Printable · GCSE Foundation · ages 14-16
Expected outcomes and fairness worksheet — GCSE Foundation
Fifteen questions on "expected outcomes and fairness" — DfE statement P2. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
Expected outcomes and fairness worksheet — GCSE Foundation
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- 1.A phone network sends automatic text alerts to customers. On average, 1,500 alerts are sent each day, and the probability that a customer replies 'STOP' to an alert is 0.18. Work out how many replies of 'STOP' the network should expect over a 30-day month.
- 2.Dice A is a fair six-sided dice. Dice B is biased so that P(6) = 0.3. Dice A is rolled 150 times and Dice B is rolled 150 times. Work out how many more sixes you would expect from Dice B than from Dice A.
- 3.A spinner is divided into sectors of 180°, 120° and 60°, coloured red, blue and green in that order. The spinner is spun 60 times. Work out how many times you would expect it to land on green.
- 4.A fair six-sided dice is rolled 120 times. Work out how many of the rolls you would expect to give a score of 3 or more.
- 5.A fair six-sided dice is rolled 90 times. Work out how many more times you would expect it to land on a number less than 4 than on a 6.
- 6.A fair six-sided dice is rolled 300 times. Work out how many more times you would expect it to land on an even number than on a six.
- 7.A raffle sells 400 tickets at 50p each. There is one prize of £45. Aisha buys 8 tickets. Work out how much money Aisha should expect to lose from playing, giving your answer in pounds.
- 8.A factory finds that the probability a randomly chosen light bulb is defective is 0.035. In a batch of 4,000 bulbs, work out how many bulbs you would expect to work correctly.
- 9.A basketball player has a free-throw success probability of 0.75. She wants to expect to score 60 successful free throws. Work out how many free throws she needs to attempt.
- 10.Two fair coins are thrown together 400 times. Work out how many of the 400 throws you would expect to give one head and one tail.
- 11.A raffle at a school fair sells 500 tickets. Mia buys 25 of the tickets. One ticket is drawn at random to win the prize. Work out the probability that Mia wins the raffle. Give your answer as a fraction in its simplest form.
- 12.A charity raffle sells 250 tickets. A group of friends buy 40 of the tickets between them. Three prizes are awarded, each to a different randomly drawn ticket, and no ticket can win more than one prize. Work out the expected number of prizes won by the group. Give your answer as a fraction in its simplest form.
- 13.On a Saturday, 2,000 customers go through the tills at a supermarket. The probability that a customer buys a reusable bag is 0.08. Work out how many customers you would expect to buy a reusable bag.
- 14.A grower knows that the probability that one of their seeds germinates is 0.6. The grower wants to expect 300 of the seeds to germinate. Work out how many seeds the grower should plant.
- 15.A fair spinner has 10 equal sections, numbered 1 to 10. The spinner is spun 150 times. Work out how many times you would expect it to land on a number greater than 7.
Answer key
- (b) 8,100 — First find the total number of alerts sent in the month: 1,500 × 30 = 45,000. Then apply the probability of a 'STOP' reply: 45,000 × 0.18 = 8,100. Stopping after finding only one day's expected replies, 1,500 × 0.18 = 270, forgets to scale up to the whole month. Multiplying the number of days by the probability instead of by the daily total of alerts gives 30 × 0.18 = 5.4, which rounds to 5. Shifting the decimal point in the probability, using 0.018 instead of 0.18, gives 45,000 × 0.018 = 810.
- (a) 20 — Dice A is fair, so its expected number of sixes is 150 × 1/6 = 25. Dice B has P(6) = 0.3, so its expected number of sixes is 150 × 0.3 = 45. The difference is 45 − 25 = 20. Adding the two expected values instead of subtracting them gives 25 + 45 = 70. Reporting Dice B's expected sixes on their own, without comparing to Dice A, gives 45. Using the fair probability 1/6 for Dice B as well as Dice A ignores the bias altogether, giving 150 × 1/6 = 25 for both dice and a difference of 0.
- (b) 10 — The probability that the spinner lands on green is its angle out of the whole circle: 60° ÷ 360° = 1/6. Expected number of times on green = 1/6 × 60 = 10. Assuming the three colours are equally likely because there are three sectors, regardless of their different angles, gives 60 ÷ 3 = 20. Using red's angle of 180° instead of green's 60° gives a probability of 1/2, so 60 × 1/2 = 30. Treating green's angle, 60°, as a percentage instead of finding its fraction of 360° gives 60 × 0.60 = 36.
- (a) 80 — Method: list the outcomes that count as a success, write the probability from them, then multiply by the number of rolls. Working: the scores of 3 or more are 3, 4, 5 and 6, which is 4 of the 6 equally likely scores, so the probability is 4/6, which cancels to 2/3. Over 120 rolls the expected number is 120 × 2 ÷ 3 = 80. Answer: about 80 of the rolls would be expected to give 3 or more. The distractors: 60 comes from reading a score of 3 or more as a score above 3 and counting only 4, 5 and 6, giving 120 × 3 ÷ 6 = 60; 40 is the expected number of rolls that are not 3 or more, 120 × 2 ÷ 6 = 40; 20 is 120 ÷ 6 and is the expected count for one single score.
- (c) 30 — The numbers less than 4 are 1, 2 and 3, so the probability of that event is 3/6, and the expected count in 90 rolls is 90 × 3/6 = 45. The probability of rolling a 6 is 1/6, and the expected count is 90 × 1/6 = 15. The difference between the two expected counts is 45 − 15 = 30. A candidate who answers 45 has given the expected count for 'less than 4' only, forgetting to subtract the other expected count. A candidate who answers 15 has given the expected count for '6' only. A candidate who answers 36 has used a dice with 5 possible numbers instead of 6, giving 90 × 3/5 = 54 and 90 × 1/5 = 18, a difference of 36.
- (c) 100 — There are 3 even numbers on a fair dice (2, 4 and 6), so the probability of landing on an even number is 3/6 = 1/2, and 300 × 1/2 = 150. The probability of landing on a six is 1/6, so 300 × 1/6 = 50. The dice is expected to land on an even number 150 − 50 = 100 more times than on a six. Writing 50 is wrong because that is just the expected number of sixes on its own, without comparing it to the expected number of evens. Writing 150 is wrong because that is just the expected number of evens on its own, without subtracting the sixes. Writing 200 is wrong because it adds the two expected frequencies together (150 + 50 = 200) instead of finding the difference between them. The dice is expected to land on an even number 100 more times than on a six.
- (c) £3.10 — Aisha's tickets cost 8 × 50p = £4.00. Her expected winnings are (8/400) × £45 = £0.90, since she holds 8 of the 400 tickets. Her expected loss is the cost minus the expected winnings: £4.00 − £0.90 = £3.10. Writing £4.00 is wrong because it is only the cost of her tickets, with no account taken of the expected winnings she might get back. Writing £0.90 is wrong because that is her expected WINNINGS, not her loss — the cost has not been subtracted. Writing £3.89 is wrong because it uses 1 ticket instead of her actual 8 tickets when working out the expected winnings: (1/400) × £45 = £0.1125, giving £4.00 − £0.11 = £3.89. Aisha should expect to lose £3.10.
- (a) 3,860 — The probability a bulb works correctly is the complement of being defective: 1 − 0.035 = 0.965. Expected number working correctly = 0.965 × 4,000 = 3,860. Using the probability of being defective instead of its complement gives 4,000 × 0.035 = 140, the expected number of DEFECTIVE bulbs, not working ones. Shifting the decimal point in the complement, using 0.0965 instead of 0.965, gives 4,000 × 0.0965 = 386. Assuming every bulb works, ignoring the 0.035 probability altogether, gives the full batch of 4,000.
- (b) 80 — To find the number of attempts needed, divide the target number of successes by the probability of success: 60 ÷ 0.75 = 80. Writing 45 is wrong because 60 × 0.75 = 45 multiplies instead of dividing — that is the number of successes expected from 60 attempts, not the number of attempts needed for 60 successes. Writing 240 is wrong because 60 ÷ 0.25 = 240 uses 0.25, the probability of MISSING, instead of 0.75, the probability of scoring. Writing 90 is wrong because it comes from misremembering 0.75 as 2/3 and dividing by that instead: 60 ÷ (2/3) = 90. She needs to attempt 80 free throws.
- (b) 200 — Method: list the equally likely outcomes for the two coins before writing any probability, then multiply by the number of throws. Working: the equally likely outcomes are head then head, head then tail, tail then head, and tail then tail, so there are 4 of them. Two of those 4 give one head and one tail, so the probability is 2/4, which is 1/2. Over 400 throws the expected number is 400 × 1 ÷ 2 = 200. Answer: about 200 of the throws would be expected to give one head and one tail. The distractors: 133 comes from treating two heads, two tails and one of each as three equally likely results and working out 400 ÷ 3 = 133.3, then rounding; 100 comes from counting only head then tail as a success, giving 400 × 1 ÷ 4 = 100; 300 is the expected number of throws that do not give two heads, 400 × 3 ÷ 4 = 300.
- (d) 1/20 — Mia has 25 of the 500 tickets, so the probability she wins is 25/500 = 1/20, dividing the top and the bottom by 25. Writing 19/20 is wrong because that is the probability she does NOT win (1 − 1/20 = 19/20), the opposite of what is asked. Writing 1/25 is wrong because it puts 1 over the number of tickets Mia holds, as though her 25 tickets were the whole raffle — the denominator has to be the 500 tickets sold, not her own share. Writing 1/10 is wrong because it treats the raffle as having only 250 tickets instead of the actual 500: 25/250 = 1/10. The probability that Mia wins is 1/20.
- (d) 12/25 — The group holds 40 of the 250 tickets, so for any one prize the probability the group wins it is 40/250 = 4/25. There are 3 prizes and the group has the same chance at each one, so the expected number won is 3 × 4/25 = 12/25. Writing 4/25 is wrong because it is the chance of winning just ONE prize, without multiplying by the 3 prizes available. Writing 4/75 is wrong because it divides by the 3 prizes instead of multiplying (4/25 ÷ 3 = 4/75), which would mean the group did worse the more prizes were on offer. Writing 64/15625 is wrong because it multiplies the single-prize probability by itself three times, (4/25)³, as though all three prizes had to be won together, instead of adding up the expected number across the three separate prizes. The expected number of prizes won by the group is 12/25.
- (a) 160 — Expected number = probability × number of trials = 0.08 × 2,000 = 160. Moving the decimal point one place too far, using 0.008 instead of 0.08, gives 2,000 × 0.008 = 16. Working out the expected number of customers who do NOT buy a bag, using the complement 1 − 0.08 = 0.92, gives 2,000 × 0.92 = 1,840. Rounding 0.08 up to 0.1 before multiplying gives 2,000 × 0.1 = 200.
- (c) 500 — Method: the expected number of successes is the number of trials multiplied by the probability, so to find the number of trials, divide the expected number by the probability. Working: let n be the number of seeds planted. Then n multiplied by 0.6 must come to 300, so n = 300 ÷ 0.6 = 500. Answer: the grower should plant 500 seeds. The distractors: 180 comes from multiplying instead of dividing, 300 × 0.6 = 180, which answers how many of 300 seeds would germinate; 750 comes from dividing by the probability of not germinating, 300 ÷ 0.4 = 750; 120 comes from multiplying by that same 0.4, 300 × 0.4 = 120.
- (c) 45 — Three of the ten numbers (8, 9 and 10) are greater than 7, so the probability is 3/10, and 150 × 3/10 = 45. Writing 105 is wrong because 150 × 7/10 = 105 uses the seven numbers that are NOT greater than 7 (1 to 7), the opposite of what is asked. Writing 60 is wrong because it counts 7, 8, 9 and 10 as four numbers greater than 7, wrongly including 7 itself: 150 × 4/10 = 60. Writing 50 is wrong because 150 ÷ 3 = 50 divides by the count of favourable numbers instead of multiplying by the correct fraction of the spinner. The expected number of spins landing on a number greater than 7 is 45.
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