Printable · GCSE Foundation · ages 14-16
Independent and dependent combined events worksheet — GCSE Foundation
Fifteen questions on "independent and dependent combined events" — DfE statement P8. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
Independent and dependent combined events worksheet — GCSE Foundation
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- 1.An ordinary fair dice is rolled twice. Work out the probability of getting at least one 5.
- 2.The probability that it rains on a given morning is 0.3. If it rains, the probability that Priya's bus is late is 0.8. If it does not rain, the probability that her bus is late is 0.1. Work out the probability that her bus is late.
- 3.The probability that Leo wakes up before his alarm is 0.35. The probability that his younger sister wakes up before her alarm, independently of Leo, is 0.28. Work out the probability that neither of them wakes up before their alarm.
- 4.Two events are independent. The first has probability 2/5 and the second has probability 1/3. Work out the probability that both events happen.
- 5.In Manchester the probability of rain on any day in November is taken to be 0.3, and whether it rains on one day is independent of whether it rains on the next. Work out the probability that it rains on both the 10th and the 11th of November.
- 6.The probability that Kofi passes his driving test on any attempt is 0.6, and each attempt is independent of the others. Work out the probability that he fails both his first two attempts.
- 7.Harry plays a game in which he flips a fair coin and rolls an ordinary fair dice. He wins a prize only if the coin shows heads and the dice shows a 6. Work out the probability that Harry wins a prize.
- 8.A drawer contains 4 red socks and 2 blue socks. Two socks are taken out at random, one after the other, without the first being put back. Work out the probability that both socks are the same colour.
- 9.The probability that a component is faulty is 1/10. Two components are tested independently. Work out the probability that at least one of the two components is faulty.
- 10.A red dice and a blue dice are rolled. Both dice are ordinary and fair. Work out the probability of getting a 1 on the red dice and a 2 on the blue dice.
- 11.The probability that Ben catches his bus on time is 0.8. The probability that Freya catches her bus on time is 0.5. The two events are independent. Work out the probability that both Ben and Freya catch their bus on time.
- 12.The probability that Isla answers a quiz question correctly is 0.8. She answers three questions, and her answers are independent of each other. Work out the probability that she answers all three correctly.
- 13.The probability that a seed fails to germinate is 1/5. Three seeds are planted independently. Work out the probability that at least one of the three seeds germinates.
- 14.The probability that Priya scores with a netball shot is 3/4. She takes two shots, and the shots are independent. Work out the probability that she scores with her first shot but misses her second shot.
- 15.A tree diagram shows two rounds of a game, played independently. On each round, the probability of losing is 0.7 and the probability of winning is 0.3. Work out the probability of losing both rounds.
Answer key
- (b) 11/36 — Method: 'at least one' is the opposite of 'none at all', so work out the probability of no 5 on either roll and take it away from 1. Working: a roll that is not a 5 has probability 5/6, and the rolls are independent, so no 5 at all has probability 5/6 × 5/6 = 25/36. Taking this from 36/36 leaves 11/36. Answer: the probability is 11/36. The distractors: 25/36 is the probability of no 5 at all, written down without the final subtraction; 12/36 comes from counting the 6 pairs with a 5 on the first roll and the 6 pairs with a 5 on the second and adding them, which counts the pair (5, 5) twice; 30/36 comes from working out 1 − 1/6 as though only one roll were made.
- (c) 0.31 — Method: a late bus can be reached along two paths of the tree. Multiply along each path, then add the paths that end in a late bus. Working: the rain path gives 0.3 × 0.8 = 0.24. No rain has probability 1 − 0.3 = 0.7, so the second path gives 0.7 × 0.1 = 0.07. Adding the two paths gives 0.24 + 0.07. Answer: the probability is 0.31. The distractors: 0.9 comes from adding the two branch probabilities 0.8 and 0.1 without first weighting them by how often it rains; 0.24 comes from following the rain path only and ignoring that the bus can also be late when it is dry; 0.27 comes from using 0.3 at the start of both paths, so that the no-rain path is given 0.3 × 0.1 instead of 0.7 × 0.1.
- (d) 0.468 — The probability that Leo does not wake up before his alarm is 1 − 0.35 = 0.65, and the probability that his sister does not is 1 − 0.28 = 0.72. Since the two events are independent, multiply the complements: 0.65 × 0.72 = 0.468. A candidate who answers 0.63 has added the two given probabilities, 0.35 + 0.28, instead of finding and multiplying the complements. A candidate who answers 0.098 has multiplied the two given probabilities directly, 0.35 × 0.28, without taking complements first. A candidate who answers 0.532 has correctly reached 0.468 but then subtracted it from 1 again by mistake.
- (c) 2/15 — Method: for independent events, the probability that both happen is the product of the two probabilities. Working: the calculation is 2/5 × 1/3. Multiplying fractions gives 2 × 1 = 2 on the top and 5 × 3 = 15 on the bottom. Answer: the probability is 2/15. The distractors: 11/15 comes from adding the two probabilities, 6/15 + 5/15, instead of multiplying them; 3/8 comes from adding the numerators and the denominators separately; 1/15 comes from subtracting one probability from the other, 6/15 − 5/15.
- (a) 0.09 — Method: two independent events that must both happen are combined by multiplying their probabilities. Working: the same probability 0.3 applies to each day, so the calculation is 0.3 × 0.3. Written as fractions this is 3/10 × 3/10 = 9/100. Answer: the probability is 0.09. The distractors: 0.6 comes from adding 0.3 and 0.3 instead of multiplying them; 0.3 comes from quoting the single-day probability, as though the second day added no further condition; 0.9 comes from multiplying 3 by 3 correctly but keeping only one decimal place in the product instead of two.
- (a) 0.16 — The probability that Kofi fails a single attempt is 1 − 0.6 = 0.4. Since the attempts are independent, the probability he fails both is 0.4 × 0.4 = 0.16. Choosing 0.36 comes from squaring the probability of PASSING instead, 0.6 × 0.6 = 0.36, which is the probability of passing both attempts, not failing both. Choosing 0.4 comes from giving the probability of failing just one attempt, forgetting to combine two attempts. Choosing 0.24 comes from multiplying the fail probability by the pass probability, 0.4 × 0.6 = 0.24, mixing up passing and failing between the two attempts.
- (b) 1/12 — Method: the coin does not affect the dice, so the two events are independent and the probability that both happen is the product of their probabilities. Working: heads has probability 1/2 and a 6 on an ordinary dice has probability 1/6. Multiplying gives 1 on the top and 2 × 6 = 12 on the bottom. Answer: the probability is 1/12. The distractors: 2/3 comes from adding 1/2 and 1/6 instead of multiplying them; 1/6 comes from using the dice alone and ignoring the condition on the coin; 1/8 comes from counting the possible results as 6 + 2 = 8 and treating the winning result as one of those eight.
- (d) 7/15 — Both socks are the same colour either if both are red or if both are blue. The probability both are red is 4/6 × 3/5 = 12/30. The probability both are blue is 2/6 × 1/5 = 2/30. Adding these gives 12/30 + 2/30 = 14/30 = 7/15. Choosing 2/5 comes from only working out the 'both red' path, 12/30, and forgetting the 'both blue' path also counts. Choosing 5/9 comes from treating the draws as if the first sock were replaced, using 4/6 × 4/6 + 2/6 × 2/6 = 20/36 = 5/9, instead of reducing the totals for the second draw. Choosing 7/18 comes from reducing the number of socks removed but not the number left to choose from, using 4/6 × 3/6 + 2/6 × 1/6 = 14/36 = 7/18, instead of 5 remaining socks for the second draw.
- (d) 19/100 — It is easier to first find the probability that NEITHER component is faulty, then subtract from 1. The probability a component is not faulty is 9/10, so the probability neither is faulty is 9/10 × 9/10 = 81/100. So the probability at least one is faulty is 1 − 81/100 = 19/100. Choosing 1/5 comes from adding the two probabilities of a fault instead, 1/10 + 1/10 = 1/5, which double-counts the case where both are faulty. Choosing 1/10 comes from giving the probability for just one component being faulty. Choosing 1/100 comes from squaring the probability of a fault directly, 1/10 × 1/10 = 1/100, which is actually the probability that BOTH are faulty, not at least one.
- (c) 1/36 — Method: the two dice do not affect each other, so the probability that both results happen is the product of the two separate probabilities. Working: a 1 on the red dice has probability 1/6 and a 2 on the blue dice has probability 1/6. Multiplying gives 1 × 1 = 1 on the top and 6 × 6 = 36 on the bottom. Answer: the probability is 1/36. The distractors: 1/18 comes from allowing either order, counting a 2 on the red dice with a 1 on the blue dice as well, although the stem names which dice shows which score; 1/3 comes from adding 1/6 and 1/6 instead of multiplying; 1/6 comes from giving the probability for one dice only and treating the other as making no difference.
- (b) 0.4 — The two events are independent, so multiply the two probabilities: 0.8 × 0.5 = 0.4. Choosing 1.3 comes from adding the two probabilities instead of multiplying, 0.8 + 0.5 = 1.3. Choosing 0.3 comes from subtracting the two probabilities, 0.8 − 0.5 = 0.3, instead of multiplying them. Choosing 0.1 comes from using the probability that Ben is NOT on time, 1 − 0.8 = 0.2, and multiplying that by Freya's probability instead, 0.2 × 0.5 = 0.1.
- (c) 0.512 — Method: independent events that must all happen are combined by multiplying their probabilities. Working: the first two questions give 0.8 × 0.8 = 0.64. Bringing in the third gives 0.64 × 0.8, and since 64 × 8 = 512 with three decimal places in the product, this is 0.512. Answer: the probability is 0.512. The distractors: 0.64 comes from multiplying only two of the three probabilities and stopping; 0.8 comes from reading 'independent' as meaning the probability never changes and writing down the single-question figure; 0.0512 comes from a place-value slip in the last multiplication, counting four decimal places instead of three.
- (d) 124/125 — The probability that a seed germinates is 1 − 1/5 = 4/5, so the probability that all three seeds fail to germinate is 1/5 × 1/5 × 1/5 = 1/125. The probability that at least one germinates is 1 − 1/125 = 124/125. Choosing 4/5 comes from giving the probability that a single seed germinates, forgetting to combine all three seeds. Choosing 64/125 comes from working out the probability that ALL three seeds germinate, 4/5 × 4/5 × 4/5 = 64/125, instead of at least one. Choosing 12/125 comes from working out the probability that EXACTLY one seed germinates, 3 × 4/5 × 1/5 × 1/5 = 12/125, instead of at least one.
- (c) 3/16 — The probability that Priya misses a shot is 1 − 3/4 = 1/4. The two shots are independent, so multiply the probability of scoring the first shot by the probability of missing the second shot: 3/4 × 1/4 = 3/16. Choosing 9/16 comes from using the probability of scoring, 3/4, for both shots instead of switching to the miss probability for the second shot, 3/4 × 3/4 = 9/16, which is the probability that she scores with both shots. Choosing 1/2 comes from subtracting the two probabilities instead of multiplying them, 3/4 − 1/4 = 1/2. Choosing 3/4 comes from giving only the probability that she scores with the first shot, forgetting to combine it with what happens on the second shot.
- (c) 0.49 — The two rounds are independent, so multiply the probability of losing each round: 0.7 × 0.7 = 0.49. Choosing 0.7 comes from giving the probability of losing just one round, forgetting there are two rounds to combine. Choosing 1.4 comes from adding the two probabilities instead of multiplying, 0.7 + 0.7 = 1.4. Choosing 0.09 comes from using the probability of WINNING instead of losing, 0.3 × 0.3 = 0.09.
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