Probability: sample spaces and Venn diagrams – Year 9 maths worksheet answers to download as PDF

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1. Answer to Question 1

   1. One outcome out of 8: \(\dfrac18\) The spinner is fair, so each of the 8 sections is equally likely and only one section shows 5.
   2. Even numbers are 2, 4, 6, 8: \(\dfrac48=\) \(\dfrac12\) Four of the eight numbers are even, so the probability simplifies to one half.
   3. Primes are 2, 3, 5, 7: \(\dfrac48=\) \(\dfrac12\) A prime number has exactly two factors; 1 is not prime. Four of the eight numbers are prime.
   4. 7 and 8: \(\dfrac28=\) \(\dfrac14\) Only 7 and 8 are greater than 6, so two outcomes out of eight, which simplifies to one quarter.

2. Answer to Question 2

   1. \(1-0.35=\) 0.65 It either rains or it does not, so the two probabilities add up to 1.
   2. \(1-\dfrac38=\) \(\dfrac58\) Subtract from 1, written as \(\dfrac88\), to get the probability that the counter is not red.
   3. \(1-0.2=\) 0.8 Assuming a draw is not possible, winning and losing are the only outcomes, so they add up to 1.
   4. \(1-\dfrac7{40}=\) \(\dfrac{33}{40}\) Write 1 as \(\dfrac{40}{40}\) and subtract \(\dfrac7{40}\).

3. Answer to Question 3

   1. \(0.3+0.45+0.15+x=1\), so \(0.9+x=1\) and \(x=\) 0.1 The four colours are the only possible outcomes, so their probabilities add up to 1.
   2. \(0.3+0.45=\) 0.75 Red and blue cannot happen at the same time, so add their probabilities.
   3. \(1-0.15=\) 0.85 Not green is the complement of green, so subtract from 1. Check: \(0.3+0.45+0.1=0.85\).
   4. \(0.45+0.1=\) 0.55 Blue and yellow are mutually exclusive, so add 0.45 and the value of \(x\) from part a).

4. Answer to Question 4

   1. Six 7s in the table: \(\dfrac6{36}=\) \(\dfrac16\) The 7s lie on a diagonal of the table, one in each row, so there are 6 of them out of 36.
   2. Doubles are \((1,1)\) to \((6,6)\): \(\dfrac6{36}=\) \(\dfrac16\) Each score can be doubled in exactly one way, giving 6 doubles out of 36 outcomes.
   3. Totals 11 and 12: \(2+1=3\), so \(\dfrac3{36}=\) \(\dfrac1{12}\) A total of 11 can happen in 2 ways and 12 in only 1 way.
   4. Totals 2 and 3: \(1+2=3\), so \(\dfrac3{36}=\) \(\dfrac1{12}\) A total of 2 can happen in 1 way and 3 in 2 ways, giving 3 outcomes out of 36.

5. Answer to Question 5

   1. H1, H2, H3, T1, T2, T3 Pair each side of the coin with each number on the spinner, so that no outcome is missed.
   2. There are 6 outcomes (\(2\times3\)). There are 2 coin outcomes and 3 spinner outcomes, so \(2\times3=6\) combinations.
   3. One outcome (H2): \(\dfrac16\) All 6 outcomes are equally likely, and only H2 matches.
   4. T1 and T3: \(\dfrac26=\) \(\dfrac13\) The odd numbers are 1 and 3, so two of the six outcomes match.

6. Answer to Question 6

   1. \(A\cap B=\{6,12\}\). \(A\cup B=\{2,3,4,6,8,9,10,12\}\). The intersection contains numbers in both sets: even multiples of 3. The union contains numbers in at least one set.
   2. \(P(A)=\dfrac6{12}=\dfrac12\). \(P(A\cap B)=\dfrac2{12}=\dfrac16\). There are 6 even numbers out of 12, and 2 numbers in the intersection.
   3. Neither: 1, 5, 7, 11. \(\dfrac4{12}=\) \(\dfrac13\) Numbers outside both circles are odd and not multiples of 3: there are four of them.
   4. \(n(A)=6\), \(n(B)=4\), \(n(A\cap B)=2\): \(6+4-2=8=n(A\cup B)\). Correct. The formula subtracts the overlap because it was counted twice, once in \(A\) and once in \(B\).

7. Answer to Question 7

   1. \(\dfrac5{200}=\) \(\dfrac1{40}\) Each ticket has the same chance, and Maya has 5 of the 200 tickets.
   2. \(\dfrac{12}{200}=\) \(\dfrac3{50}\) Ali has 12 of the 200 tickets; the fraction simplifies by dividing by 4.
   3. \(\dfrac5{200}+\dfrac{12}{200}=\) \(\dfrac{17}{200}\). Only one ticket wins, so the events are mutually exclusive. Only one ticket is drawn, so Maya and Ali cannot both win, and we may add their probabilities.
   4. \(1-\dfrac{17}{200}=\) \(\dfrac{183}{200}\) Someone else winning is the complement of Maya or Ali winning.
   5. \(200-50=\) £150 The school receives £1 for each of the 200 tickets and pays out one prize of £50.

8. Answer to Question 8

   1. \(1-0.6=\) 0.4 A day is either sunny or not sunny, so the probabilities add up to 1.
   2. \(20\times0.6=\) 12 Expected number = number of trials × probability.
   3. Yes, as long as rain is part of the 0.4: the not-sunny days are 0.4, which is more than 0.25. Rainy days must fit inside the not-sunny days, so the probability of rain cannot be more than 0.4.
   4. \(50\times0.4=\) 20 Multiply the number of Saturdays by the probability of not sunny.

9. Answer to Question 9

   1. \(P(\text{red})=\dfrac8{20}=\dfrac25\); \(P(\text{orange})=\dfrac2{20}=\dfrac1{10}\) There are 20 sweets altogether; 8 are red and 2 are orange.
   2. \(\dfrac{8+6}{20}=\dfrac{14}{20}=\) \(\dfrac7{10}\) A sweet cannot be red and green at once, so add the 8 red and 6 green sweets.
   3. \(1-\dfrac1{10}=\) \(\dfrac9{10}\) Use the complement: the probability of orange is \(\dfrac1{10}\).
   4. Now there are 6 red out of 18 sweets: \(\dfrac6{18}=\) \(\dfrac13\) After eating two red sweets there are 2 fewer red sweets and 2 fewer sweets in total.

10. Answer to Question 10

    1. \(\dfrac6{36}=\) \(\dfrac16\) There are 6 doubles among the 36 equally likely outcomes.
    2. Totals 9, 10, 11, 12 occur \(4+3+2+1=10\) times: \(\dfrac{10}{36}=\) \(\dfrac5{18}\) Count the outcomes in the table with a total of 9, 10, 11 or 12.
    3. Pairs \((2,6),(6,2),(3,4),(4,3)\): \(\dfrac4{36}=\) \(\dfrac19\) Only \(2\times6\) and \(3\times4\) give 12, and each can happen in either order.
    4. \(72\times\dfrac16=\) 12 extra turns Expected number = number of rolls × probability of a double.

11. Answer to Question 11

    1. Pizza-juice, pizza-water, pasta-juice, pasta-water, salad-juice, salad-water. Work systematically: take each main course and pair it with each drink.
    2. 6 outcomes. There are 3 main courses and 2 drinks, so \(3\times2=6\) possible lunches.
    3. One outcome: \(\dfrac16\) Each lunch is equally likely, and only one of the six is pizza with water.
    4. Two outcomes (salad-juice, salad-water): \(\dfrac26=\) \(\dfrac13\) Salad appears with each of the two drinks.
    5. Pasta-juice, pasta-water, pizza-juice, salad-juice: \(\dfrac46=\) \(\dfrac23\) Count every lunch that contains pasta or juice or both; pasta-juice is counted only once.

12. Answer to Question 12

    1. Football only \(=18-5=13\). Hockey only \(=12-5=7\). The 5 pupils who play both are inside both circles, so subtract them from each total.
    2. \(13+5+7=25\), so neither \(=30-25=\) 5. Add the three regions inside the circles and subtract from the class total.
    3. \(\dfrac5{30}=\) \(\dfrac16\) Five of the 30 pupils play both sports.
    4. \(\dfrac{25}{30}=\) \(\dfrac56\) Football or hockey means at least one sport: \(13+5+7=25\) pupils.
    5. \(\dfrac{13+7}{30}=\dfrac{20}{30}=\) \(\dfrac23\) Exactly one sport means football only or hockey only, but not both.

13. Answer to Question 13

    1. \(1-0.15=\) 0.85 Winning and not winning are complementary, so subtract 0.15 from 1.
    2. \(200\times0.15=\) 30 prizes Expected number of prizes = 200 × 0.15.
    3. \(30\times4=\) £120 Each of the 30 expected prizes is worth £4.
    4. Takings £200 minus £120 \(=\) £80 The fair takes £1 for each spin, so it takes £200 and pays out about £120.
    5. £80 is below £100, so the organiser would need about 250 spins (\(250-37.5\times4=100\)). Each spin gives an expected profit of \(1-0.15\times4=0.40\) pounds, so £100 needs 250 spins.

14. Answer to Question 14

    1. \(\dfrac{13}{52}=\) \(\dfrac14\) There are 13 hearts among the 52 cards.
    2. \(\dfrac4{52}=\) \(\dfrac1{13}\) There is one king in each of the four suits.
    3. \(\dfrac{4+4}{52}=\dfrac8{52}=\) \(\dfrac2{13}\) A card cannot be a king and a queen at once, so add the 4 kings and the 4 queens.
    4. \(1-\dfrac14=\) \(\dfrac34\) There are 13 spades, so 39 cards are not spades, which is \(\dfrac{39}{52}=\dfrac34\).
    5. \(104\times\dfrac14=\) 26 Expected number = number of trials × probability of a heart.

15. Answer to Question 15

    1. Boys \(=50\), girls \(=50\); walk \(=38\), bus \(=42\), car \(=20\). Add across each row to find the totals for boys and girls, and down each column for each type of travel.
    2. \(\dfrac{38}{100}=\) 0.38 38 of the 100 pupils walk.
    3. \(\dfrac{24}{100}=\) 0.24 Look at the cell where the girls' row meets the bus column: 24 out of 100.
    4. \(\dfrac{16}{50}=\) 0.32 This time the total is only the 50 girls, of whom 16 walk.
    5. \(\dfrac{20}{100}\times1200=\) 240 pupils The sample suggests that 20% of pupils travel by car; 20% of 1 200 is 240.

16. Answer to Question 16

    1. \(3\times4=\) 12 outcomes. Spinner A has 3 outcomes and spinner B has 4, so \(3\times4=12\).
    2. Totals of 5: 3 outcomes, so \(\dfrac3{12}=\) \(\dfrac14\) A total of 5 comes from (1,4), (2,3) and (3,2).
    3. Even totals: 2, 4, 4, 6, 4, 6, so 6 outcomes: \(\dfrac12\) Count the even totals in the grid: 6 out of 12.
    4. Totals of 6, 6 and 7: \(\dfrac3{12}=\) \(\dfrac14\) The totals greater than 5 are 6, 6 and 7.
    5. \(120\times\dfrac14=\) 30 Expected number = number of trials × probability.

17. Answer to Question 17

    1. French only \(=16\), both \(=6\), Spanish only \(=9\), neither \(=40-31=9\). Fill in the overlap first: 6 study both, so French only is \(22-6\) and Spanish only is \(15-6\).
    2. \(\dfrac9{40}\) Nine of the 40 students are outside both circles.
    3. \(\dfrac{16+9}{40}=\dfrac{25}{40}=\) \(\dfrac58\) Exactly one language means French only or Spanish only.
    4. Among the 22 French students, 6 also study Spanish: \(\dfrac6{22}=\) \(\dfrac3{11}\) The question restricts the choice to the 22 French students, so the denominator is 22.
    5. \(\dfrac9{40}\times200=\) 45 students Expected number = 200 × the probability of neither language.

18. Answer to Question 18

    1. \(P(7)=\dfrac6{36}=\dfrac16\) and \(P(\text{double})=\dfrac6{36}=\dfrac16\). A total of 7 is odd, and the total of a double is even, so they cannot happen together. Look at the 36-outcome grid for two dice. Since a double always gives an even total, a total of 7 and a double are mutually exclusive.
    2. \(\dfrac16\times4+\dfrac16\times2=\dfrac66=\) £1 Multiply each prize by its probability and add.
    3. Yes: the expected winnings equal the £1 cost, so the game is fair in the long run. A game is fair when the expected winnings equal the cost of playing.
    4. Takings \(=360\times1=\) £360. Expected payout \(=60\times4+60\times2=240+120=\) £360. Expected wins of each type: \(360\times\dfrac16=60\). The payout equals the takings, so the fair expects no profit.

19. Answer to Question 19

    1. \(2\times2\times2=\) 8 outcomes. Each coin has 2 outcomes, so three coins have \(2\times2\times2\) outcomes.
    2. Three outcomes out of 8: \(\dfrac38\) HHT, HTH and THH each have exactly two heads.
    3. Everything except HHH: \(1-\dfrac18=\) \(\dfrac78\) At least one tail is the complement of no tails, which is HHH.
    4. HHH and TTT: \(\dfrac28=\) \(\dfrac14\) All the same means all heads or all tails.
    5. \(80\times\dfrac38=\) 30 Expected number = 80 × the probability of exactly 2 heads.

20. Answer to Question 20

    1. Pizza only \(=18-x\); pasta only \(=20-x\). The \(x\) pupils who like both are inside both circles, so subtract \(x\) from each total.
    2. \((18-x)+x+(20-x)+2=30\), so \(40-x=30\). The 2 pupils who like neither are outside both circles, and all four regions add up to 30.
    3. \(x=\) 10 Collect the terms: \(40-x=30\). Check: \(8+10+10+2=30\).
    4. Pizza only \(=8\): \(\dfrac8{30}=\) \(\dfrac4{15}\) Pizza only is \(18-10=8\) pupils out of 30.
    5. Of the 20 who like pasta, 10 like pizza: \(\dfrac{10}{20}=\) \(\dfrac12\) The choice is limited to the 20 pasta lovers, and 10 of them are in the overlap.