GCSE Probability scale and relative frequency Practice Questions

Free GCSE Probability scale and relative frequency practice questions with full step-by-step worked solutions. Covers theoretical probability, equally likely outcomes, simplifying fractions, fair spinner. Practise exam-style problems and check your method.

theoretical probabilityequally likely outcomessimplifying fractionsfair spinnerP(not A) = 1 - P(A)probabilities sum to 1
GCSE Foundation70 questionsStep-by-step solutions
Question 1
1 markeasy
A bag contains 33 red counters, 55 blue counters and 22 green counters. Amina takes one counter at random. Work out the probability that the counter is red. Give your answer as a fraction in its simplest form.
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Worked solution

  1. Count how many counters there are altogether

    3+5+2=103 + 5 + 2 = 10

    There are 33 red, 55 blue and 22 green counters, so there are 1010 counters in the bag. Taking one at random makes all 1010 counters equally likely, and that is what lets us use theoretical probability.

  2. Write the theoretical probability of red

    P(red)=310P(\text{red}) = \frac{3}{10}

    Theoretical probability is the number of favourable outcomes divided by the total number of equally likely outcomes, so P(red)=310P(\text{red}) = \frac{3}{10}.

  3. State the answer

    310\frac{3}{10}

    The probability is 310\frac{3}{10}, written as a fraction in its simplest form.

Answer
310\frac{3}{10}
Question 2
1 markeasy
Five numbers are written on cards. Which of these numbers cannot be a probability?
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Worked solution

  1. Recall the two ends of the probability scale

    0P(event)10 \le P(\text{event}) \le 1

    A probability can never be smaller than 00 (an event that cannot happen) or bigger than 11 (an event that is certain). Every probability lies on the scale between them.

  2. Say why the odd one out fails

    1.4=1.41.4 = 1.4

    1.41.4 is greater than 11, and nothing can be more certain than certain.

  3. State the answer

    1.41.4

    1.41.4 cannot be a probability.

Answer
1.41.4
Question 3
2 marksintermediate
A student writes down five numbers. Which of these numbers cannot be a probability?
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Worked solution

  1. Recall the two ends of the probability scale

    0P(event)10 \le P(\text{event}) \le 1

    A probability can never be smaller than 00 (an event that cannot happen) or bigger than 11 (an event that is certain). Every probability lies on the scale between them.

  2. Write each number as a decimal so they can be compared

    75=1.4,0.08=0.08,12=0.5,0=0.0,1=1.0\frac{7}{5} = 1.4, \quad 0.08 = 0.08, \quad \frac{1}{2} = 0.5, \quad 0 = 0.0, \quad 1 = 1.0

    Fractions, decimals and percentages can all be probabilities, but they have to be written in the same form before they can be compared with 00 and 11.

  3. Test each number against the scale

    75[0,1],0.08[0,1],12[0,1],0[0,1],1[0,1]\frac{7}{5} \notin [0, 1], \quad 0.08 \in [0, 1], \quad \frac{1}{2} \in [0, 1], \quad 0 \in [0, 1], \quad 1 \in [0, 1]

    Four of the numbers lie between 00 and 11 inclusive, so each of them could be a probability. Only 75\frac{7}{5} does not.

  4. Place the possible probabilities on the scale

    0p10 \le p \le 1

    The four possible values all sit on the 00 to 11 scale, which is a quick visual check that they are legitimate probabilities.

  5. Say why the odd one out fails

    75=1.4\frac{7}{5} = 1.4

    75\frac{7}{5} is greater than 11, and nothing can be more certain than certain.

  6. State the answer

    75\frac{7}{5}

    75\frac{7}{5} cannot be a probability.

Answer
75\frac{7}{5}
Question 4
3 markshard
Event AA has a probability of 0.150.15, event BB has a probability of 25\frac{2}{5}, event CC has a probability of 70%70\% and event DD has a probability of 120\frac{1}{20}. Which list puts the events in order, starting with the least likely?
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Worked solution

  1. Say why the probabilities must be converted first

    fraction, decimal, percentagedecimal\text{fraction}, \ \text{decimal}, \ \text{percentage} \rightarrow \text{decimal}

    The probabilities are written in different forms. They cannot be put in order until they are all written in the same form, and decimals are the easiest.

  2. Write the probability of event AA as a decimal

    P(A)=0.15=0.15P(A) = 0.15 = 0.15

    0.150.15 is already a decimal, so nothing needs doing.

  3. Write the probability of event BB as a decimal

    P(B)=25=0.4P(B) = \frac{2}{5} = 0.4

    25\frac{2}{5} written as a decimal is 0.40.4. Probabilities given as fractions, decimals and percentages cannot be compared until they are all in the same form.

  4. Write the probability of event CC as a decimal

    P(C)=70%=0.7P(C) = 70\% = 0.7

    70%70\% written as a decimal is 0.70.7. Probabilities given as fractions, decimals and percentages cannot be compared until they are all in the same form.

  5. Write the probability of event DD as a decimal

    P(D)=120=0.05P(D) = \frac{1}{20} = 0.05

    120\frac{1}{20} written as a decimal is 0.050.05. Probabilities given as fractions, decimals and percentages cannot be compared until they are all in the same form.

  6. Recall what the two ends of the scale mean

    0=impossible,1=certain0 = \text{impossible}, \quad 1 = \text{certain}

    A probability of 00 means the event cannot happen and a probability of 11 means it is certain, so a bigger number always means a more likely event.

  7. Mark the events on the probability scale

    A:0.15,B:0.4,C:0.7,D:0.05A: 0.15, \quad B: 0.4, \quad C: 0.7, \quad D: 0.05

    Marking each event on the 00 to 11 scale shows the order at a glance: the further to the right, the more likely the event.

  8. Put the decimals in order, smallest first

    0.05<0.15<0.4<0.70.05 < 0.15 < 0.4 < 0.7

    Ordering the decimals from smallest to largest orders the events from least likely to most likely.

  9. Read the events off in that order

    D, \ A, \ B, \ C

    The least likely event is DD and the most likely is CC.

  10. State the answer

    D, A, B, C\text{D, A, B, C}

    In order from least likely to most likely: DD, AA, BB, CC

Answer
D, A, B, C\text{D, A, B, C}
Question 5
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.
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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)=158=38P(4) = 1 - \frac{5}{8} = \frac{3}{8}

    Everything left over belongs to 4: 158=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

    03810 \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

    01203200 \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 frequencyP(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

    1438 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

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