Senior probability problems (ages 16 to 18)

Answer: B, 15/38

1. Pairs of different numbers from 1 to 20: C(20, 2) = 190.
2. The product fails to be a multiple of 6 when it has no factor 2 or no factor 3. Count those bad pairs.
3. No factor 2: both numbers odd: C(10, 2) = 45.
4. No factor 3: both from the 14 non-multiples of 3: C(14, 2) = 91.
5. 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.
6. 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.

Answer: B, 91/36

1. Let I_k be 1 if face k appears in the three rolls and 0 if not. The number of different faces is I₁ + … + I₆.
2. Face k is missing from all three rolls with probability (5/6)³ = 125/216, so E[I_k] = 1 − 125/216 = 91/216.
3. Expectation adds, even though the I_k depend on each other: E[total] = 6 × 91/216.
4. 6 × 91/216 = 91/36 ≈ 2.53.
5. 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.

Answer: C, 5/9

1. The pair (x, y) is a random point in the unit square, so probabilities are areas.
2. The region |x − y| ≥ 1/3 is two corner triangles: y ≤ x − 1/3 and y ≥ x + 1/3.
3. Each is a right-angled isosceles triangle with legs 1 − 1/3 = 2/3, area ½ × (2/3)² = 2/9.
4. Together: 4/9.
5. 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.

Answer: C, 341/1024

1. The number of heads in 10 tosses is a multiple of 3 when it is 0, 3, 6 or 9.
2. Count sequences with k heads: C(10, k). C(10, 0) = 1, C(10, 3) = 120, C(10, 6) = 210, C(10, 9) = 10.
3. Favourable sequences: 1 + 120 + 210 + 10 = 341.
4. All sequences: 2¹⁰ = 1024.
5. Probability: 341/1024 ≈ 0.333 (C).

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.

Answer: C, 1/5

1. Distinct arrangements of BANANA (3 A, 2 N, 1 B): 6!/(3! 2!) = 60, all equally likely.
2. Arrange the non-A letters B, N, N first: 3!/2! = 3 ways.
3. They leave 4 gaps (before, between, after): _ B _ N _ N _.
4. No two As together means the three As go into three different gaps: C(4, 3) = 4 ways.
5. 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.

Answer: A, 19/217 ≈ 8.8%

1. Think of 10 000 people. 1% have the condition: 100 people; 9900 do not.
2. The test detects 95% of those with it: 95 true positives.
3. It gives a positive for 10% of those without it: 990 false positives.
4. All positives: 95 + 990 = 1085, of whom 95 have the condition.
5. 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.

Answer: 6/7

1. C(8, 3) = 56 choices, and no three corners are collinear.
2. 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).
3. 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.

Answer: 6

1. From the start (or just after a tail): toss once; heads moves to state H, tails stays: E = 1 + ½E_H + ½E.
2. From state H: toss once; heads finishes, tails sends you back to the start: E_H = 1 + ½E.
3. 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).

Answer: C, 2/3

1. The chance the point lies within distance r is the area ratio πr²/π = r².
2. So the distance R has density 2r for 0 ≤ r ≤ 1.
3. 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.

Answer: 7/75

1. 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.
2. Each choice of 3 digits gives exactly one increasing number: C(9, 3) = 84.
3. 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.

Answer: 1

1. Let X_i = 1 if person i gets their own hat, else 0. The number of matches is X₁ + … + X₁₀.
2. E[X_i] = P(person i gets their hat) = 1/10, by symmetry.
3. Expectation adds even when the events are not independent: 10 × 1/10 = 1.

Why this works: Linearity of expectation sidesteps the complicated dependence between people’s matches.

Where it leads: The answer is 1 for any number of people. The probability of no match at all tends to 1/e ≈ 0.37.

Answer: A, 35/12

1. E[X] = (1 + 2 + … + 6)/6 = 7/2.
2. E[X²] = (1 + 4 + 9 + 16 + 25 + 36)/6 = 91/6.
3. Var(X) = 91/6 − 49/4 = 182/12 − 147/12 = 35/12 (A).

Why this works: The shortcut formula E[X²] − E[X]² avoids squaring each deviation.

Where it leads: For a fair n-sided die the variance is (n² − 1)/12, the discrete version of the uniform distribution’s 1/12.

Answer: 19/144

1. P(no six) = 625/1296. P(exactly one six) = 4 × 125/1296 = 500/1296.
2. P(at least two) = 1 − 1125/1296 = 171/1296.
3. = 19/144 (about 13%).

Why this works: ‘At least two’ has three cases; its complement has two, both easy binomial probabilities.

Where it leads: This is the binomial distribution B(4, 1/6). The 17th-century gambler’s puzzles (de Méré) about sixes started probability theory.