Intermediate probability problems (ages 13 to 16)

Answer: B, 7/36

1. Two dice: 36 equally likely ordered outcomes.
2. The larger number is at most 4 when both dice are at most 4: 4 × 4 = 16 outcomes.
3. The larger number is at most 3 when both are at most 3: 3 × 3 = 9 outcomes.
4. The larger is exactly 4 in 16 − 9 = 7 outcomes: (4,1), (4,2), (4,3), (4,4), (1,4), (2,4), (3,4).
5. Probability: 7/36 (B).

Why this works: ‘Max ≤ k’ is easy (every die ≤ k), so ‘max = k’ is a difference of two easy counts.

Where it leads: ‘The maximum is at most k’ is easy; taking differences of these gives the exact distribution of the maximum.

Answer: C, 5/9

1. Cards 1–9: five odd (1, 3, 5, 7, 9) and four even (2, 4, 6, 8).
2. A sum is odd exactly when one card is odd and the other even.
3. Pairs with one of each: 5 × 4 = 20.
4. All pairs of different cards: C(9, 2) = 36.
5. Probability: 20/36 = 5/9 (C).

Why this works: Parity questions reduce to counting odd/even choices. Without replacement, count unordered pairs consistently (or ordered pairs consistently).

Where it leads: Parity arguments make many probability questions short: only odd/even matters here, not the actual numbers.

Answer: C, π/4

1. A random point in the square is equally likely to be anywhere, so probability = favourable area ÷ 4.
2. Points within 1 of a corner form a quarter disc of radius 1 at that corner, with area π × 1² ÷ 4 = π/4.
3. Neighbouring corners are 2 apart, so quarter discs of radius 1 only touch at the midpoints of the sides: they do not overlap.
4. Favourable area: 4 × π/4 = π.
5. Probability: π/4 ≈ 0.785, answer π/4 (C).

Why this works: For a point chosen uniformly, probability = favourable area ÷ total area. Always check whether the pieces overlap before adding.

Where it leads: Geometric probability: for a uniform random point, probability = area of the region / total area.

Answer: B, 2/5

1. Four coins give 2⁴ = 16 equally likely outcomes.
2. The condition ‘at least one head’ removes only TTTT, leaving 15 equally likely outcomes.
3. Exactly two heads: choose which 2 of the 4 coins, C(4, 2) = 6 outcomes, all of which have at least one head.
4. Probability: 6/15 = 2/5 (B). (Without the condition it would be 6/16 = 3/8, the trap.)

Why this works: Conditioning shrinks the set of possible outcomes. Count favourable outcomes inside the new, smaller set.

Where it leads: Conditioning shrinks the sample space; here the condition removes only TTTT.

Answer: B, 5/54

1. Three rolls give 6³ = 216 equally likely outcomes.
2. A strictly increasing outcome uses three different numbers in increasing order.
3. Every choice of three different numbers from 1–6 gives exactly one increasing order, so count choices: C(6, 3) = 20.
4. Probability: 20/216 = 5/54 (B).

Why this works: Of the 6 orders of three different numbers exactly one is increasing, so you could also say: P(all different) × 1/6 = (120/216) × (1/6).

Where it leads: For any n rolls, P(strictly increasing) = C(6, n)/6ⁿ, which is 0 once n > 6.

Answer: C, 1/6

1. The first red ball on the third draw means: blue, then blue, then red.
2. First draw blue: 6 of 10 balls, probability 6/10.
3. Second blue: 5 blue left of 9, probability 5/9.
4. Third red: 4 red of the 8 left, probability 4/8.
5. Multiply: 6/10 × 5/9 × 4/8 = 120/720 = 1/6 (C).

Why this works: Without replacement, multiply conditional probabilities: each fraction reflects what is left in the bag at that moment.

Where it leads: The first red ball’s position has a neat distribution; its expected position is 11/5 here.

Answer: 8/15

1. Pairs: C(10, 2) = 45, all equally likely.
2. The product is not a multiple of 3 only if neither card is: C(7, 2) = 21 pairs.
3. P(multiple of 3) = 1 − 21/45 = 24/45 = 8/15.

Why this works: Because 3 is prime, the product is a multiple of 3 exactly when at least one factor is; the complement is one clean count.

Where it leads: For a non-prime like 4 this fails (2 × 6 = 12). Then you need cases: one multiple of 4, or two even numbers.

Answer: C, 2/3

1. Prime scores: 2, 3, 5. Given the score is prime, these three are equally likely.
2. Odd ones: 3 and 5.
3. P(odd | prime) = 2/3 (C).

Why this works: Conditioning shrinks the sample space to the outcomes you are told happened; then count within it.

Where it leads: P(A | B) = P(A and B)/P(B) = (2/6)/(3/6). Getting the condition the right way round matters: P(prime | odd) is also 2/3 here, but usually the two differ.

Answer: 5/6 (about 83p)

1. Win £10 with probability 1/6; lose £1 with probability 5/6.
2. Expected gain = 10/6 − 5/6 = 5/6.
3. 5/6 of a pound, about 83p per game, in your favour.

Why this works: Expected value weights each outcome by its probability; it is the average gain per game over many games.

Where it leads: A fair game has expected gain 0: here the prize would need to be £5. Casinos set every game’s expected gain slightly negative for the player.

Answer: 25/216

1. The first 6 is on the third roll exactly when rolls 1 and 2 are not 6 and roll 3 is 6.
2. The rolls are independent: (5/6) × (5/6) × (1/6).
3. = 25/216.

Why this works: ‘First success on trial k’ is k − 1 failures followed by a success, so the probabilities multiply.

Where it leads: This is the geometric distribution. Summing (5/6)^k−1(1/6) over all k gives 1: a 6 turns up eventually with certainty.

Answer: D, 3/4

1. Of the 8 equally likely outcomes, the only ones without both a head and a tail are HHH and TTT.
2. So 6 outcomes succeed.
3. Probability = 6/8 = 3/4 (D).

Why this works: The complement (‘all the same’) has just two outcomes, so subtracting is quickest.

Where it leads: With n coins the answer is 1 − 2/2ⁿ. With n dice, ‘at least one of each of the six faces’ needs inclusion–exclusion: the coupon collector problem.

Answer: 2/9

1. Doubles (1, 1), …, (6, 6) always give squares: 6 outcomes.
2. Non-doubles: the product of two different numbers from 1 to 6 is a square only for 1 × 4 = 4. (Checking the other 14 pairs, such as 2 × 3 = 6 or 3 × 6 = 18, none is a square.) So (1, 4) and (4, 1): 2 outcomes.
3. Probability = 8/36 = 2/9.

Why this works: A product ab is a square when a and b have the same ‘square-free part’: 1 and 4 both have square-free part 1, every other number from 1 to 6 has its own.

Where it leads: Grouping numbers by square-free part is a key idea in problems like ‘choose numbers so that no product is a square’.

Answer: 15/28

1. Ways to choose any 3 counters: C(8, 3) = 56.
2. Ways to choose 2 reds and 1 green: C(5, 2) × C(3, 1) = 10 × 3 = 30.
3. Probability = 30/56 = 15/28.

Why this works: When counters are drawn without replacement and order does not matter, counting combinations is cleaner than a tree diagram.

Where it leads: This is the hypergeometric distribution, used for quality control and for estimating fish populations by capture–recapture.

Answer: B, 1/8

1. The square has area 4.
2. x + y < 1 (with x, y ≥ 0) is the right-angled triangle with corners (0, 0), (1, 0), (0, 1): area 1/2.
3. Probability = (1/2)/4 = 1/8 (B).

Why this works: For a uniformly random point, probabilities are proportions of area.

Where it leads: Geometric probability solves problems like ‘two people arrive at random within an hour; what is the chance they meet?’ by drawing the region in a square.

Answer: 0.32

1. Rain Monday and Tuesday: 0.3 × 0.6 = 0.18.
2. Dry Monday, rain Tuesday: 0.7 × 0.2 = 0.14.
3. Total: 0.18 + 0.14 = 0.32.

Why this works: The law of total probability: add the probabilities of every route that leads to the event.

Where it leads: Weather that depends only on yesterday is a Markov chain. In the long run the chance of rain settles to a fixed value: here 1/3.

Answer: C, 3/4

1. P(A and red) = 1/2 × 3/4 = 3/8. P(B and red) = 1/2 × 1/4 = 1/8.
2. P(red) = 3/8 + 1/8 = 1/2.
3. P(A | red) = (3/8)/(1/2) = 3/4 (C).

Why this works: Bayes’ rule: compare the routes that produce what you saw. Box A produces red three times as often, so it is three times as likely.

Where it leads: Think of 8 equally likely ‘ball draws’: 4 from each box. Red appears 3 times from A and once from B. Counting like this makes Bayes’ rule intuitive.

Answer: 6

1. Let E be the expected number of rolls.
2. After the first roll: with probability 1/6 it is a 6 (total 1 roll); with probability 5/6 you start again, having used 1 roll.
3. So E = 1 + (5/6)E, giving E/6 = 1 and E = 6.

Why this works: ‘If it fails you are back at the start’ gives an equation for E in terms of itself: first-step analysis.

Where it leads: An event with probability p takes 1/p tries on average. Collecting all six faces takes 6(1 + 1/2 + … + 1/6) = 14.7 rolls on average.

Answer: 6/25

1. Total 8: (3, 5), (4, 4), (5, 3). Total 9: (4, 5), (5, 4). Total 10: (5, 5).
2. That is 6 outcomes out of 25.
3. Probability = 6/25.

Why this works: With a small table of equally likely outcomes, a careful list of the high totals is quickest.

Where it leads: By symmetry (replace each score s by 6 − s), P(total ≥ 8) = P(total ≤ 4) = 6/25.

Answer: D, 1/3

1. LETTER has two Ts and two Es: 6!/(2! 2!) = 180 arrangements.
2. Treat TT as one block: arrange L, E, E, R and the block: 5!/2! = 60.
3. Probability = 60/180 = 1/3 (D).

Why this works: Gluing the Ts into a block counts exactly the arrangements where they are adjacent.

Where it leads: Quick check: the two Ts occupy 2 of 6 positions, C(6, 2) = 15 equally likely pairs, and 5 of those pairs are adjacent: 5/15 = 1/3.

Answer: 1/10

1. Whatever the first number a, the second must be 11 − a.
2. 11 − a is always one of 1 to 10, so exactly one of the 10 choices works.
3. Probability = 1/10.

Why this works: Fixing the first choice and asking what the second must be often avoids listing pairs.

Where it leads: For a total of 12 the answer drops to 9/100, since a = 1 leaves no partner. Totals near the middle are most likely.

Answer: 41/96

1. P(all different) = (12 × 11 × 10 × 9)/12⁴ = 11880/20736 = 55/96.
2. P(at least two share) = 1 − 55/96.
3. = 41/96, about 43%.

Why this works: ‘At least two the same’ covers many cases; ‘all different’ is one product.

Where it leads: With 5 people the chance passes 60%. With birthdays (365 days) it passes 50% at 23 people: the famous birthday paradox.

Answer: B, 1/3

1. P(first head on toss k) = (1/2)^k.
2. P(even) = 1/4 + 1/16 + 1/64 + …, a geometric series with first term 1/4 and ratio 1/4.
3. Sum = (1/4)/(1 − 1/4) = 1/3 (B).

Why this works: An infinite geometric series adds up to a/(1 − r). Alternatively, P(odd) = 2 × P(even), because each even case is half as likely as the odd case before it.

Where it leads: The second argument shows the first tosser in a ‘first head wins’ game has a 2/3 chance: going first is a real advantage.

Answer: 2/3

1. If the shorter piece is s, the longer is 1 − s, and 1 − s ≥ 2s means s ≤ 1/3.
2. The shorter piece is at most 1/3 when the break is within 1/3 of either end: in [0, 1/3] or [2/3, 1].
3. Total length of these intervals: 2/3, so the probability is 2/3.

Why this works: Translating the condition onto the break point turns it into lengths on the stick.

Where it leads: Breaking a stick at two random points gives three pieces that form a triangle with probability 1/4, a classic geometric probability result.

Answer: 3/4

1. 72 = 2³ × 3² has (3 + 1)(2 + 1) = 12 divisors.
2. Odd divisors use no factor 2: they are the divisors of 9, namely 1, 3, 9.
3. So 12 − 3 = 9 are even, and the probability is 9/12 = 3/4.

Why this works: Counting divisors through the prime factorisation makes ‘odd divisors’ simply ‘choose exponent 0 for 2’.

Where it leads: For n = 2^am with m odd, the probability is a/(a + 1). So a number’s divisors are mostly even exactly when it has a high power of 2.

Answer: B, 1/8

1. Unordered sets summing to 10: {1, 3, 6}, {1, 4, 5}, {2, 3, 5} (all different: 6 orders each) and {2, 2, 6}, {2, 4, 4}, {3, 3, 4} (one repeat: 3 orders each).
2. Outcomes: 3 × 6 + 3 × 3 = 27.
3. Probability = 27/216 = 1/8 (B).

Why this works: Counting unordered sets and then their number of orderings is safer than listing 27 ordered triples.

Where it leads: Totals 10 and 11 are the most likely with three dice (27 ways each). Galileo explained this to gamblers who had noticed it in practice.

Answer: C, 161/36

1. P(larger ≤ k) = P(both ≤ k) = k²/36. So P(larger = k) = (2k − 1)/36.
2. Expected value = Σ k(2k − 1)/36 = (1 + 6 + 15 + 28 + 45 + 66)/36.
3. = 161/36 ≈ 4.47. Answer: 161/36 (C).

Why this works: ‘The maximum is at most k’ is easy (both dice at most k); differences of these give the exact distribution.

Where it leads: The smaller score has expected value 7 − 161/36 = 91/36, because larger + smaller = sum of the dice, whose expectation is 7.