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 The table shows the drink chosen by 60 customers. A customer who chose coffee is picked at random. Find the probability that the customer is under 30.
Show worked solution
The coffee column holds 35 customers, and 21 are under 30: \(\dfrac{21}{35}\) = \(\dfrac{3}{5}\). \(\dfrac{7}{10}\) = \(\dfrac{21}{30}\) is P(coffee | under 30), the reverse; \(\dfrac{7}{20}\) = \(\dfrac{21}{60}\) is P(under 30 and coffee); \(\dfrac{7}{12}\) = \(\dfrac{35}{60}\) is P(coffee).
2 A person is chosen at random from the group shown. Find P(A | B) and P(B | A), each in simplest form.
Show worked solution
Circle B holds 7 + 21 = 28, so P(A | B) = \(\dfrac{7}{28}\) = 1/4. Circle A holds 14 + 7 = 21, so P(B | A) = \(\dfrac{7}{21}\) = 1/3.
3 For two events, P(B) = 0.4 and P(B and not A) = 0.1. Find P(A | B).
Show worked solution
P(A and B) = 0.4 − 0.1 = 0.3, so P(A | B) = 0.3 ÷ 0.4 = 0.75.
Level 2 · Core
4 In a town of 200 people, 40 are taller than 190 cm. 8 people play professional basketball, and all 8 are taller than 190 cm. Compare P(tall | player) with P(player | tall).
Show worked solution
Given a player, the whole is 8 people and all 8 are tall: 8 ÷ 8 = 1. Given tall, the whole is 40 people and 8 of them are players: \(\dfrac{8}{40}\) = \(\dfrac{1}{5}\). So P(tall | player) = 1 and P(player | tall) = \(\dfrac{1}{5}\). The third option swaps them; \(\dfrac{1}{25}\) = \(\dfrac{8}{200}\) is P(tall and player), which divides by the whole town.
5 Machine A makes 0.6 of a factory's items and machine B makes the rest. 0.05 of A's items are faulty, and 0.1 of B's items are faulty. An item is found to be faulty. Find, in simplest form, the probability that it was made by B.
Show worked solution
P(A and faulty) = 0.6 × 0.05 = 0.03 and P(B and faulty) = 0.4 × 0.1 = 0.04, so P(faulty) = 0.07. P(B | faulty) = 0.04 ÷ 0.07 = 4/7.
6 Of 90 students, 50 play a sport. P(plays music | plays sport) = \(\dfrac{2}{5}\) and P(plays music | no sport) = \(\dfrac{3}{8}\). How many of the 90 students play music?
Show worked solution
Sport and music: \(\dfrac{2}{5}\) × 50 = 20. No sport: 90 − 50 = 40, and music: \(\dfrac{3}{8}\) × 40 = 15. Total: 20 + 15 = 35.
Level 3 · Challenge
7 Three statements about two events A and B, each with a probability greater than 0:
1 If P(A) = P(B), then P(A | B) = P(B | A).
2 If P(A and B) = 0.2 and P(B | A) = 0.5, then P(A) = 0.4.
3 P(A | B) + P(B | A) = 1 for any two events.
Which statements are correct?
1 If P(A) = P(B), then P(A | B) = P(B | A).
2 If P(A and B) = 0.2 and P(B | A) = 0.5, then P(A) = 0.4.
3 P(A | B) + P(B | A) = 1 for any two events.
Which statements are correct?
Show worked solution
P(A | B) = P(A and B) ÷ P(B) and P(B | A) = P(A and B) ÷ P(A); equal bottoms give equal answers (1 ✓). P(B | A) = P(A and B) ÷ P(A), so 0.5 = 0.2 ÷ P(A) and P(A) = 0.4 (2 ✓). If A and B are the same event, both conditionals are 1 and they add to 2, so statement 3 is false (3 ✗). 1 and 2.
8 A bag has 3 red and 5 blue counters. Two are drawn without replacement. Find, in simplest form, P(the second counter is red), and then P(the first was red | the second is red).
Show worked solution
Red, red: \(\dfrac{3}{8}\) × \(\dfrac{2}{7}\) = \(\dfrac{6}{56}\). Blue, red: \(\dfrac{5}{8}\) × \(\dfrac{3}{7}\) = \(\dfrac{15}{56}\). P(second red) = \(\dfrac{21}{56}\) = 3/8. P(first red | second red) = \(\dfrac{6}{56}\) ÷ \(\dfrac{21}{56}\) = \(\dfrac{6}{21}\) = 2/7.
9 In a class, P(passes Maths) = 0.7, P(passes English) = 0.8 and P(passes both) = 0.6. A student who failed English is chosen at random. Find the probability that this student passed Maths.
Show worked solution
P(fails English) = 1 − 0.8 = 0.2. P(passes Maths and fails English) = 0.7 − 0.6 = 0.1. So P(passes Maths | fails English) = 0.1 ÷ 0.2 = 0.5.