Junior probability problems (ages 11 to 13)

Answer: C, 2/9

1. Two dice give 6 × 6 = 36 equally likely ordered outcomes.
2. Pairs differing by exactly 2, smaller first: (1, 3), (2, 4), (3, 5), (4, 6): 4 pairs.
3. Each can appear in either order, e.g. (3, 1) as well as (1, 3): 8 outcomes.
4. Probability: 8/36 = 2/9 (C).

Why this works: With two dice, list outcomes as ordered pairs so each of the 36 is equally likely. Forgetting the reversed pairs is the usual error.

Where it leads: The difference of two dice has a triangular distribution, most likely 0 or 1.

Answer: 7

1. Suppose r red balls are added. Then there are 3 + r red balls out of 8 + r in total.
2. We need (3 + r)/(8 + r) = 2/3.
3. Cross-multiply: 3(3 + r) = 2(8 + r), so 9 + 3r = 16 + 2r.
4. So r = 7.
5. Check: 10 red out of 15 is 10/15 = 2/3. ✓ Add 7 red balls.

Why this works: A probability is a fraction of a total, so changing the contents changes both the top and the bottom. Set up the fraction and cross-multiply.

Where it leads: Adding red balls pushes P(red) towards 1 but never reaches it: the probability is always less than 1 with any blue balls left.

Answer: C, 1/8

1. Two spins give 8 × 8 = 64 equally likely outcomes.
2. For a sum of 9, the second number must be 9 minus the first.
3. Whatever the first number (1 to 8), 9 minus it is also between 1 and 8, so exactly one second number works.
4. That gives 8 good outcomes: (1,8), (2,7), …, (8,1).
5. Probability: 8/64 = 1/8 (C).

Why this works: Sometimes it is quicker to see that every first outcome has exactly one good partner than to list all good pairs.

Where it leads: Totals in the middle of the range are the most likely; with two spinners 1 to 8, 9 is the most likely total.

Answer: D, 1/2

1. There are 2⁴ = 16 equally likely sequences. Count those with no HH.
2. Let g(n) be the number of good sequences of length n. A good sequence ends in T (after any good sequence of length n − 1) or in TH (after any good sequence of length n − 2).
3. So g(n) = g(n − 1) + g(n − 2), with g(1) = 2 (H, T) and g(2) = 3 (HT, TH, TT).
4. g(3) = 5 and g(4) = 8. (They are TTTT, TTTH, TTHT, THTT, HTTT, THTH, HTHT, HTTH.)
5. Probability: 8/16 = 1/2 (D).

Why this works: The same ‘look at the ending’ recurrence as for colouring squares works for coin sequences: good sequences of length n number F(n + 2), a Fibonacci number.

Where it leads: Fibonacci again: the number of length-n sequences with no HH is F_n+2, so the probability shrinks like (0.809)ⁿ.

Answer: D, 1/2

1. Let A = multiples of 3 from 1 to 50 and B = numbers containing the digit 3.
2. |A| = 16 (3, 6, …, 48).
3. B: 3, 13, 23, 43 and 30–39 (ten numbers): |B| = 14.
4. Both: 3, 30, 33, 36, 39: 5 numbers.
5. |A or B| = 16 + 14 − 5 = 25.
6. Probability: 25/50 = 1/2 (D).

Why this works: ‘Or’ means inclusion–exclusion again: count both lists and subtract the numbers that are on both.

Where it leads: ‘Or’ conditions overlap; listing the extra cases (not already counted) avoids double counting.

Answer: C, 3/8

1. There are 4! = 24 equally likely orders.
2. Card 1 must go to position 2, 3 or 4; by symmetry each choice gives the same number of good orders. Take card 1 in position 2.
3. Case card 2 in position 1: cards 3 and 4 must swap (3 in 4, 4 in 3): 1 way.
4. Case card 2 in position 3: then card 3 cannot go to 3, so 3 goes to 4 and 4 to 1: 1 way.
5. Case card 2 in position 4: 4 must avoid 4, so 4 goes to 3 and 3 to 1: 1 way.
6. 3 ways for each of 3 places for card 1: 9 good orders.
7. Probability: 9/24 = 3/8 (C).

Why this works: Arrangements where nothing is in its own place are called derangements. For n cards the probability is close to 1/e ≈ 0.37 even for small n — 3/8 = 0.375 here.

Where it leads: These are derangements: D₄ = 9 of the 24 orders. As n grows, the probability tends to 1/e.

Answer: C, 3/4

1. A product is odd only if both scores are odd.
2. P(first odd) = 1/2 and P(second odd) = 1/2, so P(both odd) = 1/4.
3. P(product even) = 1 − 1/4 = 3/4 (C).

Why this works: ‘Even’ has many cases; its opposite ‘odd’ has just one, so use P(A) = 1 − P(not A).

Where it leads: With n dice the chance of an even product is 1 − (1/2)ⁿ: very quickly almost certain.

Answer: 5/18

1. P(first red) = 5/9.
2. If the first is red, 8 counters are left and 4 are red: P(second red) = 4/8.
3. P(both red) = 5/9 × 4/8 = 20/72 = 5/18.

Why this works: Without replacement, the second draw depends on the first, so the second probability is worked out from what is left.

Where it leads: The same answer comes from counting pairs: C(5, 2)/C(9, 2) = 10/36 = 5/18. Two methods agreeing is a strong check.

Answer: 1/2

1. The 8 outcomes are HHH, HHT, HTH, THH, HTT, THT, TTH, TTT.
2. At least two heads: HHH, HHT, HTH, THH: 4 outcomes.
3. Probability = 4/8 = 1/2.
4. Symmetry check: with three tosses, either heads or tails is in the majority, and swapping H and T shows they are equally likely.

Why this works: Listing equally likely outcomes works for small experiments; symmetry gives the answer instantly.

Where it leads: With an odd number of tosses, P(more heads than tails) is always exactly 1/2. With an even number, ties make it less.

Answer: D, 4/5

1. Primes: 2, 3, 5, 7. Even numbers: 2, 4, 6, 8, 10.
2. Together: 2, 3, 4, 5, 6, 7, 8, 10 (2 is counted once). That is 8 numbers.
3. Only 1 and 9 are neither. Probability = 8/10 = 4/5 (D).

Why this works: ‘Or’ means the union: list it without double counting, or count the complement (just 1 and 9).

Where it leads: P(A or B) = P(A) + P(B) − P(A and B): here 4/10 + 5/10 − 1/10 = 8/10.

Answer: 5/12

1. Prime totals: 2, 3, 5, 7, 11.
2. Ways (out of 36): total 2: 1; total 3: 2; total 5: 4; total 7: 6; total 11: 2.
3. 1 + 2 + 4 + 6 + 2 = 15, so the probability is 15/36 = 5/12.

Why this works: The 36 ordered outcomes are equally likely; the totals are not, so count the outcomes behind each total.

Where it leads: The number of ways to make each total, 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1, is the coefficient pattern of (x + x² + … + x⁶)². Generating functions turn dice into algebra.

Answer: 17/100

1. Multiples of 4 from 1 to 100: 25.
2. Of these, the ones also divisible by 6 are the multiples of 12: 12, 24, …, 96, which is 8.
3. 25 − 8 = 17, so the probability is 17/100.

Why this works: Divisible by 4 and 6 means divisible by LCM(4, 6) = 12, not by 4 × 6 = 24.

Where it leads: In the long run the proportion of whole numbers divisible by 4 but not 6 is 1/4 − 1/12 = 1/6. Densities like this are a starting point for analytic number theory.

Answer: C, 8

1. P(blue) = 1 − 2/5 = 3/5, so 3/5 of the beads are blue.
2. 3/5 of the total is 12, so the total is 20.
3. Red beads: 20 − 12 = 8 (C).

Why this works: Probabilities of all outcomes add to 1, so knowing P(red) tells you P(blue) and hence the total.

Where it leads: If you add one more bead of each colour the probability of red changes. Can you predict whether it goes up or down?

Answer: 1/3

1. ‘A is to the left of both B and C’ means A is the leftmost card.
2. By symmetry, each of A, B and C is equally likely to be leftmost.
3. So the probability is 1/3. (Check by listing: ABC and ACB out of 6 orders.)

Why this works: Symmetry between the cards answers the question without listing: nothing makes one card more likely than another to come first.

Where it leads: With n cards, P(a given card is to the left of all the others) = 1/n. The same idea shows the chance that the last of n random numbers is the largest is 1/n.

Answer: 1/24

1. The three events are independent, so multiply.
2. P(two heads) = 1/2 × 1/2 = 1/4. P(six) = 1/6.
3. P(all) = 1/4 × 1/6 = 1/24.

Why this works: For independent events, the probability that all happen is the product of their probabilities.

Where it leads: There are 2 × 2 × 6 = 24 equally likely outcomes and exactly one is ‘HH6’: multiplying probabilities is counting in disguise.

Answer: B, 4/11

1. MATHEMATICS has 11 letters: M, A, T, H, E, M, A, T, I, C, S.
2. Vowels: A, E, A, I: 4 letters.
3. Probability = 4/11 (B).

Why this works: Each position is equally likely, so repeated letters count each time they appear.

Where it leads: If instead you chose one of the different letters (M, A, T, H, E, I, C, S) at random, the answer would be 3/8. Always check what is equally likely.

Answer: 5/12

1. P(equal) = 6/36 = 1/6, so P(not equal) = 5/6.
2. By symmetry (swap the two rolls), ‘second bigger’ and ‘first bigger’ are equally likely.
3. P(second bigger) = ½ × 5/6 = 5/12.

Why this works: Symmetry between the two rolls splits the unequal outcomes exactly in half.

Where it leads: For three rolls, P(strictly increasing) = C(6, 3)/216 = 20/216. Symmetry plus ties is a powerful pair.

Answer: 1/10

1. Digit sum 9: 18, 27, 36, 45, 54, 63, 72, 81, 90: 9 numbers.
2. There are 90 two-digit numbers (10 to 99).
3. Probability = 9/90 = 1/10.

Why this works: These are exactly the two-digit multiples of 9: a number’s digit sum and the number leave the same remainder on division by 9.

Where it leads: That remainder fact is the basis of ‘casting out nines’, an old way of checking arithmetic.

Answer: C, 1/4

1. There are 16 equally likely outcomes.
2. Total 5: (1, 4), (2, 3), (3, 2), (4, 1): 4 outcomes.
3. Probability = 4/16 = 1/4 (C).

Why this works: 5 is the middle total, so it can be made in the most ways: one for each first score.

Where it leads: The totals form a triangular distribution: 1, 2, 3, 4, 3, 2, 1 ways for totals 2 to 8. Adding more spinners makes it look more and more like a bell curve.

Answer: 1/6

1. Students who play at least one sport: 18 + 12 − 5 = 25.
2. So 30 − 25 = 5 students play neither.
3. Probability = 5/30 = 1/6.

Why this works: A Venn diagram (or inclusion–exclusion) avoids counting the overlap twice.

Where it leads: Fill in a Venn diagram from the inside out: start with ‘both’, then ‘football only’ (13) and ‘tennis only’ (7). The same method handles three sets.

Answer: 144

1. Estimated probability of point up = 18/50 = 0.36.
2. Expected number in 400 drops = 400 × 0.36.
3. = 144.

Why this works: When a probability cannot be worked out by symmetry, the relative frequency from an experiment is the best estimate.

Where it leads: The more drops, the more reliable the estimate: the law of large numbers. But the error only shrinks like 1/√n, so 4 times as many trials only halves it.

Answer: B, 1/3

1. Take the first sock: it is black or white, it does not matter which.
2. Of the 3 socks left, exactly 1 is the same colour.
3. P(match) = 1/3 (B), not 1/2.

Why this works: Focusing on the second sock, given the first, avoids listing pairs and shows why the answer is below 1/2.

Where it leads: With n pairs of different-coloured socks mixed up, P(the first two match) = 1/(2n − 1). Mixed drawers really are worse than they seem.

Answer: 3/8

1. There are 2⁴ = 16 equally likely outcomes.
2. Exactly two heads: choose which 2 of the 4 tosses are heads: 4 × 3 ÷ 2 = 6 ways (HHTT, HTHT, HTTH, THHT, THTH, TTHH).
3. Probability = 6/16 = 3/8.

Why this works: Counting the positions of the heads, rather than listing outcomes, scales to any number of tosses.

Where it leads: Exactly k heads in n tosses has probability C(n, k)/2ⁿ: the binomial distribution. Notice that 2 heads in 4 tosses is less likely than you might guess.

Answer: 5/9

1. The first die can be anything.
2. P(second differs from first) = 5/6. P(third differs from both) = 4/6.
3. P(all different) = 5/6 × 4/6 = 20/36 = 5/9.

Why this works: Building up one die at a time keeps track of how many scores are still allowed.

Where it leads: This is the ‘birthday problem’ with 6 days instead of 365. With 23 people the chance that all birthdays differ is already below 1/2.

Answer: B, 7/13

1. Red cards: 26.
2. Kings not already counted: the 2 black kings.
3. Total: 28 cards, so the probability is 28/52 = 7/13 (B).

Why this works: Adding 26 + 4 would count the two red kings twice. Add only what is new.

Where it leads: In symbols: P(R or K) = P(R) + P(K) − P(R and K) = 26/52 + 4/52 − 2/52.

Answer: 9/25

1. The product is odd only if both numbers are odd.
2. P(one person picks odd) = 3/5.
3. P(both odd) = 3/5 × 3/5 = 9/25.

Why this works: ‘Odd product’ turns into two independent conditions, one for each person.

Where it leads: P(product even) = 1 − 9/25 = 16/25. With more people choosing, an even product becomes almost certain.