Work through the levels in order. Type your answer and press Check — open the worked solution if you get stuck. Answers like -6, 3/4 or 0.25 are typed exactly as shown. A ⭐ Bonus Challenge at the end is extra credit — skip it if the rest was hard work.
Level 1 · Warm-up
1 A fair four-sided die, numbered 1 to 4, is rolled twice. What is the probability that the two numbers add to 5?
Show worked solution
Each path has probability \(\dfrac{1}{4}\) × \(\dfrac{1}{4}\) = \(\dfrac{1}{16}\). A sum of 5 comes from (1, 4), (2, 3), (3, 2) and (4, 1): four paths, so P = \(\dfrac{4}{16}\) = \(\dfrac{1}{4}\). \(\dfrac{1}{16}\) is one path; \(\dfrac{1}{8}\) counts (1, 4) and (2, 3) but not their reverses; \(\dfrac{3}{16}\) misses one of the four orders.
2 A spinner lands on A with probability \(\dfrac{1}{3}\) and on B otherwise. It is spun twice. Find P(both spins land on the same letter), in simplest form.
Show worked solution
P(AA) = \(\dfrac{1}{3}\) × \(\dfrac{1}{3}\) = \(\dfrac{1}{9}\) and P(BB) = \(\dfrac{2}{3}\) × \(\dfrac{2}{3}\) = \(\dfrac{4}{9}\). Add: 5/9.
3 A house has two separate smoke alarms. Alarm A works with probability 0.9 and alarm B works with probability 0.8, independently. Find the probability that both alarms fail.
Show worked solution
P(A fails) = 1 − 0.9 = 0.1 and P(B fails) = 1 − 0.8 = 0.2. Both fail: 0.1 × 0.2 = 0.02.
Level 2 · Core
4 A quiz has two questions, each with five options and one correct answer. Lily guesses both. What is the probability that she gets at least one right?
Show worked solution
P(wrong) = \(\dfrac{4}{5}\), so P(both wrong) = \(\dfrac{4}{5}\) × \(\dfrac{4}{5}\) = \(\dfrac{16}{25}\) and P(at least one right) = 1 − \(\dfrac{16}{25}\) = \(\dfrac{9}{25}\). \(\dfrac{16}{25}\) is P(none right); \(\dfrac{8}{25}\) is exactly one right (two paths of \(\dfrac{1}{5}\) × \(\dfrac{4}{5}\)) and leaves out "both right"; \(\dfrac{1}{25}\) is both right.
5 Each seed germinates with probability \(\dfrac{2}{3}\), independently. Three seeds are planted. Find P(exactly two germinate), in simplest form.
Show worked solution
Two G and one F come in three orders: GGF, GFG and FGG. Each path is \(\dfrac{2}{3}\) × \(\dfrac{2}{3}\) × \(\dfrac{1}{3}\) = \(\dfrac{4}{27}\), so P = 3 × \(\dfrac{4}{27}\) = \(\dfrac{12}{27}\) = 4/9.
6 A team wins its first game with probability 0.6. After a win, it wins the next game with probability 0.7. After a loss, it wins the next game with probability 0.4. Find the probability that it wins exactly one of its first two games.
Show worked solution
Win then lose: 0.6 × 0.3 = 0.18. Lose then win: 0.4 × 0.4 = 0.16. Add: 0.18 + 0.16 = 0.34.
Level 3 · Challenge
7 Three statements:
1 A fair spinner numbered 1 to 5 is spun twice. P(both numbers are even) = \(\dfrac{4}{25}\).
2 A spinner lands on red with probability \(\dfrac{1}{4}\). In two spins, P(exactly one red) = \(\dfrac{3}{16}\).
3 An archer hits the target with probability 0.9 on each shot. In two shots, P(at least one hit) = 0.99.
Which statements are correct?
1 A fair spinner numbered 1 to 5 is spun twice. P(both numbers are even) = \(\dfrac{4}{25}\).
2 A spinner lands on red with probability \(\dfrac{1}{4}\). In two spins, P(exactly one red) = \(\dfrac{3}{16}\).
3 An archer hits the target with probability 0.9 on each shot. In two shots, P(at least one hit) = 0.99.
Which statements are correct?
Show worked solution
The even numbers are 2 and 4, so P(even) = \(\dfrac{2}{5}\) and P(both even) = \(\dfrac{4}{25}\) (1 ✓). Exactly one red has two paths, each \(\dfrac{1}{4}\) × \(\dfrac{3}{4}\) = \(\dfrac{3}{16}\), so the probability is \(\dfrac{6}{16}\) = \(\dfrac{3}{8}\), not \(\dfrac{3}{16}\) (2 ✗). P(no hit) = 0.1 × 0.1 = 0.01, so P(at least one hit) = 1 − 0.01 = 0.99 (3 ✓). 1 and 3.
8 Ann scores a penalty with probability \(\dfrac{3}{4}\) and Ben with probability \(\dfrac{2}{3}\), independently. Each takes one penalty. Find, in simplest form, P(exactly one of them scores) and P(at least one of them scores).
Show worked solution
Exactly one: Ann scores and Ben misses, \(\dfrac{3}{4}\) × \(\dfrac{1}{3}\) = \(\dfrac{3}{12}\), or Ann misses and Ben scores, \(\dfrac{1}{4}\) × \(\dfrac{2}{3}\) = \(\dfrac{2}{12}\). Add: 5/12. At least one: P(neither) = \(\dfrac{1}{4}\) × \(\dfrac{1}{3}\) = \(\dfrac{1}{12}\), so P(at least one) = 1 − \(\dfrac{1}{12}\) = 11/12.
9 A spinner lands on gold with probability \(\dfrac{1}{3}\). What is the smallest number of spins for which P(at least one gold) is greater than 0.9?
Show worked solution
P(at least one gold) > 0.9 means P(no gold) = (2/3)n < 1/10, that is 10 × 2n < 3n. For n = 5: 10 × 32 = 320, but 35 = 243, so not yet. For n = 6: 10 × 64 = 640 and 36 = 729, and 640 < 729 ✓. The smallest number of spins is 6.