → Combination
✓ Permutations > Combinations for same n and r
✓ Remember: 0! = 1
✓ Practice identifying when to use each formula!
✓ Probability is always between 0 and 1 (or 0% and 100%)
✓ Theoretical: favorable/possible; Experimental: occurred/trials
✓ For conditional probability, limit to the "given" condition row/column
✓ Independent: outcome doesn't change; Dependent: outcome changes
✓ AND means multiply probabilities
✓ Counting Principle: multiply number of choices
✓ Order matters → Permutation; Order doesn't matter → Combination
✓ Permutations > Combinations for same n and r
✓ Remember: 0! = 1
✓ Practice identifying when to use each formula!