The Mathematics of Maybe: Introduction to Probability — Advanced Question Bank
A difficult, source-aligned practice resource covering randomness, the probability scale, subjective and objective estimates, experimental and theoretical probability, relative frequency, sampling, Law of Large Numbers, Gambler’s Fallacy, fair and unbiased experiments, sample spaces, events, tree diagrams, replacement versus non-replacement, multi-step experiments and introductory geometric probability.
Probability Revision Sheet
Probability Scale
0 ≤ P(E) ≤ 1
0 = impossible
1/2 = even chance
1 = certain
Experimental Probability
Experimental P(E) = frequency of E / total number of trials
Theoretical Probability
For equally likely outcomes:
P(E)=n(E)/n(S)
Sample Space & Event
S lists all possible outcomes once.
An event E is a subset of S.
Multi-step Experiments
Use systematic listing or tree diagrams. Each complete path represents one ordered outcome.
Long-run Thinking
Experimental probability may differ in small samples but tends toward the theoretical value over many fair trials.
Visual Learning — HTML Canvas
JavaScript is limited to these four diagrams. All solution accordions are native HTML <details>.
Section A — 30 MCQs
Conceptual, computational, source-inspired and reasoning-based.
Probability is primarily a measure of:
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Answer: B
Probability measures how likely an event is to occur.
Which of the following best describes a random experiment?
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Answer: C
A random experiment is repeatable, but the exact outcome of a trial cannot be predicted with certainty.
The probability of an impossible event is:
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Answer: B
Impossible events have probability 0.
The probability of a certain event is:
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Answer: D
Certain events have probability 1.
If P(E)=0.75, then E is:
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Answer: C
0.75 is greater than 0.5, so the event is more likely than not.
Which statement is correct for every event E?
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Answer: B
The probability scale runs from 0 to 1 inclusive.
A die is rolled 50 times and a 4 appears 8 times. The experimental probability of 4 is:
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Answer: B
Experimental probability=8/50=4/25=0.16.
Theoretical probability is computed directly from favourable/total outcomes only when:
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Answer: A
The chapter explicitly requires equally likely outcomes for the theoretical formula.
The sample space for tossing two fair coins is:
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Answer: C
The ordered outcomes are HH, HT, TH and TT.
When three coins are tossed and only the number of heads is recorded, a suitable sample space is:
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Answer: C
The recorded variable is the count of heads, which can be 0,1,2 or 3.
An event is:
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Answer: B
An event may contain one or several outcomes and is a subset of S.
For S={1,2,3,4,5,6}, the event 'number greater than 4' is:
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Answer: B
Only 5 and 6 are greater than 4.
A tree diagram is most useful for:
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Answer: B
Tree diagrams list paths/outcomes in multi-step experiments.
Two fair coin tosses give P(HH)=
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Answer: C
There are four equally likely outcomes and only HH is favourable.
For two fair coin tosses, P(exactly one head)=
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Answer: B
HT and TH are favourable out of four outcomes.
A fair die shows six 6s in a row. P(6 on the next roll) is:
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Answer: C
Each fair die roll is independent; previous rolls do not change the next-roll probability.
Believing that tails is 'due' after six consecutive heads is called:
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Answer: B
The chapter names this misconception Gambler's Fallacy.
As the number of fair repeated trials becomes large, experimental probability tends to:
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Answer: C
This is the Law of Large Numbers described in the chapter.
A fair coin is called unbiased because:
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Answer: B
Symmetry gives no reason for one side to be favoured.
A sample has 20 mango-lovers among 50 students. The relative frequency is:
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Answer: B
20/50=0.4.
Using the 0.4 sample estimate for a school of 1500, the estimated number preferring mango is:
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Answer: C
0.4×1500=600.
A box has 5 green and 7 red balls. If one ball is drawn and only colour is recorded, the sample space is:
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Answer: A
The possible recorded outcomes are simply Green or Red; they are not equally likely.
A bag has 3 red and 7 blue marbles. Which statement is correct?
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Answer: B
Each marble is equally likely, but colour outcomes are not; probabilities depend on counts.
The letters of PEACE are on five identical cards. P(not E) is:
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Answer: C
There are 3 non-E cards: P,A,C.
A spinner has equally likely outcomes 1 to 8. P(multiple of 3) is:
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Answer: B
Multiples of 3 are 3 and 6: 2/8=1/4.
Two dice are rolled. The probability that the sum is 7 is:
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Answer: C
Six ordered pairs sum to 7 out of 36 equally likely pairs.
Two balls are drawn without replacement from 4 red and 5 blue. P(red then blue) is:
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Answer: B
After a red is removed, 8 balls remain, 5 blue.
From the same basket, P(two blue) is:
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Answer: B
Without replacement: 5/9 then 4/8.
A dye is dropped uniformly at random on a 3 m by 2 m rectangle containing a circle of diameter 1 m. The probability it lands in the circle is:
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Answer: A
Geometric probability=area circle/area rectangle=[π(1/2)²]/6=π/24.
Three multiple-choice questions have 4 options each, one correct. A student guesses. P(exactly 2 correct) is:
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Answer: B
Choose which 2 are correct: C(3,2)(1/4)²(3/4)=3×1/16×3/4=9/64.
Section B — 15 One-Mark Questions
Core terminology, sample spaces and quick probabilities.
Define probability in one sentence.
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Probability is a measurement of the likelihood of an event.
What is randomness?
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A situation in which possible outcomes may be known but the exact outcome of an individual trial cannot be predicted in advance.
Write the probability scale.
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0≤P(E)≤1.
What is a sample space?
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The set/list of all possible outcomes of a random experiment.
What is an event?
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An event is one outcome or a group of outcomes, i.e. a subset of the sample space.
Write the sample space for one fair coin toss.
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{H,T}.
Write the sample space for one standard die roll.
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{1,2,3,4,5,6}.
Write the sample space for two coin tosses.
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{HH,HT,TH,TT}.
Write the experimental probability formula.
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Number of times the event occurred / total number of trials.
Write the theoretical probability formula for equally likely outcomes.
View answer / solution
P(E)=number of favourable outcomes / total number of possible outcomes.
A die is rolled 12 times and a 3 appears 3 times. Find the experimental probability.
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3/12=1/4.
For a fair die, find P(3).
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1/6.
A fair coin gives 6 heads in succession. Find P(tail on the next toss).
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1/2.
In a sample of 50 students, 15 like football. Find the relative frequency.
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15/50=3/10=0.3.
If S={HH,HT,TH,TT}, write the event 'at least one head'.
View answer / solution
{HH,HT,TH}.
Section C — 15 Two-Mark Questions
Short experimental, theoretical and sample-space applications.
Differentiate experimental probability from theoretical probability.
View answer / solution
Experimental probability is based on observed relative frequency from actual trials/data. Theoretical probability is calculated by reasoning under equally likely outcomes in an ideal fair setting.
A teacher samples 30 sweets: 10 red, 8 green, 7 yellow, 5 blue. Find the experimental probability of green.
View answer / solution
P(green)=8/30=4/15.
Using the same sample, estimate how many yellow sweets there are in a bag of 600 sweets.
View answer / solution
Estimated yellow proportion=7/30. Estimated count=600×7/30=140.
A sample of 40 students has 11 Arts Club preferences. Find the estimated probability and expected count in a school of 800.
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Probability=11/40=0.275. Estimated count=800×11/40=220.
A fair die is rolled. Find P(even number).
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Favourable={2,4,6}:3 outcomes out of 6. P=1/2.
A die is rolled. Find P(number greater than 4).
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Favourable={5,6}; P=2/6=1/3.
Two coins are tossed. Find P(at least one head).
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S={HH,HT,TH,TT}; favourable={HH,HT,TH}. P=3/4.
Three coins are tossed. Find P(exactly two heads).
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Favourable={HHT,HTH,THH}; total=8. P=3/8.
Ten cards numbered 1 to 10 are equally likely. Find P(even card).
View answer / solution
Even cards={2,4,6,8,10}:5 of 10, so P=1/2.
A bag contains 3 red,2 blue,1 green ball. Find P(not red).
View answer / solution
Non-red=3 balls out of 6, so P=1/2.
A random letter is chosen from PROBABILITY. Find P(B).
View answer / solution
PROBABILITY has 11 letters and 2 Bs, so P(B)=2/11.
The letters of PEACE are on identical cards. Find P(P,E or C).
View answer / solution
P,E,E,C are favourable:4 cards out of 5, so P=4/5.
A spinner has equally likely numbers 1 to 8. Find P(odd) and P(number>2).
View answer / solution
Odd={1,3,5,7}:4/8=1/2. Number>2={3,4,5,6,7,8}:6/8=3/4.
Explain why getting a 6 three times in a row does not change P(6) on the next fair die roll.
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Each die roll is independent. The die has no memory, so every roll retains P(6)=1/6.
A box has 5 green and 7 red balls. Explain why S={G,R} does not mean P(G)=P(R).
View answer / solution
Sample space lists possible recorded outcomes, not their probabilities. Since there are 5 green and 7 red equally likely balls, P(G)=5/12 and P(R)=7/12.
Section D — 15 Three-Mark Questions
Tree diagrams, surveys, dice, replacement and multi-step probability.
Explain the Law of Large Numbers in the context of rolling a fair die.
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For a fair die, theoretical P(4)=1/6. In a small number of rolls, the observed relative frequency of 4 may differ noticeably. As the number of independent rolls becomes large, the experimental proportion tends to get closer to 1/6.
A die is rolled 12 times and a 3 appears 3 times. Compare experimental and theoretical probabilities and explain the difference.
View answer / solution
Experimental P=3/12=1/4. Theoretical P=1/6. They differ because 12 trials are few and random variation is expected. With many more rolls, the experimental value should tend toward 1/6.
Construct the sample space for rolling one die and tossing one coin together, and find P(even number and H).
View answer / solution
S={(1,H),(1,T),…,(6,H),(6,T)}, so n(S)=12. Favourable={(2,H),(4,H),(6,H)}:3 outcomes. P=3/12=1/4.
A village fair offers 3 snacks (Samosa,Pakora,Bhaji) and 2 drinks (Chai,Lassi). List the sample space and find P(Samosa) if each combination is equally likely.
View answer / solution
There are 6 combinations: SC,SL,PC,PL,BC,BL. Samosa event={SC,SL}, so P=2/6=1/3.
Two fair coins are tossed. Use a tree-diagram argument to find P(one head and one tail).
View answer / solution
First toss branches H,T each with 1/2; each branches again H,T. Terminal outcomes HH,HT,TH,TT each have probability 1/4. One head and one tail={HT,TH}, so P=1/2.
Basket A has one apple and two oranges; Basket B has one banana and one mango. One fruit is selected randomly from each basket. Find P(apple and banana).
View answer / solution
P(apple from A)=1/3 and P(banana from B)=1/2. The selections are independent, so P=1/3×1/2=1/6.
A box has 3 red,4 black,2 green pens. A pen is picked, replaced, and another is picked. Find P(same colour).
View answer / solution
P(RR)=(3/9)²=1/9; P(BB)=(4/9)²=16/81; P(GG)=(2/9)²=4/81. Total=9/81+16/81+4/81=29/81.
A tyre company's 1000 records are 20 (<4000), 210 (4001–9000), 325 (9001–14000),445 (>14000). Find the three probabilities requested in the chapter.
View answer / solution
P(<4000)=20/1000=0.02. P(4000–14000)=(210+325)/1000=0.535. P(>14000)=445/1000=0.445.
A basket has 4 red and 5 blue balls. Two are drawn without replacement. Find P(R then B) and P(BB).
View answer / solution
P(R then B)=4/9×5/8=5/18. P(BB)=5/9×4/8=5/18.
Two fair dice are rolled. Find P(sum is a prime greater than 5).
View answer / solution
Possible prime sums >5 are 7 and 11. Sum 7 has 6 ordered pairs; sum 11 has 2. Total favourable=8 of 36, so P=2/9.
A bag has 4 red,3 green,2 blue balls. Two are drawn without replacement. Find P(different colours).
View answer / solution
Total unordered pairs=C(9,2)=36. Same-colour pairs=C(4,2)+C(3,2)+C(2,2)=6+3+1=10. Different=26, so P=26/36=13/18.
Three fair coins are tossed. Find P(first coin is H and exactly two heads occur in total).
View answer / solution
The favourable ordered outcomes are HHT and HTH. Total outcomes=8. Therefore P=2/8=1/4.
A four-digit number is formed using 1,2,3,4 without repetition. Find P(number is even).
View answer / solution
Total permutations=4!=24. An even number must end in 2 or 4:2 choices. Remaining digits arrange in 3!=6 ways. Favourable=12. P=12/24=1/2.
A student guesses answers to 3 questions, each with 4 options. Find P(exactly 2 correct).
View answer / solution
Choose which 2 are correct: C(3,2)=3. Probability for a chosen pattern=(1/4)²(3/4). Thus total=3×1/16×3/4=9/64.
A dye is dropped uniformly on a 3m×2m rectangle. A circle of diameter 1m lies inside it. Find the probability of landing in the circle.
View answer / solution
Rectangle area=6 m². Circle radius=1/2 m; area=π/4. Under uniform landing, probability=(π/4)/6=π/24.
Section E — 10 Four-Mark Questions
Full reasoning, probability models, independence and geometric probability.
A box has balls numbered 1 to 4. Two draws are made with replacement. Write the sample space and its size; then find P(the two numbers are different).
View answer / solution
With replacement, S={(i,j):i,j∈{1,2,3,4}}, so n(S)=16. Equal-number outcomes are (1,1),(2,2),(3,3),(4,4):4. Different outcomes=12. P=12/16=3/4.
Repeat the previous experiment without replacement. Write the sample space size and find P(the second number is greater than the first).
View answer / solution
Without replacement there are 4×3=12 ordered outcomes. For each unordered pair {a,b}, exactly one order has second>first. There are C(4,2)=6 such favourable outcomes. P=6/12=1/2.
A bag contains 4 red and 5 blue balls. Two are drawn without replacement. Find P(one of each colour), P(same colour), and verify the probabilities sum to 1.
View answer / solution
One of each=P(RB)+P(BR)=4/9·5/8+5/9·4/8=10/18=5/9. Same colour=P(RR)+P(BB)=4/9·3/8+5/9·4/8=12/72+20/72=4/9. Sum=5/9+4/9=1.
Two fair dice are rolled. Find the probability that the sum is a prime greater than 5 and compare it with the probability that the sum is 7.
View answer / solution
Prime sums >5 are 7 and 11:8 favourable pairs out of 36, so 2/9. Sum 7 has 6 favourable pairs, so 1/6. Difference=2/9−1/6=1/18; prime>5 is more likely.
A fair coin is tossed 8 times and gives 8 heads. A student says tails is now more likely. Identify the fallacy and give the correct next-toss probability. Then distinguish this from the probability of getting 9 heads in a row from the start.
View answer / solution
The claim is Gambler's Fallacy. The next toss is independent, so P(T)=1/2. But before any tosses, P(9 heads in a row)=(1/2)^9=1/512. Conditional on already observing 8 heads, only the ninth toss remains, so P(H next)=1/2.
A biased-looking bag contains 3 red and 7 blue marbles. Explain why the colour outcomes are not equally likely, yet individual marble outcomes may be considered equally likely. Use this to find P(red).
View answer / solution
If the draw is fair, each of 10 individual marbles is equally likely. But the colour event Red contains 3 elementary outcomes while Blue contains 7, so the colour outcomes are not equally likely. Hence P(red)=3/10.
A random sample of 50 students gives mango preference 20, apple 15, banana 10, grapes 5. Estimate counts for a school of 1500 and explain one limitation.
View answer / solution
Estimated proportions are 0.4,0.3,0.2,0.1. Estimated counts:600 mango,450 apple,300 banana,150 grapes. Limitation: the sample may be too small or not representative; a larger, less biased sample can improve confidence.
For three fair coins, compare these two experiments: A records the full sequence; B records only number of heads. Write each sample space and explain why the four outcomes of B are not equally likely.
View answer / solution
A: {HHH,HHT,HTH,THH,HTT,THT,TTH,TTT}, 8 equally likely outcomes. B:{0,1,2,3} heads. Their probabilities are 1/8,3/8,3/8,1/8 respectively, so B's four outcomes are not equally likely. Thus one cannot use '1/4 each' merely because the sample space has four recorded values.
A spinner has equally likely labels 1 to 8. Find P(odd), P(>2), P(<9), P(multiple of 3), and explain which event is certain.
View answer / solution
P(odd)=4/8=1/2. P(>2)=6/8=3/4. P(<9)=8/8=1, so this event is certain. P(multiple of 3)=2/8=1/4.
A rectangle is 3m×2m and contains a circle of diameter 1m. If a point is selected uniformly at random, derive the geometric probability of landing inside the circle and outside the circle.
View answer / solution
Rectangle area=6. Circle area=π(1/2)²=π/4. P(inside)=π/24. P(outside)=1−π/24=(24−π)/24.
Section F — 5 Case Studies
Sampling, long-run frequency, non-replacement, guessing and geometric probability.
Case Study 1 — School Fruit Survey
A class sample of 50 students gives preferences: mango 20, apple 15, banana 10 and grapes 5. The school has 1500 students.
- a) Find the estimated probability of mango preference.
- b) Estimate the number preferring mango in the school.
- c) Estimate the number preferring banana.
- d) State one way to improve the reliability of the estimate.
View case-study solutions
a) Find the estimated probability of mango preference. 20/50=0.4.
b) Estimate the number preferring mango in the school. 0.4×1500=600.
c) Estimate the number preferring banana. 10/50×1500=300.
d) State one way to improve the reliability of the estimate. Use a larger and more representative sample across different classes/grades.
Case Study 2 — Fair Die Experiment
A fair die is rolled 60 times. The number 4 appears 7 times.
- a) Find the experimental probability of 4.
- b) Find the theoretical probability of 4.
- c) Are the two values required to be exactly equal?
- d) What is expected if the experiment is repeated for very many trials?
View case-study solutions
a) Find the experimental probability of 4. 7/60.
b) Find the theoretical probability of 4. 1/6.
c) Are the two values required to be exactly equal? No. Random variation can make experimental probability differ, especially in a finite sample.
d) What is expected if the experiment is repeated for very many trials? The experimental relative frequency should tend to move closer to 1/6.
Case Study 3 — Two Draws Without Replacement
A basket contains 4 red and 5 blue balls. One ball is drawn and laid aside; then a second is drawn.
- a) Find P(R then B).
- b) Find P(B then R).
- c) Find P(one of each colour).
- d) Why do second-draw probabilities change?
View case-study solutions
a) Find P(R then B). 4/9×5/8=5/18.
b) Find P(B then R). 5/9×4/8=5/18.
c) Find P(one of each colour). 10/18=5/9.
d) Why do second-draw probabilities change? The first ball is not replaced, so the composition and total number of balls change.
Case Study 4 — Multiple-Choice Guessing
A student guesses on 3 questions. Each question has 4 options and exactly one correct option.
- a) Find P(correct on one question).
- b) Find P(wrong on one question).
- c) Find P(exactly two correct).
- d) Find P(all three correct).
View case-study solutions
a) Find P(correct on one question). 1/4.
b) Find P(wrong on one question). 3/4.
c) Find P(exactly two correct). C(3,2)(1/4)²(3/4)=9/64.
d) Find P(all three correct). (1/4)³=1/64.
Case Study 5 — Geometric Probability
A dye is dropped uniformly at random on a 3 m by 2 m rectangular sheet. A circular target of diameter 1 m lies completely inside it.
- a) Find the area of the rectangle.
- b) Find the area of the target.
- c) Find P(hit the target).
- d) Find P(miss the target).
View case-study solutions
a) Find the area of the rectangle. 6 m².
b) Find the area of the target. π/4 m².
c) Find P(hit the target). π/24.
d) Find P(miss the target). 1−π/24=(24−π)/24.
Special Exemplar Challenge — 15 Questions
Non-routine complements, counting, dice, coins and replacement problems.
A fair die is rolled once. Find P(prime number), P(composite number), and P(neither prime nor composite).
View worked solution
Prime={2,3,5}:1/2. Composite={4,6}:1/3. Neither={1}:1/6.
Two coins are tossed. Find P(at least one tail) without listing favourable outcomes directly.
View worked solution
Use complement: P(at least one tail)=1−P(HH)=1−1/4=3/4.
Three coins are tossed. Find P(at most one head).
View worked solution
Outcomes with 0 or 1 head: TTT,HTT,THT,TTH =4 of 8. P=1/2.
A number is selected from 1 to 20. Find P(multiple of 3 or 5).
View worked solution
Multiples of 3:6; of 5:4; common multiples of 15:1. Favourable=6+4−1=9, so P=9/20.
A number is selected from 1 to 30. Find P(prime).
View worked solution
Primes:2,3,5,7,11,13,17,19,23,29 →10. P=10/30=1/3.
Two dice are rolled. Find P(sum=8).
View worked solution
Favourable ordered pairs:(2,6),(3,5),(4,4),(5,3),(6,2):5. P=5/36.
Two dice are rolled. Find P(sum≥10).
View worked solution
Sums 10,11,12 have 3+2+1=6 outcomes. P=6/36=1/6.
Two dice are rolled. Find P(product is even).
View worked solution
Complement: product odd only when both dice are odd:3×3=9 outcomes. P(even product)=1−9/36=3/4.
From PEACE, find P(vowel) and P(E | the chosen card is a vowel) conceptually.
View worked solution
Vowel cards are E,A,E:3/5. Conditional among vowels, two of the three are E, so 2/3.
A bag has 2 red,3 green,5 blue balls. One is drawn. Find P(not blue).
View worked solution
Non-blue=5 out of 10, so P=1/2.
A bag has 4 red,3 green,2 blue balls. Two are drawn without replacement. Find P(both same colour).
View worked solution
[C(4,2)+C(3,2)+C(2,2)]/C(9,2)=(6+3+1)/36=5/18.
Two draws are made with replacement from balls numbered 1,2,3,4. Find P(sum=5).
View worked solution
There are 16 ordered pairs; favourable (1,4),(2,3),(3,2),(4,1):4. P=1/4.
Two draws are made without replacement from 1,2,3,4. Find P(sum=5).
View worked solution
There are 12 ordered outcomes; favourable same four ordered pairs:4. P=1/3.
A point is selected uniformly from a 10 cm by 10 cm square. A circle of radius 5 cm is inscribed. Find P(point lies in the circle).
View worked solution
Area ratio=25π/100=π/4.
A student guesses 4 true/false questions. Find P(exactly 3 correct).
View worked solution
C(4,3)(1/2)^4=4/16=1/4.
Extra Olympiad / HOTS Challenge — 20 Questions
Conditional reasoning, advanced counting, repeated trials and geometric probability.
Three fair coins are tossed. Given that at least one head occurs, find the probability of exactly two heads.
View worked solution
Conditioned sample excludes TTT, leaving 7 equally likely full-sequence outcomes. Exactly two heads has 3 outcomes, so probability=3/7.
Two dice are rolled. Given that the sum is 8, find P(at least one die is 3).
View worked solution
Sum-8 outcomes are (2,6),(3,5),(4,4),(5,3),(6,2). Two contain a 3, so probability=2/5.
Two dice are rolled. Find P(the larger number is 4).
View worked solution
Ordered outcomes with max=4 are all pairs from {1,2,3,4} minus all pairs from {1,2,3}:16−9=7. P=7/36.
Two dice are rolled. Find P(the numbers are different).
View worked solution
Complement of doubles. P=1−6/36=5/6.
Two dice are rolled. Find P(sum is prime).
View worked solution
Prime sums possible:2,3,5,7,11. Counts=1+2+4+6+2=15. P=15/36=5/12.
Three coins are tossed. Let X be number of heads. Find P(X is even).
View worked solution
X=0 or 2. Counts=1+C(3,2)=1+3=4 of 8, so P=1/2.
Four fair coins are tossed. Find P(exactly two heads).
View worked solution
C(4,2)/16=6/16=3/8.
Four fair coins are tossed. Find P(at least three heads).
View worked solution
[C(4,3)+C(4,4)]/16=(4+1)/16=5/16.
A number is chosen uniformly from 1 to 100. Find P(it is divisible by 2 or 5 but not both).
View worked solution
Divisible by 2:50; by 5:20; by both(10):10. Exactly one=50+20−2·10=50. Probability=1/2.
A number is chosen from 1 to 60. Find P(it is divisible by 4 or 6).
View worked solution
Multiples of 4=15; of 6=10; of lcm12=5. Union=20. P=20/60=1/3.
A bag has 5 red,4 blue,3 green balls. Two are drawn without replacement. Find P(at least one red).
View worked solution
Use complement: no red means both from 7 non-red balls. P=1−C(7,2)/C(12,2)=1−21/66=45/66=15/22.
From the same bag, find P(the two balls have different colours).
View worked solution
Total pairs=C(12,2)=66. Same-colour=C(5,2)+C(4,2)+C(3,2)=10+6+3=19. Different=47. P=47/66.
A box contains 2 red and 3 blue balls. Two balls are drawn with replacement. Find P(exactly one red).
View worked solution
P(RB)+P(BR)=2/5·3/5+3/5·2/5=12/25.
The same box is used without replacement. Find P(exactly one red).
View worked solution
P=2/5·3/4+3/5·2/4=6/20+6/20=3/5.
A student guesses 5 multiple-choice questions with 4 choices each. Find P(exactly one correct).
View worked solution
C(5,1)(1/4)(3/4)^4=5·1/4·81/256=405/1024.
A student guesses 5 such questions. Find P(at least one correct).
View worked solution
Complement all wrong:1−(3/4)^5=1−243/1024=781/1024.
A point is chosen uniformly inside a square of side 2r. A circle of radius r is inscribed. Find P(point is outside the circle).
View worked solution
Square area=4r², circle=πr². P(outside)=(4−π)/4=1−π/4.
A point is uniformly selected in an annulus between concentric radii r and 2r. Find probability it lies within distance 3r/2 of the centre.
View worked solution
Desired annular area=π[(3r/2)²−r²]=5πr²/4. Total annulus=π(4r²−r²)=3πr². Probability=(5/4)/3=5/12.
A fair coin is tossed repeatedly until the first head. Find P(first head occurs on toss 3).
View worked solution
The sequence must be TTH. Probability=(1/2)^3=1/8.
A fair die is rolled repeatedly until a 6 appears. Find P(first 6 occurs on roll 4).
View worked solution
First three must be non-6 and fourth is 6: (5/6)^3(1/6)=125/1296.
High-Level Thinking Laboratory — 10 Questions
Students analyse assumptions, sample-space design, independence, bias and long-run reasoning.
Why does a sample space need enough detail to match the question being asked?
View worked solution
A coarse sample space can merge outcomes that the question needs to distinguish. For example, {Rain,No Rain} is enough for 'Will it rain?' but not for comparing drizzle, light rain and heavy rain.
Why can two outcomes listed in a sample space fail to be equally likely?
View worked solution
A sample space records what can happen, not necessarily equal probabilities. For example, colour outcomes {R,B} from a bag with 3 red and 7 blue are not equally likely.
Why is 'experimental probability should equal theoretical probability after 12 trials' an incorrect claim?
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Experimental probability fluctuates randomly. The Law of Large Numbers describes long-run tendency, not exact equality after a fixed small number of trials.
A coin lands heads 10 times in a row. Explain the difference between 'the probability of 11 heads from the start' and 'the probability the next toss is head now'.
View worked solution
Before tossing, P(11 consecutive heads)=1/2^11. After 10 heads have already occurred, the next fair toss is independent, so P(head next)=1/2.
Why is P(E)=favourable/total dangerous if applied without checking equally likely outcomes?
View worked solution
Counting outcomes works only if elementary outcomes have equal probabilities. If recorded categories have unequal weights, simple counting can give the wrong answer.
Why does replacement matter in multi-step probability?
View worked solution
Replacement restores the original composition, making successive draw probabilities unchanged. Without replacement, totals and category counts change after each draw.
Why is a larger sample usually more useful but not automatically unbiased?
View worked solution
A larger sample reduces random fluctuation, but a systematically unrepresentative selection method can remain biased regardless of size.
Why can geometric probability be treated as an area ratio in the dye-drop question?
View worked solution
Uniform random landing means equal-area regions are equally likely. Therefore probability is proportional to area, so favourable area/total area gives the probability.
Why is {0,1,2,3} a valid sample space for 'number of heads in three tosses' even though these four outcomes are not equally likely?
View worked solution
A sample space only needs to list all possible recorded outcomes once. Equal likelihood is not a requirement for being a sample space.
How does a tree diagram help prevent omitted or duplicated outcomes?
View worked solution
Each complete root-to-leaf path corresponds to one ordered outcome, so systematic branching makes it easier to enumerate every possible sequence exactly once.
Exam Strategy & Common Traps
| Topic | Best first move | Common trap |
|---|---|---|
| Sample space | Identify exactly what the experiment records. | Using an over-detailed or under-detailed sample space. |
| Theoretical probability | Check that elementary outcomes are equally likely before counting. | Assuming listed categories are equally likely. |
| Experimental probability | Use observed frequency ÷ total trials. | Replacing observed frequency with theoretical expectation. |
| Tree diagrams | Label every branch and multiply probabilities along a path. | Adding probabilities along one path instead of multiplying. |
| Without replacement | Update both numerator and denominator after the first draw. | Reusing first-draw probabilities. |
| At least / not | Consider a complement when it is simpler. | Omitting cases. |
| Gambler’s Fallacy | Ask whether trials are independent. | Believing an outcome is “due”. |
| Sampling | Check both sample size and representativeness. | Thinking a large biased sample must be reliable. |

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