Class 7 Mathematics Ch 13 Mastery Guide: Probability, Sample Space, Events, Complements & Tree Diagrams (FBISE)
🗺️ Teacher & Parent Roadmap: Unit 13 (Probability)
Probability is the mathematical language of uncertainty, chance, and risk assessment. In everyday life, we encounter statements like "It will probably rain today" or "There is a 50-50 chance of winning the coin toss". In Grade 7 Mathematics (FBISE Curriculum), students move from intuitive guesses to precise mathematical calculation. This unit covers foundational concepts including Experiments, Sample Spaces, Events & Favorable Outcomes, the Probability Scale ($0 \le P \le 1$), Equally Likely Outcomes, Mutually Exclusive vs. Non-Mutually Exclusive Events, the Complement Rule ($P(\text{not } E) = 1 - P(E)$), and multi-stage Tree Diagrams.
- Define Experiment, Outcome, and construct the complete Sample Space ($S$) and count its elements $n(S)$.
- Quantify chance using the Probability Scale: Impossible ($P=0$), Unlikely ($0 < P < 0.5$), Even Chance ($P=0.5$), Likely ($0.5 < P < 1$), and Certain ($P=1$).
- Apply the classical probability formula: $P(E) = \frac{n(E)}{n(S)} = \frac{\text{favourable outcomes}}{\text{total possible outcomes}}$.
- Differentiate between Mutually Exclusive Events ($A \cap B = \emptyset$) and non-mutually exclusive events.
- Calculate the Complement of an Event using $P(E') = P(\text{not } E) = 1 - P(E)$.
- Construct and interpret multi-stage Tree Diagrams for combined events with and without replacement.
- Probability > 1 or < 0 Error: A probability can NEVER be less than 0 or greater than 1 (or greater than 100%). If your answer is $\frac{5}{4}$ or $-0.2$, check your work immediately!
- Mutually Exclusive Confusion: Assuming two events cannot happen together when they can (e.g. drawing a Queen and drawing a Spade — the Queen of Spades is both!).
- Deck of Cards Ignorance: Forgetting standard card counts: 52 total cards, 4 suits of 13 each, 26 red cards, 26 black cards, 12 face/picture cards (4 Kings, 4 Queens, 4 Jacks), 4 Aces.
- Tree Diagram Stage Branches: Forgetting to multiply possibilities across stages (e.g., 3 coin tosses give $2 \times 2 \times 2 = 8$ outcomes, not $2 + 2 + 2 = 6$).
🎲 Core Concepts & Visual Foundations
2.1 The Probability Scale & Likelihood of Events
Probability is the numerical measure of how likely an event is to occur. The value of probability always lies between $0$ and $1$ inclusive ($0 \le P(E) \le 1$).
- Impossible Event ($P = 0$): An event that can NEVER happen (e.g., rolling an 8 on a standard 6-sided die, the sun rising in the west).
- Unlikely Event ($0 < P < 0.5$ or $0\% < P < 50\%$): Has a small chance of happening (e.g., winning a lottery with 1 ticket out of 10,000).
- Even Chance / Equally Likely ($P = 0.5 = \frac{1}{2} = 50\%$): Has exactly equal chances of happening or not happening (e.g., tossing a fair coin and getting Heads).
- Likely Event ($0.5 < P < 1$ or $50\% < P < 100\%$): Has a good chance of occurring (e.g., picking a colored marble from a bag containing 9 blue and 1 red marble).
- Certain Event ($P = 1 = 100\%$): An event that is GUARANTEED to happen (e.g., the day after Monday will be Tuesday, rolling a number less than 7 on a standard die).
2.2 Experiment, Outcomes, Sample Space ($S$), and Events
Understanding probability terminology is key to solving all problems correctly:
An activity or trial that produces observable results (e.g., rolling a die, tossing a coin, drawing a card).
A single possible result of an experiment (e.g., getting a '3' on a die, getting 'Heads').
The set of ALL possible outcomes of an experiment. The total count is written as $n(S)$.
A subset of the sample space containing the outcomes favorable to a specific condition. Total count is $n(E)$.
| Random Experiment | Sample Space ($S$) | Total Outcomes $n(S)$ |
|---|---|---|
| Tossing 1 fair coin | $S = \{H, T\}$ | $n(S) = 2$ |
| Rolling 1 fair six-sided die | $S = \{1, 2, 3, 4, 5, 6\}$ | $n(S) = 6$ |
| Tossing 2 fair coins | $S = \{HH, HT, TH, TT\}$ | $n(S) = 2^2 = 4$ |
| Tossing 3 fair coins | $S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$ | $n(S) = 2^3 = 8$ |
| Standard 52-card deck | 4 Suits: 13 Hearts ♥, 13 Diamonds ♦, 13 Clubs ♣, 13 Spades ♠ | $n(S) = 52$ |
2.3 The Classical Probability Formula & Complement Rule
2.4 Mutually Exclusive vs. Non-Mutually Exclusive Events
Two events $A$ and $B$ are Mutually Exclusive (Disjoint) if they cannot happen at the same time ($A \cap B = \emptyset$). If they can occur simultaneously, they are Non-Mutually Exclusive.
2.5 Tree Diagrams for Multi-Stage Combined Events
A Tree Diagram is a powerful visual branching structure used to systematically display all possible outcomes of a multi-stage experiment. The total number of outcomes is found by multiplying the number of options at each stage: $$\text{Total Outcomes} = n_1 \times n_2 \times n_3 \times \dots$$
📝 Step-by-Step Solved Textbook Exercises
Exercise 13.1 (Probability Basics, Sample Space, Equally Likely Events & Complements)
Q1. Describe the probability of the following events using words (Impossible, Unlikely, Even chance, Likely, Certain):
- (a) The Sun will rise in the west tomorrow:
Solution: Impossible ($P = 0$), because astronomically the Sun always rises in the east due to Earth's rotation. - (b) Tossing a fair coin and getting a Head:
Solution: Even chance ($P = \frac{1}{2} = 0.5 = 50\%$), since Heads and Tails are equally likely. - (c) Rolling a 7 on a standard six-sided die:
Solution: Impossible ($P = 0$), as the faces are only numbered 1 through 6. - (d) Winning a nationwide lottery when buying 1 ticket out of 1,000,000 sold:
Solution: Unlikely ($P = \frac{1}{1000000} \approx 0.000001$), extremely small chance. - (e) The day after Tuesday will be Wednesday:
Solution: Certain ($P = 1 = 100\%$), as the weekly sequence is fixed.
Q2. Write down the sample space $S$ and the total number of outcomes $n(S)$ for the following experiments:
- (a) Tossing a single coin:
$$S = \{H, T\}, \quad n(S) = 2$$ - (b) Rolling a fair six-sided die:
$$S = \{1, 2, 3, 4, 5, 6\}, \quad n(S) = 6$$ - (c) Choosing a letter at random from the word "MATH":
$$S = \{M, A, T, H\}, \quad n(S) = 4$$ - (d) Selecting one primary color:
$$S = \{\text{Red}, \text{Blue}, \text{Yellow}\}, \quad n(S) = 3$$
Q3. A box contains 4 red marbles, 5 green marbles, and 6 blue marbles. A marble is drawn at random. Find the probability of getting:
(a) A red marble: $n(\text{Red}) = 4 \implies P(\text{Red}) = \frac{4}{15}$
(b) A green marble: $n(\text{Green}) = 5 \implies P(\text{Green}) = \frac{5}{15} = \frac{1}{3}$
(c) A blue marble: $n(\text{Blue}) = 6 \implies P(\text{Blue}) = \frac{6}{15} = \frac{2}{5}$
(d) Not a red marble: Using the complement rule: $$P(\text{Not Red}) = 1 - P(\text{Red}) = 1 - \frac{4}{15} = \frac{15 - 4}{15} = \frac{11}{15}$$
Q4. A fair six-sided die is rolled. Find the probability of getting:
(a) The number 4: Favourable outcome $E = \{4\}$, $n(E) = 1 \implies P(4) = \frac{1}{6}$
(b) An even number: $E = \{2, 4, 6\}$, $n(E) = 3 \implies P(\text{Even}) = \frac{3}{6} = \frac{1}{2}$
(c) A number greater than 4: $E = \{5, 6\}$, $n(E) = 2 \implies P(>4) = \frac{2}{6} = \frac{1}{3}$
(d) A prime number: $E = \{2, 3, 5\}$, $n(E) = 3 \implies P(\text{Prime}) = \frac{3}{6} = \frac{1}{2}$ (Note: 1 is not prime!)
(e) A number less than 1: $E = \emptyset$, $n(E) = 0 \implies P(<1) = \frac{0}{6} = 0$ (Impossible event).
Q5. A bag contains 8 red hats, 4 blue hats, and 6 black hats. A hat is drawn at random. Total = 18. Find the probability that the hat is:
(a) Red: $P(\text{Red}) = \frac{8}{18} = \frac{4}{9}$
(b) Blue: $P(\text{Blue}) = \frac{4}{18} = \frac{2}{9}$
(c) Black: $P(\text{Black}) = \frac{6}{18} = \frac{1}{3}$
(d) Not blue: $P(\text{Not Blue}) = 1 - P(\text{Blue}) = 1 - \frac{2}{9} = \frac{7}{9}$
Q6. The letters of the word "CHANCE" are written on separate identical cards and placed in a box. A card is drawn at random. Find the probability of drawing:
(a) The letter 'C': Appears 2 times $\implies P(\text{'C'}) = \frac{2}{6} = \frac{1}{3}$
(b) A vowel: Vowels in CHANCE are {A, E}, $n = 2 \implies P(\text{Vowel}) = \frac{2}{6} = \frac{1}{3}$
(c) A consonant: Consonants are {C, H, N, C}, $n = 4 \implies P(\text{Consonant}) = \frac{4}{6} = \frac{2}{3}$
(d) The letter 'Z': Does not exist in the word $\implies P(\text{'Z'}) = \frac{0}{6} = 0$.
Q7. An 8-sector circular spinner is numbered 1 through 8. The pointer is spun. Find the probability of landing on:
(a) An odd number: $\{1, 3, 5, 7\} \implies n = 4 \implies P(\text{Odd}) = \frac{4}{8} = \frac{1}{2}$
(b) A multiple of 3: $\{3, 6\} \implies n = 2 \implies P(\text{Multiple of 3}) = \frac{2}{8} = \frac{1}{4}$
(c) A number divisible by 5: $\{5\} \implies n = 1 \implies P(\text{Divisible by 5}) = \frac{1}{8}$
(d) A number less than 9: All 8 numbers are $<9 \implies n = 8 \implies P(<9) = \frac{8}{8} = 1$ (Certain).
Q8. A card is drawn from a well-shuffled standard pack of 52 playing cards. Find the probability of getting:
(a) A King: There are 4 Kings $\implies P(\text{King}) = \frac{4}{52} = \frac{1}{13}$
(b) A Heart: There are 13 Hearts $\implies P(\text{Heart}) = \frac{13}{52} = \frac{1}{4}$
(c) A black card: There are 26 black cards (13 Spades + 13 Clubs) $\implies P(\text{Black}) = \frac{26}{52} = \frac{1}{2}$
(d) A red Queen: 2 red Queens (Queen of Hearts + Queen of Diamonds) $\implies P(\text{Red Queen}) = \frac{2}{52} = \frac{1}{26}$
(e) Not an Ace: 4 Aces in the deck $\implies P(\text{Not Ace}) = 1 - \frac{4}{52} = \frac{48}{52} = \frac{12}{13}$
Q9. Slips of paper numbered 1 to 20 are placed in a jar. One slip is drawn at random. Find the probability that the number is:
(a) A prime number: $\{2, 3, 5, 7, 11, 13, 17, 19\} \implies 8 \text{ numbers} \implies P(\text{Prime}) = \frac{8}{20} = \frac{2}{5}$
(b) A multiple of 4: $\{4, 8, 12, 16, 20\} \implies 5 \text{ numbers} \implies P(\text{Multiple of 4}) = \frac{5}{20} = \frac{1}{4}$
(c) Divisible by both 2 and 3 (multiples of 6): $\{6, 12, 18\} \implies 3 \text{ numbers} \implies P = \frac{3}{20}$
(d) Greater than 15: $\{16, 17, 18, 19, 20\} \implies 5 \text{ numbers} \implies P(>15) = \frac{5}{20} = \frac{1}{4}$
Q10. In a prize lottery, 300 tickets are sold. If Ali buys 15 tickets, calculate the probability that:
(a) Ali wins the prize: $$P(\text{Win}) = \frac{15}{300} = \frac{1}{20} = 0.05 \quad (5\%)$$ (b) Ali does not win the prize: $$P(\text{Does not win}) = 1 - P(\text{Win}) = 1 - \frac{1}{20} = \frac{19}{20} = 0.95 \quad (95\%)$$
Q11. A fair six-sided die is rolled. Let event $A$ be rolling a number less than 3, and event $B$ be rolling a number greater than 4:
Event $A = \{1, 2\} \implies n(A) = 2 \implies P(A) = \frac{2}{6} = \frac{1}{3}$.
Event $B = \{5, 6\} \implies n(B) = 2 \implies P(B) = \frac{2}{6} = \frac{1}{3}$.
(a) Find $P(A)$ and $P(B)$: $P(A) = \frac{1}{3}, \quad P(B) = \frac{1}{3}$.
(b) Are events $A$ and $B$ mutually exclusive? Explain:
Yes! $A \cap B = \{1, 2\} \cap \{5, 6\} = \emptyset$. A single roll of a die cannot simultaneously show a number less than 3 AND greater than 4.
(c) Find $P(A \text{ or } B)$: $$P(A \cup B) = P(A) + P(B) = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$$
Q12. A loaded die is rolled. If the probability of getting a 6 is $\frac{1}{3}$, what is the probability of NOT getting a 6?
Q13. A bag contains 2 red balls, 3 green balls, and 4 yellow balls. A ball is drawn at random, its color noted, and it is replaced. Then another ball is drawn:
(a) Find the probability of drawing a red ball on the first draw: $$P(\text{Red}) = \frac{2}{9}$$ (b) Are drawing a red ball and drawing a yellow ball on the first draw mutually exclusive? Explain:
Yes. When drawing a single ball, it cannot be both red and yellow at the same time. The two outcomes are completely disjoint ($A \cap B = \emptyset$).
Exercise 13.2 (Tree Diagrams, Combined Events & Sample Spaces)
Q1. A school cafeteria offers 2 main dishes (Rice $R$, Pasta $P$) and 3 beverages (Juice $J$, Milk $M$, Water $W$). Draw a tree diagram to list all possible meal combinations. Find the total number of combinations:
Tree Diagram Breakdown:
• Branch 1 (Rice $R$) $\to$ $RJ$ (Rice & Juice), $RM$ (Rice & Milk), $RW$ (Rice & Water)
• Branch 2 (Pasta $P$) $\to$ $PJ$ (Pasta & Juice), $PM$ (Pasta & Milk), $PW$ (Pasta & Water)
Sample Space: $S = \{RJ, RM, RW, PJ, PM, PW\}, \quad n(S) = 6$.
Q2. Zara has 3 shirts (Red $R$, Blue $B$, White $W$) and 2 skirts (Black $K$, Grey $G$). Draw a tree diagram showing all outfit choices and determine the total number of distinct outfits:
Branches & Sample Space:
$S = \{RK, RG, BK, BG, WK, WG\}, \quad n(S) = 6$.
Q3. Ali ($A$), Babar ($B$), and Camran ($C$) stand in a queue at a ticket counter. Draw a tree diagram to show all possible line-up orders:
Possible Orders:
1. $A$ first $\to ABC, ACB$
2. $B$ first $\to BAC, BCA$
3. $C$ first $\to CAB, CBA$
$$S = \{ABC, ACB, BAC, BCA, CAB, CBA\}, \quad n(S) = 6$$
Q4. A fast-food restaurant serves Burgers (Chicken $C$, Beef $B$) and Drinks (Soda $S$, Tea $T$, Coffee $K$). Draw a tree diagram and find the probability that a randomly chosen order is a Chicken burger with Soda:
$S = \{CS, CT, CK, BS, BT, BK\}, \quad n(S) = 6$.
Favourable outcome: $\{CS\} \implies n(CS) = 1$.
$$P(\text{Chicken Burger with Soda}) = \frac{1}{6}$$
Q5. A drawer contains 2 black socks ($B$) and 2 white socks ($W$). Two socks are pulled out one after another WITH REPLACEMENT. Draw a tree diagram and find the probability of getting a matching pair:
Favourable outcomes for a matching pair: $E = \{BB, WW\} \implies n(E) = 2$.
$$P(\text{Matching Pair}) = \frac{2}{4} = \frac{1}{2} = 0.50$$
Q6. A bag contains 1 black ball ($B$) and 1 white ball ($W$). A ball is drawn, its color recorded, and returned to the bag. A second ball is then drawn. Draw a tree diagram and determine the probability of getting at least one white ball:
Favourable event (at least one $W$): $E = \{BW, WB, WW\} \implies n(E) = 3$.
$$P(\text{At least one white ball}) = \frac{3}{4} = 0.75$$
Q7. Bag 1 contains counters numbered 1 and 2. Bag 2 contains counters numbered 3 and 4. One counter is drawn from each bag. Draw a tree diagram, list the sample space, and calculate the probability that the sum of the numbers is odd:
• $(1, 3) \to 1 + 3 = 4$ (Even)
• $(1, 4) \to 1 + 4 = 5$ (Odd)
• $(2, 3) \to 2 + 3 = 5$ (Odd)
• $(2, 4) \to 2 + 4 = 6$ (Even)
Odd sums occur in 2 outcomes: $\{(1, 4), (2, 3)\} \implies n(E) = 2$.
$$P(\text{Sum is Odd}) = \frac{2}{4} = \frac{1}{2}$$
Q8. An ice cream parlor offers Vanilla ($V$), Chocolate ($C$), and Mango ($M$) flavors. Two friends each independently choose one flavor. Draw a tree diagram, list the sample space, and find the probability that both friends choose the same flavor:
$$S = \{VV, VC, VM, CV, CC, CM, MV, MC, MM\}, \quad n(S) = 9$$ Favourable outcomes (same flavor): $E = \{VV, CC, MM\} \implies n(E) = 3$.
$$P(\text{Same Flavor}) = \frac{3}{9} = \frac{1}{3}$$
Review Exercise 13 (Comprehensive Chapter Review & Problem Solving)
Q1. Multiple Choice Questions (Textbook Review):
- The probability of an impossible event is: (c) $0$
- The probability of a certain event is: (d) $1$
- If $P(E) = 0.35$, then $P(\text{not } E)$ is: (b) $0.65$ (Since $1 - 0.35 = 0.65$)
- A die is rolled once. The probability of getting an odd number is: (a) $\frac{1}{2}$ ($\{1,3,5\} \to \frac{3}{6} = \frac{1}{2}$)
- In a standard pack of 52 cards, the number of picture (face) cards is: (c) $12$ (4 Kings, 4 Queens, 4 Jacks)
- Two events that cannot occur at the same time are called: (b) Mutually exclusive events
- The total number of outcomes when 3 coins are tossed simultaneously is: (d) $8$ ($2^3 = 8$)
- Which of the following cannot be the probability of an event? (a) $1.25$ (Probabilities can never exceed 1)
- The sample space for tossing a coin is: (a) $\{H, T\}$
- If a spinner has 4 equal red sections out of 8 total sections, the probability of red is: (b) $\frac{1}{2}$ ($\frac{4}{8} = \frac{1}{2}$)
Q2. A six-sided die is rolled. Find the probability of getting:
(a) A multiple of 3: $\{3, 6\} \implies n = 2 \implies P = \frac{2}{6} = \frac{1}{3}$
(b) A number less than 7: All faces $\{1,2,3,4,5,6\} \implies n = 6 \implies P = \frac{6}{6} = 1$ (Certain).
Q3. A box contains 36 discs: 5 marked $V$, 12 marked $W$, and the rest unmarked. A disc is chosen at random. Find the probability that the disc is:
Unmarked discs $= 36 - (5 + 12) = 36 - 17 = 19$.
(a) Marked $V$: $P(V) = \frac{5}{36}$
(b) Marked $W$: $P(W) = \frac{12}{36} = \frac{1}{3}$
(c) Unmarked: $P(\text{Unmarked}) = \frac{19}{36}$
Q4. From a standard pack of 52 playing cards, a card is drawn at random. Find the probability that the card is:
Q5. A bag contains 24 red balls and $x$ white balls. If the probability of drawing a white ball is $\frac{3}{7}$, find the value of $x$:
$$P(\text{White}) = \frac{x}{24 + x} = \frac{3}{7}$$ Cross-multiplying: $$7x = 3(24 + x) \implies 7x = 72 + 3x \implies 7x - 3x = 72 \implies 4x = 72 \implies x = \frac{72}{4} = 18$$ Answer: There are $18$ white balls in the bag.
Q6. A basket contains 30 apples: 14 red and 16 green. How many red apples must be removed so that the probability of picking a green apple becomes $\frac{2}{3}$?
Green apples remain unchanged $= 16$.
New total apples $= 30 - y$.
$$P(\text{Green}) = \frac{16}{30 - y} = \frac{2}{3}$$ Cross-multiplying: $$16 \times 3 = 2(30 - y) \implies 48 = 60 - 2y \implies 2y = 60 - 48 \implies 2y = 12 \implies y = 6$$ Answer: $6$ red apples must be removed.
Q7. A card is drawn from a well-shuffled pack of 52 cards. Find the probability that the card is:
(a) A Diamond: $13 \text{ Diamonds} \implies P = \frac{13}{52} = \frac{1}{4}$
(b) An Ace: $4 \text{ Aces} \implies P = \frac{4}{52} = \frac{1}{13}$
(c) The Ace of Diamonds: Exactly 1 such card $\implies P = \frac{1}{52}$
Q8. A 10-sided die (numbered 1 to 10) is rolled. Find the probability of:
(a) Getting the number 1: $P(1) = \frac{1}{10}$
(b) Getting an even number: $\{2, 4, 6, 8, 10\} \implies n = 5 \implies P = \frac{5}{10} = \frac{1}{2}$
(c) Getting a number less than 4 OR greater than 7:
Numbers $<4$: $\{1, 2, 3\}$; Numbers $>7$: $\{8, 9, 10\}$.
Combined set: $\{1, 2, 3, 8, 9, 10\} \implies n = 6$.
$$P(<4 \text{ or } >7) = \frac{6}{10} = \frac{3}{5}$$
Q9. A regular hexagonal spinner is numbered 1 to 6. Find the probability of:
(a) Spinning an odd number: $\{1, 3, 5\} \implies P = \frac{3}{6} = \frac{1}{2}$
(b) Spinning a number greater than or equal to 4: $\{4, 5, 6\} \implies P = \frac{3}{6} = \frac{1}{2}$
Q10. Two friends order ice cream from a menu offering Vanilla, Strawberry, and Chocolate. Draw a tree diagram and find the probability that they choose DIFFERENT flavors:
Outcomes where they pick the same flavor $= \{VV, SS, CC\} \implies n = 3$.
Using the complement rule: $$P(\text{Different Flavors}) = 1 - P(\text{Same Flavor}) = 1 - \frac{3}{9} = \frac{6}{9} = \frac{2}{3}$$
Q11. In a class of 40 students, 20 study Mathematics, 25 study Statistics, and 10 study both. Are the events "studies Mathematics" and "studies Statistics" mutually exclusive? Explain:
Here, $n(M \cap S) = 10 \neq 0$ (10 students study both subjects simultaneously).
Conclusion: No, the events are NOT mutually exclusive because there is an overlap of 10 students.
Q12. From a well-shuffled pack of 52 cards, determine if the following pairs of events are mutually exclusive and find the probability of their union:
• Mutually Exclusive? Yes, no single card can be both a King and a Queen ($K \cap Q = \emptyset$).
• Probability: $P(K \cup Q) = P(K) + P(Q) = \frac{4}{52} + \frac{4}{52} = \frac{8}{52} = \frac{2}{13}$.
(b) Drawing a Queen or drawing a Spade:
• Mutually Exclusive? No, because the Queen of Spades belongs to both events ($Q \cap \text{Spade} = \{\text{Queen of Spades}\}$).
• Probability: $$P(Q \cup \text{Spade}) = P(Q) + P(\text{Spade}) - P(Q \cap \text{Spade}) = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13}$$
💡 Solved In-Text Check Points
- Check Point (p. 252 - Colored Pencils): A pencil box has 2 blue ($B_1, B_2$), 1 yellow ($Y$), and 2 black ($K_1, K_2$) pencils. Sample space $S = \{B_1, B_2, Y, K_1, K_2\}, \quad n(S) = 5$.
- Check Point (p. 253 - 2-Digit Primes < 50): $S = \{11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47\}$, $n(S) = 11$. Primes with tens digit 3: $E = \{31, 37\} \implies n(E) = 2 \implies P(E) = \frac{2}{11}$.
- Check Point (p. 255 - Number Cube): (a) $P(\text{prime}) = \frac{|\{2,3,5\}|}{6} = \frac{3}{6} = \frac{1}{2}$. (b) $P(\text{number } > 3) = \frac{|\{4,5,6\}|}{6} = \frac{3}{6} = \frac{1}{2}$.
- Check Point (p. 256 - Marbles): 5 black, 3 white, 4 yellow (Total 12). $P(\text{Black}) = \frac{5}{12}$, $P(\text{White}) = \frac{3}{12} = \frac{1}{4}$, $P(\text{Yellow}) = \frac{4}{12} = \frac{1}{3}$. Most likely = Black ($\frac{5}{12} \approx 41.7\%$).
- Check Point (p. 257 - Letters of MATHEMATICS): Total 11 letters. $P = \text{vowels }\{A, E, A, I\}$, $Q = \text{letter } A$, $R = \text{consonants }\{M, T, H, M, T, C, S\}$. Events $P$ and $R$ are mutually exclusive ($P \cap R = \emptyset$). Events $P$ and $Q$ are NOT mutually exclusive ($Q \subseteq P$).
- Check Point (p. 258 - Eye Color Survey): 100 students: Brown 34, Blue 30, Green 21, Hazel 15. $P(\text{Green}) = \frac{21}{100} = 0.21 \implies P(\text{Not Green}) = 1 - 0.21 = \frac{79}{100} = 0.79$.
- Check Point (p. 259 - Playing Cards Spade): $P(\text{Spade}) = \frac{13}{52} = \frac{1}{4} \implies P(\text{Not Spade}) = 1 - \frac{1}{4} = \frac{3}{4}$.
- Check Point (p. 262 - 3 Coin Tosses): Sample space $S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}, \quad n(S) = 2^3 = 8$.
📌 Unit 13 Quick Revision & Formula Sheet
- Range: $0 \le P(E) \le 1$
- Impossible Event: $P(E) = 0$
- Certain Event: $P(E) = 1$
- Sum of All Outcomes: $\sum P(E_i) = 1$
- Equally Likely: $P(E) = \frac{n(E)}{n(S)}$
- Complement: $P(E') = 1 - P(E)$
- Mutually Exclusive: $P(A \cup B) = P(A) + P(B)$
- Non-Mutually Exclusive: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$
- 1 Die: $n(S) = 6$
- $k$ Coins: $n(S) = 2^k$
- Deck of Cards: Total 52, 4 Suits (13 each), 26 Red, 26 Black, 12 Face Cards, 4 Aces
❓ Frequently Asked Questions — Class 7 Mathematics Chapter 13
More Chapter Notes for Class 7 (FBISE)
MathematicsTest Your Knowledge on Chapter 13: Class 7 Mathematics Ch 13 Mastery Guide: Probability, Sample Space, Events, Complements & Tree Diagrams (FBISE)
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