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 bag has 4 orange and 2 purple marbles. Two are drawn without replacement. What is the probability that both are orange?
Show worked solution
P(O, O) = \(\dfrac{4}{6}\) × \(\dfrac{3}{5}\) = \(\dfrac{12}{30}\) = \(\dfrac{2}{5}\). \(\dfrac{4}{9}\) is with replacement; \(\dfrac{8}{15}\) = \(\dfrac{4}{6}\) × \(\dfrac{4}{5}\) drops only the total; \(\dfrac{1}{3}\) = \(\dfrac{4}{6}\) × \(\dfrac{3}{6}\) drops only the number of orange.
2 A bag has 1 black and 3 white counters. A counter is drawn, replaced, and a second counter is drawn. Find P(one of each colour), in simplest form.
Show worked solution
Black then white: \(\dfrac{1}{4}\) × \(\dfrac{3}{4}\) = \(\dfrac{3}{16}\). White then black: \(\dfrac{3}{4}\) × \(\dfrac{1}{4}\) = \(\dfrac{3}{16}\). Add: \(\dfrac{6}{16}\) = 3/8.
3 A hat holds 5 tickets. One ticket wins a prize and the other 4 do not. Two tickets are drawn without replacement. Find, as a decimal, the probability that neither ticket wins.
Show worked solution
P(lose, lose) = \(\dfrac{4}{5}\) × \(\dfrac{3}{4}\) = \(\dfrac{12}{20}\) = \(\dfrac{3}{5}\) = 0.6.
Level 2 · Core
4 A bag has 2 red and 4 blue counters. Two are drawn without replacement. Which is more likely, two of the same colour or one of each, and by how much?
Show worked solution
P(two the same) = \(\dfrac{2}{6}\) × \(\dfrac{1}{5}\) + \(\dfrac{4}{6}\) × \(\dfrac{3}{5}\) = \(\dfrac{2}{30}\) + \(\dfrac{12}{30}\) = \(\dfrac{14}{30}\). P(one of each) = \(\dfrac{16}{30}\). So One of each, by \(\dfrac{1}{15}\), since \(\dfrac{2}{30}\) = \(\dfrac{1}{15}\). "Two the same, by \(\dfrac{1}{9}\)" is the with-replacement result (\(\dfrac{5}{9}\) against \(\dfrac{4}{9}\)); "Two the same, by \(\dfrac{1}{15}\)" swaps the two events.
5 A bag has 2 red, 3 green and 5 yellow counters. Two are drawn without replacement. Find P(both are the same colour), in simplest form.
Show worked solution
Red, red: 2 × 1 = 2. Green, green: 3 × 2 = 6. Yellow, yellow: 5 × 4 = 20. Each over 10 × 9 = 90, so P = \(\dfrac{28}{90}\) = 14/45.
6 A bag holds 12 counters, and 4 of them are red. Two counters are drawn without replacement. The probability that both are red can be written as \(\dfrac{1}{k}\). Find k.
Show worked solution
P(R, R) = \(\dfrac{4}{12}\) × \(\dfrac{3}{11}\) = \(\dfrac{12}{132}\) = \(\dfrac{1}{11}\), so k = 11.
Level 3 · Challenge
7 Three statements:
1 A bag has 3 red and 7 white counters. Two are drawn. P(the second counter is red) = \(\dfrac{3}{10}\), whether or not the first is replaced.
2 A bag has 4 red and 4 white counters. Two are drawn without replacement. P(two the same colour) = \(\dfrac{1}{2}\).
3 For any bag holding both colours, P(one of each) is higher without replacement than with replacement.
Which statements are correct?
1 A bag has 3 red and 7 white counters. Two are drawn. P(the second counter is red) = \(\dfrac{3}{10}\), whether or not the first is replaced.
2 A bag has 4 red and 4 white counters. Two are drawn without replacement. P(two the same colour) = \(\dfrac{1}{2}\).
3 For any bag holding both colours, P(one of each) is higher without replacement than with replacement.
Which statements are correct?
Show worked solution
Without replacement, P(second red) = \(\dfrac{3}{10}\) × \(\dfrac{2}{9}\) + \(\dfrac{7}{10}\) × \(\dfrac{3}{9}\) = \(\dfrac{6}{90}\) + \(\dfrac{21}{90}\) = \(\dfrac{27}{90}\) = \(\dfrac{3}{10}\), the same as with replacement (1 ✓). P(two the same) = 2 × \(\dfrac{4}{8}\) × \(\dfrac{3}{7}\) = \(\dfrac{24}{56}\) = \(\dfrac{3}{7}\), not \(\dfrac{1}{2}\) (2 ✗). With r red, w white and n in all, one of each is 2 × \(\dfrac{r}{n}\) × \(\dfrac{w}{n}\) with replacement and 2 × \(\dfrac{r}{n}\) × \(\dfrac{w}{n - 1}\) without; the smaller bottom makes the second one bigger (3 ✓). 1 and 3.
8 A bag has 5 red and 4 green counters. Three are drawn without replacement. Find, in simplest form, P(all three are red) and P(exactly two are red).
Show worked solution
All red: \(\dfrac{5}{9}\) × \(\dfrac{4}{8}\) × \(\dfrac{3}{7}\) = \(\dfrac{60}{504}\) = 5/42. Exactly two red: each order has top 5 × 4 × 4 = 80 (for example RRG is \(\dfrac{5}{9}\) × \(\dfrac{4}{8}\) × \(\dfrac{4}{7}\)), and there are 3 orders, so \(\dfrac{240}{504}\) = 10/21.
9 A bag has 5 red counters and some blue counters. Two counters are drawn without replacement, and P(both red) = \(\dfrac{5}{14}\). How many blue counters are in the bag?
Show worked solution
20 ÷ (n(n − 1)) = \(\dfrac{5}{14}\), so n(n − 1) = 56 = 8 × 7 and n = 8. There are 8 − 5 = 3 blue counters. Check: \(\dfrac{5}{8}\) × \(\dfrac{4}{7}\) = \(\dfrac{20}{56}\) = \(\dfrac{5}{14}\) ✓.