Probability and Compound Events: practice solutions, Grade 7 – download the PDF

1 Likely or unlikely? ★★★

1. Every face is less than 7, so the event is certain: \( 1 \).
2. A die has no 7, so it is impossible: \( 0 \).
3. Heads and tails are equally likely: \( 0.5 \).
4. \( \dfrac{1}{20} = 0.05 \).
5. \( 1 - \dfrac{1}{20} = \dfrac{19}{20} = 0.95 \).

2 Marbles in a bag ★★★

There are \( 4 + 6 + 5 = 15 \) marbles.

1. \( P(\text{green}) = \dfrac{6}{15} = \dfrac{2}{5} \).
2. \( P(\text{yellow}) = \dfrac{5}{15} = \dfrac{1}{3} \).
3. \( P(\text{not red}) = \dfrac{11}{15} \), or \( 1 - \dfrac{4}{15} = \dfrac{11}{15} \).

3 The eight-section spinner ★★★

1. Even: 2, 4, 6, 8. \( \dfrac{4}{8} = \dfrac{1}{2} \).
2. Greater than 5: 6, 7, 8. \( \dfrac{3}{8} \).
3. Prime: 2, 3, 5, 7 (1 is not prime). \( \dfrac{4}{8} = \dfrac{1}{2} \).
4. Multiples of 3: 3 and 6. \( \dfrac{2}{8} = \dfrac{1}{4} \).

4 Using the complement ★★★

1. \( 1 - 0.35 = 0.65 \).
2. \( 1 - \dfrac{3}{8} = \dfrac{8}{8} - \dfrac{3}{8} = \dfrac{5}{8} \).

5 Paper cup experiment ★★★

\( \dfrac{9}{50} = 0.18 = 18\% \). Since the cup is not a symmetric object, we could not have found this number by counting equally likely outcomes.

6 True or false? ★★★

1. False. Probabilities are never greater than 1.
2. True. \( \dfrac{7}{7} = 1 \).
3. True. \( P(\text{not } A) = 1 - 0.5 = 0.5 \).
4. False. The smallest probability is 0 (impossible), never negative.

7 Two coins ★★★

1. HH, HT, TH, TT: 4 equally likely outcomes.
2. HT and TH: \( \dfrac{2}{4} = \dfrac{1}{2} \).
3. HH, HT, TH: \( \dfrac{3}{4} \).

8 Lunch model ★★★

1. The total must be 1: \( 0.4 + 0.25 + 0.1 = 0.75 \), so the salad has \( 1 - 0.75 = 0.25 \).
2. \( 0.4 + 0.1 = 0.5 \).
3. No. The outcomes do not all have the same probability (pizza has 0.4, soup has 0.1).

9 Rolling a 4 ★★★

1. \( \dfrac{25}{120} = \dfrac{5}{24} \approx 0.208 \).
2. \( \dfrac{1}{6} \approx 0.167 \).
3. \( \dfrac{1}{6} \times 600 = 100 \).
4. \( \dfrac{5}{24} \times 600 = 125 \).

The two predictions differ because 120 rolls is a small sample.

10 Defective light bulbs ★★★

Defective: \( 0.04 \times 2{,}500 = 100 \) bulbs.

Working: \( 2{,}500 - 100 = 2{,}400 \) bulbs (also \( 0.96 \times 2{,}500 \)).

About 100 bulbs are defective.

11 Two dice and a table ★★★

1. The highlighted cells: (3,6), (4,5), (5,4), (6,3). \( \dfrac{4}{36} = \dfrac{1}{9} \).
2. Sums of 2 or 3: (1,1), (1,2), (2,1). \( \dfrac{3}{36} = \dfrac{1}{12} \).
3. Half of the cells have an even sum (18 of 36). \( \dfrac{18}{36} = \dfrac{1}{2} \).

12 Outfits ★★★

1. \( 3 \times 4 \times 2 = 24 \) outfits.
2. Each shirt is equally likely: \( \dfrac{1}{3} \). (Check: 8 of the 24 outfits have the red shirt, and \( \dfrac{8}{24} = \dfrac{1}{3} \).)
3. \( \dfrac{1}{24} \).

13 Unequal sections ★★★

1. Red \( \dfrac{180}{360} = \dfrac{1}{2} \); blue \( \dfrac{90}{360} = \dfrac{1}{4} \); green \( \dfrac{60}{360} = \dfrac{1}{6} \); yellow \( \dfrac{30}{360} = \dfrac{1}{12} \).
2. \( \dfrac{6}{12} + \dfrac{3}{12} + \dfrac{2}{12} + \dfrac{1}{12} = \dfrac{12}{12} = 1 \).
3. \( \dfrac{3}{12} + \dfrac{2}{12} = \dfrac{5}{12} \).

14 Find the mistakes ★★★

1. Each flip is independent. The coin has no memory, so \( P(\text{tails}) \) is still \( \dfrac{1}{2} \).
2. The colors are not equally likely. The correct value is \( \dfrac{2}{10} = \dfrac{1}{5} \).

15 Coin and spinner ★★★

1. H1, H2, H3, H4, T1, T2, T3, T4: \( 2 \times 4 = 8 \) outcomes.
2. H3, H4: \( \dfrac{2}{8} = \dfrac{1}{4} \).
3. T1, T3: \( \dfrac{2}{8} = \dfrac{1}{4} \).
4. All 4 heads outcomes plus T4: \( \dfrac{5}{8} \). (Do not count H4 twice.)

16 Three coin flips ★★★

1. HHH, HHT, HTH, HTT, THH, THT, TTH, TTT: 8 outcomes.
2. HHT, HTH, THH: \( \dfrac{3}{8} \).
3. Only HHH has no tail: \( 1 - \dfrac{1}{8} = \dfrac{7}{8} \).
4. HHH and TTT: \( \dfrac{2}{8} = \dfrac{1}{4} \).

17 Letter and digit ★★★

1. \( 3 \times 5 = 15 \).
2. B1, B3, B5: \( \dfrac{3}{15} = \dfrac{1}{5} \).
3. A4, A5: \( \dfrac{2}{15} \).
4. A with any digit gives 5 outcomes; B5 and C5 add 2 more. \( \dfrac{7}{15} \).

18 Cereal box simulation ★★★

1. 4, 3, 6, 1, 3, 7, 2, 4, 1, 5.
2. Total: \( 4 + 3 + 6 + 1 + 3 + 7 + 2 + 4 + 1 + 5 = 36 \). Average: \( 36 \div 10 = 3.6 \) boxes. This is close to the 4 boxes we might guess from “1 out of 4”.
3. Trials 2, 4, 5, 7 and 9: 5 trials out of 10, so \( \dfrac{5}{10} = 0.5 \).

19 Is the game fair? ★★★

1. The product is odd only when both dice are odd: \( 3 \times 3 = 9 \) of 36 outcomes. \( P(\text{odd}) = \dfrac{9}{36} = \dfrac{1}{4} \) and \( P(\text{even}) = \dfrac{27}{36} = \dfrac{3}{4} \).
2. No. Ava is three times as likely to win.
3. Ava: \( 27 \times 1 = 27 \) points. Ben: \( 9 \times 3 = 27 \) points. With 3 points the game is fair.

20 Bike to school ★★★

1. \( \dfrac{28}{80} = 0.35 \).
2. \( 0.35 \times 1{,}200 = 420 \) students.
3. A random sample represents the whole school. If you only asked the track team, the result would be biased.

21 Changing the bag ★★★

1. Bag A: \( \dfrac{3}{8} = 0.375 \). Bag B: \( \dfrac{4}{12} = \dfrac{1}{3} \approx 0.333 \). Bag A is better.
2. Test adding 4 red marbles: Bag B then has 8 red out of 16, and \( \dfrac{8}{16} = \dfrac{1}{2} \). So 4 red marbles must be added. (With 3 added: \( \dfrac{7}{15} \), too small; with 5: \( \dfrac{9}{17} \), too big.)

22 At least one six ★★★

1. Each die has 5 faces that are not 6: \( 5 \times 5 = 25 \) outcomes.
2. \( \dfrac{25}{36} \).
3. \( 1 - \dfrac{25}{36} = \dfrac{11}{36} \). (Check by listing: six outcomes with a 6 on the first die plus five more with a 6 on the second, which is 11.)