Hard GCSE Probability scale and relative frequency Questions

Challenging, exam-style GCSE Probability scale and relative frequency questions with worked solutions. Stretch yourself on the hardest P(not A) = 1 - P(A), theoretical probability, simplifying fractions, relative frequency problems.

P(not A) = 1 - P(A)theoretical probabilitysimplifying fractionsrelative frequencyexpected numberprobabilities sum to 1
GCSE Foundation34 questionsStep-by-step solutions
Question 1
6 markschallenging
A biased spinner can land on 11, 22, 33 or 44 only. The probability that it lands on 11 is 18\frac{1}{8}, the probability that it lands on 22 is 14\frac{1}{4} and the probability that it lands on 33 is 14\frac{1}{4}. The spinner is spun 320320 times. Work out an estimate for the number of times it lands on 44.
Show worked solution

Worked solution

  1. Recall that the probabilities of all the outcomes add up to 1

    ∑P(outcome)=1\sum P(\text{outcome}) = 1

    The spinner lands on exactly one of the four numbers every time, so the four probabilities must add up to exactly 11.

  2. Write the given probabilities over a common denominator

    18+28+28\frac{1}{8} + \frac{2}{8} + \frac{2}{8}

    The lowest common denominator of the three given fractions is 88, so rewrite each of them in that denominator before adding.

  3. Add the given probabilities

    18+14+14=58=58\frac{1}{8} + \frac{1}{4} + \frac{1}{4} = \frac{5}{8} = \frac{5}{8}

    The three given probabilities come to 58\frac{5}{8}.

  4. Subtract from 1 to find the missing probability

    P(4)=1−58=38P(4) = 1 - \frac{5}{8} = \frac{3}{8}

    Everything left over belongs to 4: 1−58=381 - \frac{5}{8} = \frac{3}{8}.

  5. Check the four probabilities add to 1

    18+14+14+38=1\frac{1}{8} + \frac{1}{4} + \frac{1}{4} + \frac{3}{8} = 1

    Adding the answer back on gives exactly 11, which confirms it.

  6. Check the missing probability is on the 0 to 1 scale

    0≤38≤10 \le \frac{3}{8} \le 1

    38\frac{3}{8} lies between 00 (impossible) and 11 (certain), so it is a possible probability.

  7. Write the missing probability as a decimal

    38=0.375\frac{3}{8} = 0.375

    As a decimal the probability is 0.3750.375, which makes it easy to place on the probability scale and easy to multiply.

  8. Write down the rule for the expected number of successes

    expected number=P(event)×number of trials\text{expected number} = P(\text{event}) \times \text{number of trials}

    Over many trials, an event with probability pp happens about pp of the time, so the expected number is the probability times the number of trials.

  9. Substitute the probability and the number of spins

    38×320\frac{3}{8} \times 320

    The probability is 38\frac{3}{8} and the spinner is spun 320320 times.

  10. Work out the expected number

    38×320=120\frac{3}{8} \times 320 = 120

    320÷8×3=120320 \div 8 \times 3 = 120.

  11. Check the expected number is sensible

    0≤120≤3200 \le 120 \le 320

    The spinner cannot land on 4 more than 320320 times, and 120120 is well inside that range.

  12. Say what kind of answer this is

    estimate, not a guarantee\text{estimate, not a guarantee}

    This is an ESTIMATE. In a real set of 320320 spins the actual count would usually be near 120120 but not exactly 120120.

  13. Link the estimate back to relative frequency

    relative frequency→P(event)\text{relative frequency} \to P(\text{event})

    If the spinner really were spun 320320 times, the relative frequency of 4 would be close to 38\frac{3}{8} - and closer still if it were spun more times. That is exactly how relative frequency estimates probability.

  14. Note the mistake to avoid

    14≠38 in general\frac{1}{4} \ne \frac{3}{8} \text{ in general}

    Assuming each of the four numbers has probability 14\frac{1}{4} would only be right for a FAIR spinner. This spinner is biased, so the probabilities have to be read from the question.

  15. State the answer

    120120

    An estimate for the number of times the spinner lands on 4 is 120120.

Answer
120120
Question 2
6 markschallenging
A biased spinner can land on 11, 22, 33 or 44 only. The probability that it lands on 11 is 15\frac{1}{5}, the probability that it lands on 22 is 310\frac{3}{10} and the probability that it lands on 33 is 14\frac{1}{4}. The spinner is spun 200200 times. Work out an estimate for the number of times it lands on 44.
Show worked solution

Worked solution

  1. Recall that the probabilities of all the outcomes add up to 1

    ∑P(outcome)=1\sum P(\text{outcome}) = 1

    The spinner lands on exactly one of the four numbers every time, so the four probabilities must add up to exactly 11.

  2. Write the given probabilities over a common denominator

    420+620+520\frac{4}{20} + \frac{6}{20} + \frac{5}{20}

    The lowest common denominator of the three given fractions is 2020, so rewrite each of them in that denominator before adding.

  3. Add the given probabilities

    15+310+14=1520=34\frac{1}{5} + \frac{3}{10} + \frac{1}{4} = \frac{15}{20} = \frac{3}{4}

    The three given probabilities come to 34\frac{3}{4}.

  4. Subtract from 1 to find the missing probability

    P(4)=1−34=14P(4) = 1 - \frac{3}{4} = \frac{1}{4}

    Everything left over belongs to 4: 1−34=141 - \frac{3}{4} = \frac{1}{4}.

  5. Check the four probabilities add to 1

    15+310+14+14=1\frac{1}{5} + \frac{3}{10} + \frac{1}{4} + \frac{1}{4} = 1

    Adding the answer back on gives exactly 11, which confirms it.

  6. Check the missing probability is on the 0 to 1 scale

    0≤14≤10 \le \frac{1}{4} \le 1

    14\frac{1}{4} lies between 00 (impossible) and 11 (certain), so it is a possible probability.

  7. Write the missing probability as a decimal

    14=0.25\frac{1}{4} = 0.25

    As a decimal the probability is 0.250.25, which makes it easy to place on the probability scale and easy to multiply.

  8. Write down the rule for the expected number of successes

    expected number=P(event)×number of trials\text{expected number} = P(\text{event}) \times \text{number of trials}

    Over many trials, an event with probability pp happens about pp of the time, so the expected number is the probability times the number of trials.

  9. Substitute the probability and the number of spins

    14×200\frac{1}{4} \times 200

    The probability is 14\frac{1}{4} and the spinner is spun 200200 times.

  10. Work out the expected number

    14×200=50\frac{1}{4} \times 200 = 50

    200÷4×1=50200 \div 4 \times 1 = 50.

  11. Check the expected number is sensible

    0≤50≤2000 \le 50 \le 200

    The spinner cannot land on 4 more than 200200 times, and 5050 is well inside that range.

  12. Say what kind of answer this is

    estimate, not a guarantee\text{estimate, not a guarantee}

    This is an ESTIMATE. In a real set of 200200 spins the actual count would usually be near 5050 but not exactly 5050.

  13. Link the estimate back to relative frequency

    relative frequency→P(event)\text{relative frequency} \to P(\text{event})

    If the spinner really were spun 200200 times, the relative frequency of 4 would be close to 14\frac{1}{4} - and closer still if it were spun more times. That is exactly how relative frequency estimates probability.

  14. Note the mistake to avoid

    14≠14 in general\frac{1}{4} \ne \frac{1}{4} \text{ in general}

    Assuming each of the four numbers has probability 14\frac{1}{4} would only be right for a FAIR spinner. This spinner is biased, so the probabilities have to be read from the question.

  15. State the answer

    5050

    An estimate for the number of times the spinner lands on 4 is 5050.

Answer
5050
Question 3
5 markschallenging
A biased spinner is spun 400400 times. The results are: 11 came up 9292 times, 22 came up 118118 times, 33 came up 8686 times and 44 came up 104104 times. Work out the relative frequency of a 22. Give your answer as a decimal.
Show worked solution

Worked solution

  1. Add the frequencies to find how many trials there were

    92+118+86+104=40092 + 118 + 86 + 104 = 400

    The number of trials is the total of the whole frequency table: 92+118+86+104=40092 + 118 + 86 + 104 = 400. Dividing by the number of OUTCOMES instead of the number of TRIALS is the classic mistake here.

  2. Read off how many times the result was 2

    frequency of 2=118\text{frequency of } 2 = 118

    The table shows 118118 of the 400400 trials gave 2.

  3. Write down the rule for relative frequency

    relative frequency=number of successesnumber of trials\text{relative frequency} = \frac{\text{number of successes}}{\text{number of trials}}

    Relative frequency (experimental probability) comes from the results of an experiment, not from counting equally likely outcomes.

  4. Substitute the results into the rule

    relative frequency=118400\text{relative frequency} = \frac{118}{400}

    Spinner landing on 2 happened 118118 times in 400400 trials.

  5. Simplify the relative frequency

    118400=59200\frac{118}{400} = \frac{59}{200}

    Dividing top and bottom by 22 gives 59200\frac{59}{200}.

  6. Write the relative frequency as a decimal

    59200=0.295\frac{59}{200} = 0.295

    118÷400=0.295118 \div 400 = 0.295, which is the relative frequency of 2.

  7. Check the relative frequency is a possible probability

    0≤0.295≤10 \le 0.295 \le 1

    A relative frequency can never exceed 11, because the event cannot happen more times than the experiment was carried out. 0.2950.295 passes.

  8. Check the relative frequencies of every outcome add to 1

    92400+118400+86400+104400=1\frac{92}{400} + \frac{118}{400} + \frac{86}{400} + \frac{104}{400} = 1

    Every trial produced exactly one of the outcomes, so the relative frequencies of all the outcomes must add up to 11 - a quick check that the table has been read correctly.

  9. Work out the theoretical probability for a fair one

    P(2)=14=14P(\text{2}) = \frac{1}{4} = \frac{1}{4}

    If spinner were fair, all 44 outcomes would be equally likely, so the theoretical probability of 2 would be 14=0.25\frac{1}{4} = 0.25.

  10. Compare the relative frequency with the theoretical probability

    0.295−0.25≈0.0450.295 - 0.25 \approx 0.045

    The relative frequency 0.2950.295 and the theoretical probability 0.250.25 differ by about 0.0450.045. Relative frequency is only an estimate, so some difference is completely normal.

  11. Work out how many you would expect if it were fair

    14×400=100\frac{1}{4} \times 400 = 100

    A fair spinner would be expected to give 2 about 100100 times in 400400 trials, compared with the 118118 actually recorded.

  12. Say what relative frequency is for

    relative frequency≈P(event)\text{relative frequency} \approx P(\text{event})

    The relative frequency ESTIMATES the probability. It is the only tool available when the outcomes are not equally likely, because then there is no theoretical probability to calculate.

  13. Say how the estimate is improved

    more trials⇒better estimate\text{more trials} \Rightarrow \text{better estimate}

    As the number of trials increases, the relative frequency settles down towards the true probability. A handful of trials tells you very little.

  14. Note the mistake to avoid

    1184≠118400\frac{118}{4} \ne \frac{118}{400}

    Dividing by 44 (the number of possible outcomes) instead of 400400 (the number of trials) is wrong: relative frequency always divides by the number of trials.

  15. State the answer

    0.2950.295

    The relative frequency of 2 is 0.2950.295.

Answer
0.2950.295
Question 4
5 markschallenging
A biased dice is rolled 300300 times. The results are: 11 came up 4242 times, 22 came up 5454 times, 33 came up 4848 times, 44 came up 6060 times, 55 came up 4545 times and 66 came up 5151 times. Work out the relative frequency of a 22. Give your answer as a decimal.
Show worked solution

Worked solution

  1. Add the frequencies to find how many trials there were

    42+54+48+60+45+51=30042 + 54 + 48 + 60 + 45 + 51 = 300

    The number of trials is the total of the whole frequency table: 42+54+48+60+45+51=30042 + 54 + 48 + 60 + 45 + 51 = 300. Dividing by the number of OUTCOMES instead of the number of TRIALS is the classic mistake here.

  2. Read off how many times the result was 2

    frequency of 2=54\text{frequency of } 2 = 54

    The table shows 5454 of the 300300 trials gave 2.

  3. Write down the rule for relative frequency

    relative frequency=number of successesnumber of trials\text{relative frequency} = \frac{\text{number of successes}}{\text{number of trials}}

    Relative frequency (experimental probability) comes from the results of an experiment, not from counting equally likely outcomes.

  4. Substitute the results into the rule

    relative frequency=54300\text{relative frequency} = \frac{54}{300}

    Dice landing on 2 happened 5454 times in 300300 trials.

  5. Simplify the relative frequency

    54300=950\frac{54}{300} = \frac{9}{50}

    Dividing top and bottom by 66 gives 950\frac{9}{50}.

  6. Write the relative frequency as a decimal

    950=0.18\frac{9}{50} = 0.18

    54÷300=0.1854 \div 300 = 0.18, which is the relative frequency of 2.

  7. Check the relative frequency is a possible probability

    0≤0.18≤10 \le 0.18 \le 1

    A relative frequency can never exceed 11, because the event cannot happen more times than the experiment was carried out. 0.180.18 passes.

  8. Check the relative frequencies of every outcome add to 1

    42300+54300+48300+60300+45300+51300=1\frac{42}{300} + \frac{54}{300} + \frac{48}{300} + \frac{60}{300} + \frac{45}{300} + \frac{51}{300} = 1

    Every trial produced exactly one of the outcomes, so the relative frequencies of all the outcomes must add up to 11 - a quick check that the table has been read correctly.

  9. Work out the theoretical probability for a fair one

    P(2)=16=16P(\text{2}) = \frac{1}{6} = \frac{1}{6}

    If dice were fair, all 66 outcomes would be equally likely, so the theoretical probability of 2 would be 16=0.1667…\frac{1}{6} = 0.1667\ldots.

  10. Compare the relative frequency with the theoretical probability

    0.18−0.1667…≈0.0133…0.18 - 0.1667\ldots \approx 0.0133\ldots

    The relative frequency 0.180.18 and the theoretical probability 0.1667…0.1667\ldots differ by about 0.0133…0.0133\ldots. Relative frequency is only an estimate, so some difference is completely normal.

  11. Work out how many you would expect if it were fair

    16×300=50\frac{1}{6} \times 300 = 50

    A fair dice would be expected to give 2 about 5050 times in 300300 trials, compared with the 5454 actually recorded.

  12. Say what relative frequency is for

    relative frequency≈P(event)\text{relative frequency} \approx P(\text{event})

    The relative frequency ESTIMATES the probability. It is the only tool available when the outcomes are not equally likely, because then there is no theoretical probability to calculate.

  13. Say how the estimate is improved

    more trials⇒better estimate\text{more trials} \Rightarrow \text{better estimate}

    As the number of trials increases, the relative frequency settles down towards the true probability. A handful of trials tells you very little.

  14. Note the mistake to avoid

    546≠54300\frac{54}{6} \ne \frac{54}{300}

    Dividing by 66 (the number of possible outcomes) instead of 300300 (the number of trials) is wrong: relative frequency always divides by the number of trials.

  15. State the answer

    0.180.18

    The relative frequency of 2 is 0.180.18.

Answer
0.180.18
Question 5
5 markschallenging
A random sample of 7575 batteries is taken from a large crate of batteries. 66 of the batteries in the sample are faulty. The crate contains 15001500 batteries altogether. Estimate the number of faulty batteries in the crate.
Show worked solution

Worked solution

  1. Read the results of the sample

    sample=75,faulty=6\text{sample} = 75, \quad faulty = 6

    66 of the 7575 batteries in the sample were faulty.

  2. Write down the rule for relative frequency

    relative frequency=number of successesnumber of trials\text{relative frequency} = \frac{\text{number of successes}}{\text{number of trials}}

    Relative frequency (experimental probability) comes from the results of an experiment, not from counting equally likely outcomes.

  3. Substitute the results into the rule

    relative frequency=675\text{relative frequency} = \frac{6}{75}

    A battery being faulty happened 66 times in 7575 trials.

  4. Simplify the relative frequency

    675=225\frac{6}{75} = \frac{2}{25}

    Dividing top and bottom by 33 gives 225\frac{2}{25}.

  5. Write the relative frequency as a decimal

    225=0.08\frac{2}{25} = 0.08

    As a decimal the relative frequency is about 0.080.08, so a little under 8\% of the sample.

  6. Use the sample proportion as an estimate of the probability

    P(faulty)≈225P(faulty) \approx \frac{2}{25}

    The sample is random, so the proportion of faulty batteries in it is the best estimate of the probability that a randomly chosen battery from the whole crate is faulty.

  7. Write down the rule for the expected number

    expected number=P(event)×population size\text{expected number} = P(\text{event}) \times \text{population size}

    Multiplying the estimated probability by the size of the whole population estimates how many members of it have the property.

  8. Substitute the estimate and the size of the whole population

    225×1500\frac{2}{25} \times 1500

    The whole crate contains 15001500 batteries.

  9. Work out the estimate

    225×1500=120\frac{2}{25} \times 1500 = 120

    1500÷25×2=1201500 \div 25 \times 2 = 120.

  10. Check the answer by scaling the proportion

    675=1201500\frac{6}{75} = \frac{120}{1500}

    66 out of 7575 is the same proportion as 120120 out of 15001500, so the two fractions are equivalent and the answer is consistent.

  11. Check the answer is sensible

    0≤120≤15000 \le 120 \le 1500

    There cannot be more than 15001500 batteries with the property, and 120120 is well inside that range.

  12. Check the estimated probability lies on the 0 to 1 scale

    0≤225≤10 \le \frac{2}{25} \le 1

    A relative frequency is always between 00 and 11, so it is a legitimate estimate of a probability.

  13. Say why this is an estimate and not an exact count

    estimate, not a count\text{estimate, not a count}

    A different random sample would give a slightly different proportion, so this is an estimate. Only counting the whole population would give the exact number.

  14. Say how to make the estimate more reliable

    larger sample⇒better estimate\text{larger sample} \Rightarrow \text{better estimate}

    The larger the random sample, the closer its relative frequency settles to the true probability, and so the more reliable the estimate.

  15. State the answer

    120120

    An estimate for the number of faulty batteries in the crate is 120120.

Answer
120120

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