Year 7 probability basics worksheet answers to download as PDF

These Year 7 probability basics worksheet answers give a full worked solution to all 20 questions of the probability basics worksheet. Try each question yourself first, then check your method step by step.

← Back to the worksheet

Answer to Question 1
1. A normal dice only has the numbers 1 to 6, all less than 7, so this will always happen: certain (probability 1).
2. Heads and tails are equally likely: even chance (probability 0.5).
3. February is only one month out of twelve: unlikely (about \(\dfrac{1}{12}\)).
4. There is no 7 on a normal dice: impossible (probability 0).
5. Rain at some point in a whole month is normal in the UK: likely, very close to certain but not completely.
Answer to Question 2
1. There are \(5+3+2=10\) counters and 5 are red: \(\dfrac{5}{10}=\)\(\dfrac{1}{2}\).
2. 3 of the 10 counters are blue: \(\dfrac{3}{10}\).
3. P(green) \(=\dfrac{2}{10}\), so P(not green) \(=1-\dfrac{2}{10}=\dfrac{8}{10}=\)\(\dfrac{4}{5}\).
4. There are no yellow counters, so the event is impossible: 0.
5. \(\dfrac{1}{2}=1\div2=\)0.5, and \(0.5\times100=\)50%.
Answer to Question 3
1. The four probabilities must add up to 1: \(1-0.3-0.25-0.2=\)0.25.
2. \(P(C)=1-0.4-0.35=\)0.25.
3. Being late and not being late cover every possibility, so P(not late) \(=1-0.12=\)0.88.
4. Number of red = probability × total: \(20\times0.35=\)7 red counters.
Answer to Question 4
1. Each coin can land H or T, so the outcomes are HH, HT, TH, TT: 4 equally likely outcomes.
2. Only HH gives two heads: \(\dfrac{1}{4}\).
3. HT and TH both give one head and one tail: \(\dfrac{2}{4}=\)\(\dfrac{1}{2}\).
4. “At least one head” means HH, HT or TH: \(\dfrac{3}{4}\). Or \(1-P(TT)=1-\dfrac{1}{4}\).
Answer to Question 5
The grid shows all \(6\times6=36\) equally likely totals. For each event, count the cells that match and write the count over 36, then simplify.
1. 36.
2. \(\dfrac{6}{36}=\dfrac{1}{6}\)
3. Totals 10, 11, 12: \(3+2+1=6\) outcomes, so \(\dfrac{6}{36}=\dfrac{1}{6}\)
4. Primes 2, 3, 5, 7, 11 occur \(1+2+4+6+2=15\) times, so \(\dfrac{15}{36}=\dfrac{5}{12}\).
Answer to Question 6
1. P(six) \(=\dfrac{1}{6}\), so expect \(\dfrac{1}{6}\times120=\)20 sixes.
2. Expected number = probability × number of trials: \(0.15\times200=\)30 late buses.
3. \(0.3\times50=\)15 times blue.
4. Each spin or roll is random, so the results vary from one experiment to another. The expected number is what happens on average over many repeats, not a guarantee.
Answer to Question 7
There are \(1+4+15=20\) prize-winning tickets out of 200 tickets, and every ticket is equally likely to be drawn.
1. \(\dfrac{1}{200}\)
2. \(\dfrac{1+4+15}{200}=\dfrac{20}{200}=\)\(\dfrac{1}{10}\)
3. \(1-0.1=0.9\), so 90%.
4. Dan has more tickets. Aisha expects \(5\times0.1=0.5\) prizes and Dan expects \(12\times0.1=1.2\) prizes, so Dan has the better chance.
Answer to Question 8
1. It either rains or it does not, so P(no rain on Saturday) \(=1-0.7=\)0.3.
2. P(no rain on Sunday) \(=1-0.4=\)0.6.
3. 0.7 is bigger than 0.4, so rain is more likely on Saturday.
4. Expected number = probability × trials: \(20\times0.7=\)14 rainy Saturdays.
Answer to Question 9
1. Three of the eight sections show 3: \(\dfrac{3}{8}\).
2. The even numbers are 2, 2 and 4: \(\dfrac{3}{8}\).
3. Greater than 2 means 3, 3, 3, 4 or 5: \(\dfrac{5}{8}\).
4. The primes are 2, 3 and 5 (1 and 4 are not prime). Sections 2, 2, 3, 3, 3, 5 make 6: \(\dfrac{6}{8}=\)\(\dfrac{3}{4}\).
Answer to Question 10
1. \(\dfrac{18}{30}=\)\(\dfrac{3}{5}\)
2. \(1-\dfrac{4}{30}=\)\(\dfrac{13}{15}\)
3. \(28\times0.6=16.8\), so about 17 pupils.
4. After the first pupil is chosen there are only 29 pupils left, and the numbers in each group may have changed.
In part (c) the probability \(\dfrac{3}{5}=0.6\) gives 16.8, which is not a whole number of pupils, so we round to about 17.
Answer to Question 11
1. 13 of the 52 cards are hearts: \(\dfrac{13}{52}=\)\(\dfrac{1}{4}\).
2. There is one king in each suit, so 4 kings: \(\dfrac{4}{52}=\)\(\dfrac{1}{13}\).
3. Hearts and diamonds are red, \(2\times13=26\) cards: \(\dfrac{26}{52}=\)\(\dfrac{1}{2}\).
4. There is only one king of hearts: \(\dfrac{1}{52}\).
5. P(spade) \(=\dfrac{1}{4}\), so P(not a spade) \(=1-\dfrac{1}{4}=\)\(\dfrac{3}{4}\).
Answer to Question 12
1. Red \(\dfrac{26}{80}=0.325\), blue \(\dfrac{18}{80}=0.225\), green \(\dfrac{22}{80}=0.275\), yellow \(\dfrac{14}{80}=0.175\).
2. If fair, each colour has probability 0.25.
3. The relative frequencies are close to 0.25 but not equal. With only 80 spins this could be chance, so more trials are needed.
4. \(400\times0.325=\)130 times.
Answer to Question 13
1. There are \(6+4+2=12\) socks and 6 are black: \(\dfrac{6}{12}=\)\(\dfrac{1}{2}\).
2. P(white) \(=\dfrac{4}{12}=\dfrac{1}{3}\), so P(not white) \(=\)\(\dfrac{2}{3}\).
3. After one black sock is removed, 11 socks remain and 5 are black: \(\dfrac{5}{11}\).
4. 6 of the 12 socks are already black, so P(black) is exactly \(\dfrac{1}{2}\). She should add none.
Answer to Question 14
1. \(\dfrac{16}{40}\); divide by 8: \(\dfrac{2}{5}\).
2. P(slow) \(=\dfrac{4}{40}=\dfrac{1}{10}\), so P(not slow) \(=\)\(\dfrac{9}{10}\).
3. P(dance) \(=\dfrac{8}{40}=\dfrac{1}{5}\), so in 10 songs expect \(10\times\dfrac{1}{5}=\)2 dance songs.
4. \(40-20=20\) songs are left, and \(12-6=6\) of them are rock: \(\dfrac{6}{20}=\)\(\dfrac{3}{10}\).
Answer to Question 15
1. Total 7 occurs 6 times out of 36: \(\dfrac{1}{6}\).
2. Total 5 occurs 4 times and total 9 occurs 4 times: \(\dfrac{8}{36}=\)\(\dfrac{2}{9}\).
3. \(1-\dfrac{6}{36}-\dfrac{8}{36}=\)\(\dfrac{22}{36}=\dfrac{11}{18}\).
4. The game is not fair because Ben has the higher probability (\(\dfrac{8}{36}>\dfrac{6}{36}\)). For a fair game Ben could win on totals of 2, 3 or 4 instead, which also has 6 outcomes out of 36 (1 + 2 + 3).
Answer to Question 16
1. The three probabilities must add up to 1: \(0.2+x+3x=1\).
2. Collect: \(4x+0.2=1\), so \(4x=0.8\) and \(x=\)0.2. Then P(blue) = 0.2 and P(green) = 0.6.
3. Red \(0.2\times40=\)8, blue \(0.2\times40=\)8, green \(0.6\times40=\)24 (total 40).
4. Now 10 red out of 42 counters: \(\dfrac{10}{42}=\)\(\dfrac{5}{21}\).
Answer to Question 17
1. Outcomes are not equally likely, so you cannot say 1/2. The probability is tiny.
2. Probabilities of mutually exclusive outcomes must add to at most 1 (0.6 + 0.5 = 1.1 is impossible if only one can happen).
3. Six rolls is too few to decide. The probability of no sixes in 6 rolls is about 33%.
4. \(\dfrac{3}{3+5}=\)\(\dfrac{3}{8}\).
Answer to Question 18
1. Bag A: \(\dfrac{4}{10}=0.4\). Bag B: \(\dfrac{3}{5}=0.6\).
2. Bag B.
3. \(\dfrac{7}{15}\), about 0.467.
4. The average of 0.4 and 0.6 is 0.5. This is not the same because the bags have different sizes.
Combined bag: \(4+3=7\) mints out of \(10+5=15\) sweets. Bag A has more sweets, so its lower probability counts for more, pulling the combined value below 0.5.
Answer to Question 19
1. 2 coin results × 6 dice results = 12 equally likely outcomes, as in the table.
2. Heads with 2, 4 or 6: H2, H4, H6, 3 outcomes: \(\dfrac{3}{12}=\)\(\dfrac{1}{4}\).
3. Tails with 5 or 6: T5, T6: \(\dfrac{2}{12}=\)\(\dfrac{1}{6}\).
4. All 6 heads outcomes, plus T6 (H6 is already counted): \(6+1=7\) outcomes, so \(\dfrac{7}{12}\).
Answer to Question 20
1. 11 of the 12 months are different from the first pupil’s: \(\dfrac{11}{12}\).
2. Each new pupil must avoid the months already used: \(\dfrac{12}{12}\times\dfrac{11}{12}\times\dfrac{10}{12}\times\dfrac{9}{12}=\dfrac{11880}{20736}\approx\)0.573.
3. “At least two share” is the opposite of “all different”: \(1-0.573=\)0.427.
4. No, about 43%. But with only 4 pupils and 12 months it is surprisingly close to a half; with 5 pupils it would be more than 60%.

How to use these Year 7 probability basics worksheet answers

First, try every question on the worksheet without looking. Then compare your working with the solution line by line, not only the final answer. When your answer is different, find the first step that changes, because that is where the mistake is. After that, read the matching part of the probability basics lesson again. Finally, check that the method is secure with the quiz or the test.

Also, the Year 7 probability basics worksheet answers are available as a PDF, so you can mark printed work quickly. Moreover, all the pages of this chapter work together, so you can move between them at any time. For example, many pupils read the lesson on Monday, practise on Wednesday and test themselves on Friday. In short, these Year 7 probability basics worksheet answers show every step, so nothing is left unexplained.