NCERT Ganita Manjari · Grade 9 · Part I
Chapter 7: The Mathematics of Maybe
Introduction to Probability — Complete Exercise Solutions
🎲 🪙 📊 🌟
Experimental P = occurrences ÷ trials
Theoretical P = favourable ÷ possible
Scale: 0 (impossible) → 1 (certain)
0 ≤ P(E) ≤ 1
Practice Progress0 / 0 opened
Select an exercise card below to start
Exercise Set
7.1 Probability Scale
1 Question · 4 parts
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Exercise Set
7.2 Experimental & Theoretical
6 Questions
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Exercise Set
7.3 Sample Spaces & Events
3 Questions
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Exercise Set
7.4 Tree Diagrams
2 Questions
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End of Chapter
All Exercises (Q1–Q16)
16 Questions incl. ★ Challenges
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Exercise Set 7.1 — The Probability Scale
1
Rank the following events on a scale from 0 (Impossible) to 1 (Certain). Label each event and give reasons.
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Fig: The Probability Scale (0 to 1)
| Part | Event | Label | Probability | Reason |
|---|---|---|---|---|
| (i) | Next Monday after Sunday | Certain | P = 1 | This is always true by the calendar — Monday always follows Sunday. |
| (ii) | Snow in Mumbai in July | Impossible | P ≈ 0 | Mumbai has a tropical climate. Snow has never been recorded there. |
| (iii) | Elephant walks through classroom | Impossible | P ≈ 0 | Wild elephants entering urban classrooms is practically impossible under normal conditions. |
| (iv) | Greet at least one friend at school | More Likely | P close to 1 | If you attend school, it is very likely you will greet at least one friend among your classmates. |
Events placed on the probability scale
Exercise Set 7.2 — Measuring Probability Objectively
1
A teacher samples 30 sweets: 10 red, 8 green, 7 yellow, 5 blue. Find probability of green; estimate yellows in 600 total.
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Sample of 30 sweets by colour
(i) Probability of picking a green sweet
Green sweets in sample = 8, Total sample = 30
$P(\text{green}) = \dfrac{8}{30} = \dfrac{4}{15} \approx 0.267$
$P(\text{green}) = \dfrac{8}{30} = \dfrac{4}{15} \approx 0.267$
Answer$P(\text{green}) = \dfrac{4}{15} \approx 0.267\ (26.7\%)$
(ii) Estimate number of yellow sweets in 600
$P(\text{yellow}) = \dfrac{7}{30}$
Estimated count $= \dfrac{7}{30} \times 600 = 7 \times 20 = 140$
Estimated count $= \dfrac{7}{30} \times 600 = 7 \times 20 = 140$
AnswerApproximately 140 yellow sweets in the bag of 600.
2
Survey of 40 students: 14 Science, 11 Arts, 9 Sports, 6 Debate. School has 800 students. Find P(Arts) and estimate Sports students.
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(i) P(Arts Club)
Arts Club students = 11, Total sample = 40
$P(\text{Arts}) = \dfrac{11}{40} = 0.275$
$P(\text{Arts}) = \dfrac{11}{40} = 0.275$
Answer$P(\text{Arts Club}) = \dfrac{11}{40} = 0.275\ (27.5\%)$
(ii) Estimate Sports Club students in whole school (800)
$P(\text{Sports}) = \dfrac{9}{40}$
Estimated count $= \dfrac{9}{40} \times 800 = 9 \times 20 = 180$
Estimated count $= \dfrac{9}{40} \times 800 = 9 \times 20 = 180$
AnswerApproximately 180 students prefer the Sports Club.
3
Toss a coin 20 times. Record heads/tails. Answer questions about experimental and theoretical probability.
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🪙 Activity Question: Parts (i) and (ii) depend on your actual experiment results. Sample answers below assume 11 heads and 9 tails.
(i) How many times heads?
Depends on your experiment. Example: Say you got 11 heads.
AnswerRecord your own count (e.g., 11 heads out of 20)
(ii) How many times tails?
Example: $20 – 11 = 9$ tails
AnswerRecord your own count (e.g., 9 tails out of 20)
(iii) Experimental probability of heads
$P_{\text{exp}}(\text{heads}) = \dfrac{\text{number of heads}}{20}$
Using example (11 heads): $P_{\text{exp}} = \dfrac{11}{20} = 0.55$
Using example (11 heads): $P_{\text{exp}} = \dfrac{11}{20} = 0.55$
Answer$P_{\text{exp}}(\text{heads}) = \dfrac{\text{your heads count}}{20}$
(iv) Probability of tails on the next toss
Key concept: Each toss is independent. Past results don’t affect the next toss (Gambler’s Fallacy!).
$P(\text{tails on next toss}) = \dfrac{1}{2} = 0.5$
This is the theoretical probability — it never changes.
$P(\text{tails on next toss}) = \dfrac{1}{2} = 0.5$
This is the theoretical probability — it never changes.
Answer$P(\text{tails}) = \dfrac{1}{2} = 0.5$, regardless of past results.
4
Toss a paper cup 100 times. Record — bottom, top, or side. Assign probabilities using experimental probability.
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☕ Activity Question: Results depend on your actual experiment. A paper cup is not symmetric, so outcomes are NOT equally likely.
Method: After 100 tosses, record counts for each position.
If in your experiment: bottom = 35, top = 15, side = 50:
$P(\text{bottom}) = \dfrac{35}{100} = 0.35$
$P(\text{top}) = \dfrac{15}{100} = 0.15$
$P(\text{side}) = \dfrac{50}{100} = 0.50$
Check: All probabilities must add up to 1: $0.35 + 0.15 + 0.50 = 1$ ✓
If in your experiment: bottom = 35, top = 15, side = 50:
$P(\text{bottom}) = \dfrac{35}{100} = 0.35$
$P(\text{top}) = \dfrac{15}{100} = 0.15$
$P(\text{side}) = \dfrac{50}{100} = 0.50$
Check: All probabilities must add up to 1: $0.35 + 0.15 + 0.50 = 1$ ✓
Note: Unlike coins or dice, a paper cup has unequal probabilities for each outcome. Experimental probability is the only valid method here.
Answer$P(\text{each outcome}) = \dfrac{\text{count for that position}}{100}$
5
What is the probability of getting an even number when rolling a fair 6-sided die?
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Blue = Even numbers {2,4,6}; Red = Odd numbers {1,3,5}
Sample space S = {1, 2, 3, 4, 5, 6}, $n(S) = 6$
Even numbers = {2, 4, 6} → 3 favourable outcomes
$P(\text{even}) = \dfrac{3}{6} = \dfrac{1}{2}$
Even numbers = {2, 4, 6} → 3 favourable outcomes
$P(\text{even}) = \dfrac{3}{6} = \dfrac{1}{2}$
Answer$P(\text{even number}) = \dfrac{1}{2} = 0.5\ (50\%)$
6
A die is rolled 12 times and ‘3’ appears 3 times. Find experimental and theoretical probability of rolling ‘3’. Why might they differ?
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(i) Experimental probability of rolling ‘3’
Event occurred = 3 times, Total trials = 12
$P_{\text{exp}}(3) = \dfrac{3}{12} = \dfrac{1}{4} = 0.25\ (25\%)$
$P_{\text{exp}}(3) = \dfrac{3}{12} = \dfrac{1}{4} = 0.25\ (25\%)$
Answer$P_{\text{exp}}(3) = \dfrac{1}{4} = 0.25$
(ii) Theoretical probability of rolling ‘3’
Favourable outcomes = 1 (only the face showing 3)
Total outcomes = 6
$P_{\text{theo}}(3) = \dfrac{1}{6} \approx 0.167\ (16.7\%)$
Total outcomes = 6
$P_{\text{theo}}(3) = \dfrac{1}{6} \approx 0.167\ (16.7\%)$
Answer$P_{\text{theo}}(3) = \dfrac{1}{6} \approx 0.167$
(iii) Why different? What happens with more trials?
Why different? With only 12 trials, there is natural random variation. A small sample rarely matches theoretical predictions exactly.
Law of Large Numbers: As trials increase, experimental probability converges towards theoretical:
Law of Large Numbers: As trials increase, experimental probability converges towards theoretical:
Law of Large Numbers: More trials → closer to theoretical value
AnswerWith more trials, experimental P(3) approaches $\dfrac{1}{6} \approx 0.167$ (Law of Large Numbers).
Exercise Set 7.3 — Sample Spaces & Events
1
When a single 6-sided die is rolled, what is the total number of possible outcomes in the sample space?
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A standard die has faces numbered 1, 2, 3, 4, 5, 6.
Sample space $S = \{1,\ 2,\ 3,\ 4,\ 5,\ 6\}$
Sample size $n(S) = 6$
Sample space $S = \{1,\ 2,\ 3,\ 4,\ 5,\ 6\}$
Sample size $n(S) = 6$
AnswerTotal possible outcomes = 6
2
Write the sample space S for: (i) Rolling a die and tossing a coin (ii) Random integer between –5 and +5 (iii) Drawing from 5 green and 7 red balls.
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(i) Rolling a die AND tossing a coin
Die outcomes: {1,2,3,4,5,6}; Coin: {H,T}
Combine each die outcome with each coin outcome:
Combine each die outcome with each coin outcome:
Answer$S = \{(1,H),(1,T),(2,H),(2,T),(3,H),(3,T),(4,H),(4,T),(5,H),(5,T),(6,H),(6,T)\}$; $n(S)=12$
(ii) Random integer between –5 and +5 (inclusive)
Integers from –5 to +5: –5, –4, –3, –2, –1, 0, 1, 2, 3, 4, 5
Answer$S = \{-5,-4,-3,-2,-1,0,1,2,3,4,5\}$; $n(S) = 11$
(iii) Box with 5 green and 7 red balls — one drawn at random
The outcome tells us only the colour of the drawn ball.
$S = \{\text{Green},\ \text{Red}\}$; $n(S) = 2$
$S = \{\text{Green},\ \text{Red}\}$; $n(S) = 2$
Note: If we label each individual ball, $n(S)=12$, but for colour-based questions $n(S)=2$ suffices.
Answer$S = \{\text{Green, Red}\}$; $n(S) = 2$
3
Village fair: 3 snacks (Samosa, Pakora, Bhaji) and 2 drinks (Chai, Lassi). List sample space and the event ‘selecting Samosa’.
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(i) Sample space — all snack + drink combinations
All 6 possible combinations (3 snacks × 2 drinks)
Answer$S = \{(\text{Sa,C}),(\text{Sa,L}),(\text{Pa,C}),(\text{Pa,L}),(\text{Bh,C}),(\text{Bh,L})\}$; $n(S)=6$
(ii) Event: ‘Selecting Samosa as a snack’
Pick all outcomes where Samosa is chosen: Samosa+Chai and Samosa+Lassi
Answer$E = \{(\text{Samosa, Chai}),\ (\text{Samosa, Lassi})\}$; $n(E)=2$
Exercise Set 7.4 — Tree Diagrams
1
Basket A: 1 apple, 2 oranges. Basket B: 1 banana, 1 mango. Pick one from each. Draw tree diagram, list sample space, find P(apple and banana).
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(i) Tree Diagram
Tree diagram: Choosing one fruit from each basket
(ii) Sample Space
Answer$S = \{(\text{Apple,Banana}),\ (\text{Apple,Mango}),\ (\text{Orange,Banana}),\ (\text{Orange,Mango})\}$; $n(S)=4$
(iii) P(Apple and Banana)
$P(\text{Apple from A}) = \dfrac{1}{3}$
$P(\text{Banana from B}) = \dfrac{1}{2}$
Since both picks are independent:
$P(\text{Apple AND Banana}) = \dfrac{1}{3} \times \dfrac{1}{2} = \dfrac{1}{6}$
$P(\text{Banana from B}) = \dfrac{1}{2}$
Since both picks are independent:
$P(\text{Apple AND Banana}) = \dfrac{1}{3} \times \dfrac{1}{2} = \dfrac{1}{6}$
Answer$P(\text{Apple and Banana}) = \dfrac{1}{6} \approx 0.167$
2
Box: 3 red, 4 black, 2 green pens. You and a friend each pick one (with replacement). Tree diagram and P(same colour).
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(i) Possible outcomes — Tree diagram
Tree diagram — 9 outcomes (R,R),(R,B),…(G,G)
(ii) P(both pick same colour)
Total = 9 pens; picks are with replacement, so probabilities stay constant.
$P(R,R) = \dfrac{3}{9} \times \dfrac{3}{9} = \dfrac{9}{81}$
$P(B,B) = \dfrac{4}{9} \times \dfrac{4}{9} = \dfrac{16}{81}$
$P(G,G) = \dfrac{2}{9} \times \dfrac{2}{9} = \dfrac{4}{81}$
$P(\text{same colour}) = \dfrac{9+16+4}{81} = \dfrac{29}{81} \approx 0.358$
$P(R,R) = \dfrac{3}{9} \times \dfrac{3}{9} = \dfrac{9}{81}$
$P(B,B) = \dfrac{4}{9} \times \dfrac{4}{9} = \dfrac{16}{81}$
$P(G,G) = \dfrac{2}{9} \times \dfrac{2}{9} = \dfrac{4}{81}$
$P(\text{same colour}) = \dfrac{9+16+4}{81} = \dfrac{29}{81} \approx 0.358$
Answer$P(\text{same colour}) = \dfrac{29}{81} \approx 0.358\ (35.8\%)$
End-of-Chapter Exercises (Q1–Q16)
1
Fill in the blanks: (i) impossible event (ii) set of all outcomes (iii) certain event (iv) fair coin heads probability.
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| Part | Fill in the blank |
|---|---|
| (i) | P(impossible event) = 0 |
| (ii) | The set of all possible outcomes is called the sample space |
| (iii) | P(certain event) = 1 |
| (iv) | P(heads on fair coin) = $\mathbf{\dfrac{1}{2}}$ |
2
50 students surveyed; 15 like football. Find the relative frequency.
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The term needed is relative frequency.
Relative frequency $= \dfrac{15}{50} = \dfrac{3}{10} = 0.3$
Relative frequency $= \dfrac{15}{50} = \dfrac{3}{10} = 0.3$
AnswerRelative frequency = $\dfrac{15}{50} = \dfrac{3}{10} = 0.3$ (or 30%)
3
Which experiments have equally likely outcomes? Explain for each.
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| Experiment | Equally Likely? | Reason |
|---|---|---|
| (i) Car starts or not | ❌ No | Cars usually start. The two outcomes are not equally likely. |
| (ii) Fair coin toss | ✅ Yes | A fair coin is symmetric. P(H) = P(T) = 1/2. |
| (iii) Fair 6-sided die | ✅ Yes | Each face equally likely. P(1)=P(2)=…=P(6)=1/6. |
| (iv) 3 red, 7 blue marbles | ❌ No | P(blue) = 7/10 > P(red) = 3/10. Not equal. |
| (v) Baby — boy or girl | ≈ Yes | Approximately equally likely (~50:50), though not exactly equal biologically. |
4
Write sample space and calculate probability for 5 situations.
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(i) Two coins — P(at least one head)
$S = \{HH, HT, TH, TT\}$, $n(S)=4$
At least one head: $E=\{HH,HT,TH\}$, $n(E)=3$
$P(\text{at least one head}) = \dfrac{3}{4}$
At least one head: $E=\{HH,HT,TH\}$, $n(E)=3$
$P(\text{at least one head}) = \dfrac{3}{4}$
Answer$P = \dfrac{3}{4} = 0.75$
(ii) Cards 1–10 — P(even number)
$S=\{1,2,…,10\}$, $n(S)=10$
Even: $E=\{2,4,6,8,10\}$, $n(E)=5$
$P(\text{even}) = \dfrac{5}{10} = \dfrac{1}{2}$
Even: $E=\{2,4,6,8,10\}$, $n(E)=5$
$P(\text{even}) = \dfrac{5}{10} = \dfrac{1}{2}$
Answer$P = \dfrac{1}{2} = 0.5$
(iii) Die — P(number greater than 4)
$S=\{1,2,3,4,5,6\}$; Greater than 4: $E=\{5,6\}$
$P = \dfrac{2}{6} = \dfrac{1}{3}$
$P = \dfrac{2}{6} = \dfrac{1}{3}$
Answer$P = \dfrac{1}{3} \approx 0.333$
(iv) Bag: 3 red, 2 blue, 1 green — P(not red)
Total = 6 balls; Not red = blue + green = 2+1 = 3
$P(\text{not red}) = \dfrac{3}{6} = \dfrac{1}{2}$
$P(\text{not red}) = \dfrac{3}{6} = \dfrac{1}{2}$
Answer$P(\text{not red}) = \dfrac{1}{2} = 0.5$
(v) Three coins — P(exactly two heads)
$S=\{HHH,HHT,HTH,THH,HTT,THT,TTH,TTT\}$, $n(S)=8$
Exactly 2 heads: $E=\{HHT,HTH,THH\}$, $n(E)=3$
$P(\text{exactly 2 heads}) = \dfrac{3}{8}$
Exactly 2 heads: $E=\{HHT,HTH,THH\}$, $n(E)=3$
$P(\text{exactly 2 heads}) = \dfrac{3}{8}$
Answer$P = \dfrac{3}{8} = 0.375$
5
A bag has 3 candies: strawberry, lemon, mint. P(strawberry)?
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$S = \{\text{Strawberry, Lemon, Mint}\}$, $n(S)=3$
$P(\text{strawberry}) = \dfrac{1}{3}$
$P(\text{strawberry}) = \dfrac{1}{3}$
Answer$P(\text{strawberry}) = \dfrac{1}{3} \approx 0.333$
6
Child: 2 shirts (red, blue) × 3 pants (jeans, khakis, shorts). List all outfit combinations in a table.
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| Outfit # | Shirt | Pants |
|---|---|---|
| 1 | Red | Jeans |
| 2 | Red | Khakis |
| 3 | Red | Shorts |
| 4 | Blue | Jeans |
| 5 | Blue | Khakis |
| 6 | Blue | Shorts |
AnswerTotal = $2 \times 3 = \mathbf{6}$ outfit combinations
7
Tyre company data (1000 cases). Find P(less than 4000 km), P(4000–14000 km), P(more than 14000 km).
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| Range | Cases | Probability |
|---|---|---|
| < 4000 km | 20 | $20/1000 = 0.02$ |
| 4001–9000 km | 210 | $210/1000 = 0.21$ |
| 9001–14000 km | 325 | $325/1000 = 0.325$ |
| > 14000 km | 445 | $445/1000 = 0.445$ |
(i) P(less than 4000 km)
Answer$P = \dfrac{20}{1000} = 0.02\ (2\%)$
(ii) P(between 4000 and 14000 km)
Cases in 4001–9000 + 9001–14000 = $210 + 325 = 535$
Answer$P = \dfrac{535}{1000} = 0.535\ (53.5\%)$
(iii) P(more than 14000 km)
Answer$P = \dfrac{445}{1000} = 0.445\ (44.5\%)$
8
Letters of ‘PEACE’ on cards. Find P(P,E or C) and P(not E).
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P E A C E — 5 cards (note: 2 × E)
(i) P(card is P, E, or C)
$S = \{P,E,A,C,E\}$, $n(S) = 5$
Favourable: P(1) + E(2) + C(1) = 4 cards
$P(P\ \text{or}\ E\ \text{or}\ C) = \dfrac{4}{5}$
Favourable: P(1) + E(2) + C(1) = 4 cards
$P(P\ \text{or}\ E\ \text{or}\ C) = \dfrac{4}{5}$
Answer$P = \dfrac{4}{5} = 0.8$
(ii) P(not E)
Not E: cards P, A, C → 3 cards
$P(\text{not E}) = \dfrac{3}{5}$
$P(\text{not E}) = \dfrac{3}{5}$
Answer$P(\text{not E}) = \dfrac{3}{5} = 0.6$
9
★ Spinner with numbers 1–8 (equally likely). Find P(8), P(odd), P(>2), P(<9), P(multiple of 3).
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Spinner: Numbers 1–8, equal sectors
$S = \{1,2,3,4,5,6,7,8\}$; $n(S) = 8$
| Part | Event | Favourable | Probability |
|---|---|---|---|
| (i) | Points at 8 | {8} → 1 | $\dfrac{1}{8} = 0.125$ |
| (ii) | Odd number | {1,3,5,7} → 4 | $\dfrac{4}{8} = \dfrac{1}{2} = 0.5$ |
| (iii) | Greater than 2 | {3,4,5,6,7,8} → 6 | $\dfrac{6}{8} = \dfrac{3}{4} = 0.75$ |
| (iv) | Less than 9 | {1,2,3,4,5,6,7,8} → 8 | $\dfrac{8}{8} = 1$ (Certain) |
| (v) | Multiple of 3 | {3,6} → 2 | $\dfrac{2}{8} = \dfrac{1}{4} = 0.25$ |
10
★ Basket: 4 red, 5 blue balls. Draw 2 without replacement. Tree diagram. P(red then blue), P(2 blue).
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✓ marks the required outcomes
(i) P(red ball then blue ball)
$P(\text{R then B}) = \dfrac{4}{9} \times \dfrac{5}{8} = \dfrac{20}{72} = \dfrac{5}{18}$
Answer$P(\text{R then B}) = \dfrac{5}{18} \approx 0.278$
(ii) P(drawing 2 blue balls)
$P(\text{B then B}) = \dfrac{5}{9} \times \dfrac{4}{8} = \dfrac{20}{72} = \dfrac{5}{18}$
Answer$P(\text{2 blue}) = \dfrac{5}{18} \approx 0.278$
11
★ Throw a pair of 6-sided dice. Write an event with P=0 and an outcome with P=1.
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When two dice are thrown, minimum sum = 2 and maximum sum = 12.
Event with P = 0 (Impossible):
Getting a sum of 1. (Impossible because minimum is $1+1=2$)
Outcome with P = 1 (Certain):
Getting a sum between 2 and 12. (This always happens no matter what.)
Event with P = 0 (Impossible):
Getting a sum of 1. (Impossible because minimum is $1+1=2$)
Outcome with P = 1 (Certain):
Getting a sum between 2 and 12. (This always happens no matter what.)
AnswerP = 0: “Sum = 1” | P = 1: “Sum is between 2 and 12”
12
★ Complex probability: 5 parts — two dice, coloured balls, three coins, 4-digit numbers, MCQ test.
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(i) Two dice — P(sum is prime > 5)
Total outcomes = 36. Primes > 5 up to 12: 7 and 11.
Sum = 7: {(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)} → 6 ways
Sum = 11: {(5,6),(6,5)} → 2 ways
Total favourable = 8
Sum = 7: {(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)} → 6 ways
Sum = 11: {(5,6),(6,5)} → 2 ways
Total favourable = 8
Answer$P = \dfrac{8}{36} = \dfrac{2}{9} \approx 0.222$
(ii) Bag: 4 red, 3 green, 2 blue — 2 drawn without replacement — P(different colours)
Total = 9; $C(9,2) = 36$ pairs
Same colour: $C(4,2)+C(3,2)+C(2,2) = 6+3+1 = 10$
Different colours = $36-10 = 26$
$P(\text{different}) = \dfrac{26}{36} = \dfrac{13}{18}$
Same colour: $C(4,2)+C(3,2)+C(2,2) = 6+3+1 = 10$
Different colours = $36-10 = 26$
$P(\text{different}) = \dfrac{26}{36} = \dfrac{13}{18}$
Answer$P = \dfrac{13}{18} \approx 0.722$
(iii) Three coins — P(first coin = H AND exactly 2 heads total)
Outcomes with first coin H and exactly 2 total H:
H H T, H T H → 2 outcomes
Total = 8
H H T, H T H → 2 outcomes
Total = 8
Answer$P = \dfrac{2}{8} = \dfrac{1}{4} = 0.25$
(iv) 4-digit numbers using {1,2,3,4} no repetition — P(even)
Total arrangements = $4! = 24$
Even number: last digit must be 2 or 4 → 2 choices for last digit
Remaining 3 digits arranged in $3! = 6$ ways
Favourable = $2 \times 6 = 12$
Even number: last digit must be 2 or 4 → 2 choices for last digit
Remaining 3 digits arranged in $3! = 6$ ways
Favourable = $2 \times 6 = 12$
Answer$P = \dfrac{12}{24} = \dfrac{1}{2} = 0.5$
(v) 3 MCQ, 4 options each, 1 correct — P(exactly 2 correct by guessing)
$P(\text{correct}) = \dfrac{1}{4}$; $P(\text{wrong}) = \dfrac{3}{4}$
$P(\text{exactly 2 correct}) = \binom{3}{2}\!\left(\dfrac{1}{4}\right)^2\!\left(\dfrac{3}{4}\right)^1 = 3 \times \dfrac{1}{16} \times \dfrac{3}{4} = \dfrac{9}{64}$
$P(\text{exactly 2 correct}) = \binom{3}{2}\!\left(\dfrac{1}{4}\right)^2\!\left(\dfrac{3}{4}\right)^1 = 3 \times \dfrac{1}{16} \times \dfrac{3}{4} = \dfrac{9}{64}$
Answer$P = \dfrac{9}{64} \approx 0.141$
13
★ Box: 4 balls numbered 1–4. Tree diagrams for (i) with replacement (ii) without replacement. Sample space sizes.
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(i) With replacement — tree diagram
Answer$n(S) = 4 \times 4 = 16$ outcomes
(ii) Without replacement
First draw: 4 choices. Second draw: 3 remaining choices (can’t repeat).
$S = \{(1,2),(1,3),(1,4),(2,1),(2,3),(2,4),(3,1),(3,2),(3,4),(4,1),(4,2),(4,3)\}$
$S = \{(1,2),(1,3),(1,4),(2,1),(2,3),(2,4),(3,1),(3,2),(3,4),(4,1),(4,2),(4,3)\}$
Answer$n(S) = 4 \times 3 = 12$ outcomes
(iii) Sizes of the two sample spaces
AnswerWith replacement: $n(S) = 16$; Without replacement: $n(S) = 12$
14
★ Simultaneous toss of a coin and drawing of a card from cards 1–6. List all sample space elements.
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Coin: {H, T}; Cards: {1,2,3,4,5,6}
Combine each coin result with each card:
Combine each coin result with each card:
| Coin | Card | |||||
|---|---|---|---|---|---|---|
| H | (H,1) | (H,2) | (H,3) | (H,4) | (H,5) | (H,6) |
| T | (T,1) | (T,2) | (T,3) | (T,4) | (T,5) | (T,6) |
Answer$S = \{(H,1),(H,2),…,(H,6),(T,1),(T,2),…,(T,6)\}$; $n(S) = 12$
15
★ Three coins — number of heads recorded. Which list is a valid sample space? (i){1,2,3} (ii){0,1,2} (iii){0,1,2,3,4} (iv){0,1,2,3}
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Three coins can give 0, 1, 2, or 3 heads — these are the only possible values.
| List | Valid? | Reason |
|---|---|---|
| (i) {1,2,3} | ❌ | Missing 0 heads (TTT). Incomplete. |
| (ii) {0,1,2} | ❌ | Missing 3 heads (HHH). Incomplete. |
| (iii) {0,1,2,3,4} | ❌ | 4 heads is impossible with 3 coins. Extra element. |
| (iv) {0,1,2,3} | ✅ | All values 0,1,2,3 are possible and none is repeated. Correct! |
AnswerList (iv) {0,1,2,3} is the valid sample space.
16
★ A dye is dropped on a rectangle (3m × 2m) containing a circle of diameter 1m. P(landing inside circle)?
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Rectangle 3m × 2m with circle (diameter 1m)
Area of rectangle $= 3 \times 2 = 6\ \text{m}^2$
Radius of circle $= \dfrac{1}{2} = 0.5\ \text{m}$
Area of circle $= \pi r^2 = \pi(0.5)^2 = \dfrac{\pi}{4}\ \text{m}^2 \approx 0.785\ \text{m}^2$
$P(\text{landing in circle}) = \dfrac{\text{Area of circle}}{\text{Area of rectangle}} = \dfrac{\pi/4}{6} = \dfrac{\pi}{24}$
Radius of circle $= \dfrac{1}{2} = 0.5\ \text{m}$
Area of circle $= \pi r^2 = \pi(0.5)^2 = \dfrac{\pi}{4}\ \text{m}^2 \approx 0.785\ \text{m}^2$
$P(\text{landing in circle}) = \dfrac{\text{Area of circle}}{\text{Area of rectangle}} = \dfrac{\pi/4}{6} = \dfrac{\pi}{24}$
Answer$P = \dfrac{\pi}{24} \approx 0.131\ (13.1\%)$