Probability: outcomes and sample spaces – Year 8 maths test answers to download as PDF

← Back to the test

1. Answer to Question 1 (4 marks)

   1. There are \(4+6+10=20\) counters and 6 of them are blue. \(P(\text{blue})=\dfrac{6}{20}\), which simplifies to \(\dfrac{3}{10}\) by dividing top and bottom by 2. (1 mark)
   2. Ten counters are green, so \(P(\text{green})=\dfrac{10}{20}\). Then \(P(\text{not green})=1-\dfrac{10}{20}=\)\(\dfrac12\). (1 mark)
   3. Let n be the number of counters. Then \(\dfrac{5}{n}=\dfrac{1}{4}\) (1 mark). Since 5 is a quarter of n, \(n=5\times4=\)20 (1 mark).

   Check (b) by counting: the 4 red and 6 blue counters are "not green", which is 10 out of 20. Check (c): \(\dfrac{5}{20}=\dfrac14\).

2. Answer to Question 2 (4 marks)

   1. "Late" and "not late" are the only possibilities, so their probabilities add to 1: \(1-0.15=\)0.85. (1 mark)
   2. Expected number = probability × number of journeys: \(0.15\times200=\)30 late trains. (1 mark)
   3. No (1 mark). The expected number in 20 journeys is \(0.15\times20=3\), but this is only an average over many sets of journeys. In any particular 20 journeys the number of late trains could be 1, 2, 4 or another value because of chance, so "exactly 3" cannot be guaranteed (1 mark).

   Over a much larger number of journeys the proportion of late trains will usually get close to 15%.

3. Answer to Question 3 (4 marks)

   1. List each coin in order: HH, HT, TH, TT. HT and TH are different outcomes, because the first and second coins can be told apart. (1 mark)
   2. Each of the 4 coin outcomes can be paired with any of the 6 die scores, giving \(4\times6=\)24 equally likely outcomes. (1 mark)
   3. Two heads happens in 1 way (HH) and an even score in 3 ways (2, 4 or 6), so there are \(1\times3=3\) favourable outcomes (1 mark). The probability is \(\dfrac{3}{24}=\)\(\dfrac18\) (1 mark).

   Check with independent events: \(P(HH)=\dfrac14\) and \(P(\text{even})=\dfrac12\), and \(\dfrac14\times\dfrac12=\dfrac18\).

4. Answer to Question 4 (4 marks)

   1. A 6 by 6 sample space has 36 equally likely products. The pairs with product 12 are (2, 6), (6, 2), (3, 4) and (4, 3): 4 outcomes (1 mark). The probability is \(\dfrac{4}{36}=\)\(\dfrac19\) (1 mark).
   2. A product is odd only when both scores are odd; one even score makes the product even. Each die has 3 odd scores (1, 3, 5), so there are \(3\times3=9\) odd products (1 mark). The probability is \(\dfrac{9}{36}=\)\(\dfrac14\) (1 mark).

   Drawing the full multiplication grid is the safest way to avoid missing pairs. The probability of an even product is therefore \(1-\dfrac14=\dfrac34\), which is why even products appear far more often.

5. Answer to Question 5 (4 marks)

   1. Fill the Venn diagram from the middle: 5 play both. Football only is \(18-5=13\) and tennis only is \(12-5=7\) (1 mark for the regions). At least one sport: \(13+5+7=25\), so neither \(=30-25=\)5 (1 mark).
   2. Seven pupils play tennis only, out of 30 in the class: \(\dfrac{7}{30}\). (1 mark)
   3. Now only the 18 football players are considered, and 5 of them also play tennis: \(\dfrac{5}{18}\). (1 mark)

   Adding 18 and 12 to get 30 would count the 5 pupils who play both twice, which wrongly suggests nobody plays neither sport. In part (c) the denominator changes because we already know the pupil plays football.

Total : 20 marks