20 original competition-style problems: dice, cards, areas and expected values. Try each one before opening the hints; the second hint gives more away, and the full solution explains why the method works and where the idea leads.
13 free with full solutions. Problems marked ‘With a plan’ show the question to everyone; their hints, answer checking and full solutions are included with every A Level, IB, IGCSE and CBSE plan. See plans.
For teachers: project, add to a worksheet or set as homework
Press Project on any problem to show it full screen with a timer, the hints, the answer and the worked solution one step at a time (arrow keys move between problems; Space reveals the next step; F full screen; Esc closes). Switch on the ‘Add to worksheet’ buttons, pick problems, then print them from the worksheet builder or set them as homework for a class, with the full solutions as the mark scheme. Free problems are free for every class; problems marked ‘With a plan’ can be set by teachers with a plan or school licence. Ready-made sessions: maths club packs.
Two different numbers are chosen at random from 1 to 20. What is the probability that their product is a multiple of 6?
Hint
Count the bad pairs: those with no factor 2 or no factor 3 (inclusion–exclusion).
Second hint
Bad pairs: both odd, or neither a multiple of 3, minus both odd and neither a multiple of 3.
Full worked solution
Answer: B, 15/38
Pairs of different numbers from 1 to 20: C(20, 2) = 190.
The product fails to be a multiple of 6 when it has no factor 2 or no factor 3. Count those bad pairs.
No factor 2: both numbers odd: C(10, 2) = 45.
No factor 3: both from the 14 non-multiples of 3: C(14, 2) = 91.
Neither factor: both from the 7 numbers coprime to 6 (1, 5, 7, 11, 13, 17, 19): C(7, 2) = 21. Bad pairs: 45 + 91 − 21 = 115.
Good pairs: 190 − 115 = 75, probability 75/190 = 15/38 (B).
Why this works: ‘Divisible by 6’ fails when a factor 2 or a factor 3 is missing; counting the failures with inclusion–exclusion is cleaner than counting successes directly.
Where it leads: 6 = 2 × 3 is not prime, so ‘the product is a multiple of 6’ needs a 2 and a 3 from anywhere in the pair: inclusion–exclusion handles it.
A fair die is rolled three times. What is the expected number of different numbers that appear?
Hint
For each face, what is the probability that it appears at least once? Add these up.
Second hint
P(a given face appears) = 1 − (5/6)3 = 91/216; multiply by 6.
Full worked solution
Answer: B, 91/36
Let Ik be 1 if face k appears in the three rolls and 0 if not. The number of different faces is I1 + … + I6.
Face k is missing from all three rolls with probability (5/6)3 = 125/216, so E[Ik] = 1 − 125/216 = 91/216.
Expectation adds, even though the Ik depend on each other: E[total] = 6 × 91/216.
6 × 91/216 = 91/36 ≈ 2.53.
Answer: 91/36 (B).
Why this works: Linearity of expectation: the expected value of a sum is the sum of expected values, even when the parts are dependent. Indicator variables make counts easy.
Where it leads: Linearity of expectation works even though the events ‘face k appears’ are not independent.
Two numbers x and y are chosen independently and uniformly at random between 0 and 1. What is the probability that they differ by less than 1/3?
Hint
Draw the unit square. The points with |x − y| ≥ 1/3 form two triangles.
Second hint
The two triangles have legs 2/3, so their total area is (2/3)2 = 4/9.
Full worked solution
Answer: C, 5/9
The pair (x, y) is a random point in the unit square, so probabilities are areas.
The region |x − y| ≥ 1/3 is two corner triangles: y ≤ x − 1/3 and y ≥ x + 1/3.
Each is a right-angled isosceles triangle with legs 1 − 1/3 = 2/3, area ½ × (2/3)2 = 2/9.
Together: 4/9.
P(|x − y| < 1/3) = 1 − 4/9 = 5/9 (C).
Why this works: Two independent uniform choices are one random point in a square, so probabilities become areas; the complement here is two easy triangles.
Where it leads: Meeting problems (two people arriving at random within an hour) use exactly this picture.
Why this works: Counts of the form ‘k is a multiple of 3’ split the binomial coefficients into three nearly equal groups (a roots-of-unity filter shows why), which is why the answer is close to 1/3.
Where it leads: Roots of unity filter: (210 + (1 + ω)10 + (1 + ω2)10)/3 gives the same 341 without adding binomials.
The letters of BANANA are arranged in a random order (all distinct-looking arrangements equally likely). What is the probability that no two As are next to each other?
Hint
Arrange B, N, N first, then place the As in the gaps.
Second hint
B, N, N can be arranged in 3 ways, making 4 gaps; choose 3 of the 4 gaps for the As. Out of 60 arrangements in all.
Full worked solution
Answer: C, 1/5
Distinct arrangements of BANANA (3 A, 2 N, 1 B): 6!/(3! 2!) = 60, all equally likely.
Arrange the non-A letters B, N, N first: 3!/2! = 3 ways.
They leave 4 gaps (before, between, after): _ B _ N _ N _.
No two As together means the three As go into three different gaps: C(4, 3) = 4 ways.
Good arrangements: 3 × 4 = 12, probability 12/60 = 1/5 (C).
Why this works: ‘No two together’ is the gap method: place the other letters first, then choose different gaps for the ones that must be separated.
Where it leads: The gaps method handles ‘no two together’ for any number of identical items.
1% of a population has a condition. A test detects it in 95% of people who have it, but also gives a positive result for 10% of people who do not. A randomly chosen person tests positive. What is the probability that they have the condition?
Hint
Imagine 10 000 people and count the positives of each kind.
Second hint
Of 10 000 people: 100 have it (95 test positive); 9900 do not (990 test positive).
Full worked solution
Answer: A, 19/217 ≈ 8.8%
Think of 10 000 people. 1% have the condition: 100 people; 9900 do not.
The test detects 95% of those with it: 95 true positives.
It gives a positive for 10% of those without it: 990 false positives.
All positives: 95 + 990 = 1085, of whom 95 have the condition.
Probability: 95/1085 = 19/217 ≈ 8.8% (A).
Why this works: Bayes’ theorem in natural frequencies: when a condition is rare, false positives from the large healthy group can outnumber true positives.
Where it leads: Bayes’ theorem: with a rare condition, most positive tests are false positives unless the test is extremely specific.
Three different corners of a cube are chosen at random. What is the probability that they form a right-angled triangle?
Hint
Any three corners of a cube form a triangle. What shapes can it be?
Second hint
Every triangle is either right-angled or equilateral (three face diagonals). Count the equilateral ones.
Full worked solution
Answer: 6/7
C(8, 3) = 56 choices, and no three corners are collinear.
A triangle with all three sides face diagonals is equilateral: there are 8 (one cut off at each corner). Every other triangle has a right angle (it contains an edge perpendicular to the plane or face containing the rest).
P(right-angled) = (56 − 8)/56 = 48/56 = 6/7.
Why this works: Classifying the triangles by the lengths of their sides (edge, face diagonal, space diagonal) shows only one type lacks a right angle, and that type is easy to count.
Where it leads: Side lengths available are 1, √2 and √3, and the possible triangles are (1, 1, √2), (1, √2, √3) and (√2, √2, √2). Checking Pythagoras confirms the first two are right-angled.
A fair coin is tossed until two heads in a row appear. What is the expected number of tosses?
Hint
Let E be the expected number from the start, and EH the expected number still needed when the last toss was a head.
Second hint
E = 1 + ½EH + ½E and EH = 1 + ½ × 0 + ½E.
Full worked solution
Answer: 6
From the start (or just after a tail): toss once; heads moves to state H, tails stays: E = 1 + ½EH + ½E.
From state H: toss once; heads finishes, tails sends you back to the start: EH = 1 + ½E.
Substitute: E = 1 + ½(1 + ½E) + ½E = 3/2 + ¾E, so E = 6.
Why this works: Tracking only the part of the history that matters (did the last toss show a head?) gives a small system of equations.
Where it leads: Waiting for HT takes only 4 tosses on average, although HH and HT are equally likely in any two given tosses: overlaps change waiting times (Conway’s leading numbers).
A point is chosen uniformly at random inside a circle of radius 1. What is its expected distance from the centre?
Hint
P(distance ≤ r) = r2 (area of the smaller disc over the whole).
Second hint
So the density of the distance is 2r on [0, 1].
Full worked solution
Answer: C, 2/3
The chance the point lies within distance r is the area ratio πr2/π = r2.
So the distance R has density 2r for 0 ≤ r ≤ 1.
E[R] = ∫01 r × 2r dr = 2/3. Answer: 2/3 (C).
Why this works: Most of a disc’s area is near its edge, so the average distance is more than 1/2.
Where it leads: In a ball of radius 1 in n dimensions the expected distance is n/(n + 1): in high dimensions almost all the volume is near the surface.
A three-digit number from 100 to 999 is chosen at random. What is the probability that its digits are strictly increasing from left to right (like 258)?
Hint
Strictly increasing digits cannot include 0. Why?
Second hint
Choose 3 of the digits 1 to 9; they can be arranged in increasing order in exactly one way.
Full worked solution
Answer: 7/75
A 0 could only be the first digit (it is the smallest), which is not allowed. So the digits are three different digits from 1 to 9.
Each choice of 3 digits gives exactly one increasing number: C(9, 3) = 84.
Probability = 84/900 = 7/75.
Why this works: ‘Increasing’ fixes the order, so counting numbers is counting sets of digits.
Where it leads: Strictly decreasing digits allow 0 (at the end): C(10, 3) = 120 numbers. The asymmetry comes only from the rule about leading zeros.
Kai has £2. He repeatedly bets £1 on a fair coin toss (winning or losing £1 each time) and stops when he has £0 or £5. What is the probability that he reaches £5?
Hints, answer check and full worked solution. Included with every A Level, IB, IGCSE and CBSE plan.
A bag has 3 white and 2 black balls. Balls are drawn one at a time without replacement until the first black ball appears. What is the expected number of draws (including the black one)?
Hints, answer check and full worked solution. Included with every A Level, IB, IGCSE and CBSE plan.
A counter starts at 0 on a number line. Each second it moves 1 to the left or 1 to the right, each with probability 1/2. What is the probability that it is back at 0 after 6 seconds?
Hints, answer check and full worked solution. Included with every A Level, IB, IGCSE and CBSE plan.
IB Math Revision