Probability

50 maths questions with worked answers with answers for OCR 11 Probability in OCR GCSE Maths (spec J560). Practise with flashcards, a topic quiz, a random quiz across the course, 5 crosswords, or read every question and answer below. Not official OCR material.

1. Question 1 · Probability scale · OCR 11.01a

   An event has probability 0. What does this mean?

   Answer:

   It is impossible.
   An event with probability 1 is certain.

2. Question 2 · Basic probability · OCR 11.01a

   A bag has 4 red, 5 blue and 3 green counters. Find P(blue).

   Answer:

   12 counters in total.
   512.

3. Question 3 · Basic probability · OCR 11.01a

   A fair dice is rolled. Find P(a number greater than 4).

   Answer:

   5 or 6: 2 outcomes out of 6.
   26 = 13.

4. Question 4 · Basic probability · OCR 11.01a

   A card is picked from 1–20. Find P(a multiple of 3).

   Answer:

   3, 6, 9, 12, 15, 18: 6 outcomes.
   620 = 310.

5. Question 5 · Expected outcomes · OCR 11.01a

   P(a train is late) = 0.08. How many of 250 trains are expected to be late?

   Answer:

   Expected = probability × number of trials.
   0.08 × 250 = 20.

6. Question 6 · Expected outcomes · OCR 11.01a

   A fair dice is rolled 90 times. How many fours are expected?

   Answer:

   P(4) = 16.
   90 ÷ 6 = 15.

7. Question 7 · Basic probability · OCR 11.01a

   A letter is chosen at random from the word PROBABILITY. Find P(B).

   Answer:

   There are 11 letters and 2 Bs.
   211.

8. Question 8 · Relative frequency · OCR 11.01b

   A drawing pin lands point up 36 times in 120 throws. Estimate the probability it lands point up.

   Answer:

   Relative frequency = 36120.
   0.3.

9. Question 9 · Relative frequency · OCR 11.01b

   A biased dice shows a six 45 times in 300 rolls. Estimate how many sixes in 1000 rolls.

   Answer:

   Relative frequency = 45300 = 0.15.
   0.15 × 1000 = 150.

10. Question 10 · Frequency · OCR 11.01b

    A coin lands heads 18 times in 40 tosses. Find the relative frequency of heads.

    Answer:

    Relative frequency = number of heads ÷ number of tosses.
    1840 = 0.45.

11. Question 11 · Frequency · OCR 11.01b

    A spinner gives 12 red results in 60 spins. Estimate P(red).

    Answer:

    Relative frequency = red results ÷ total spins.
    1260 = 15 = 0.2.

12. Question 12 · Frequency · OCR 11.01b

    Of 80 customers, 50 buy tea. How many do not buy tea?

    Answer:

    Subtract those who buy tea from the total.
    80 − 50 = 30 customers.

13. Question 13 · Relative frequency · OCR 11.01c

    Why does more trials give a better estimate of probability?

    Answer:

    Random variation evens out over many trials.
    Relative frequency gets closer to the true probability.

14. Question 14 · Frequency · OCR 11.01c

    Why might 100 coin tosses not produce exactly 50 heads?

    Answer:

    The expected count for a fair coin is 100 × 12 = 50.
    The actual result can vary by chance, so it need not be exactly 50.

15. Question 15 · Fairness · OCR 11.01d

    A coin lands heads 7 times out of 10. Is it biased? Explain.

    Answer:

    10 trials is too few to decide.
    More trials are needed; relative frequency gets more reliable as trials increase.

16. Question 16 · Equally likely · OCR 11.01d

    Why is P(a 3) = 16 on a fair dice?

    Answer:

    The six outcomes are equally likely.
    One of them is a 3.

17. Question 17 · Frequency · OCR 11.01d

    A fair die is rolled 120 times. Find the expected number of sixes.

    Answer:

    The probability of a six is 16.
    Expected number = 120 × 16 = 20.

18. Question 18 · Frequency · OCR 11.01d

    A biased spinner has P(red) = 0.3. In 200 spins, how many reds are expected?

    Answer:

    Expected frequency = probability × number of spins.
    0.3 × 200 = 60.

19. Question 19 · Combined outcomes · OCR 11.01d

    A fair coin is tossed twice. Find P(two heads).

    Answer:

    P(heads) = 12 on each independent toss.
    P(two heads) = 12 × 12 = 14.

20. Question 20 · Combined outcomes · OCR 11.01d

    Two fair dice are rolled. Find P(two sixes).

    Answer:

    There are 36 equally likely ordered outcomes; only (6, 6) has two sixes.
    P = 136.

21. Question 21 · Combined outcomes · OCR 11.01d

    A fair die and fair coin are used independently. Find P(4 and heads).

    Answer:

    P(4) = 16 and P(heads) = 12.
    The events are independent, so multiply: 16 × 12 = 112.

22. Question 22 · Listing · OCR 11.02a

    How many two-digit codes can be made using the digits 1, 2, 3 and 4 if digits can repeat?

    Answer:

    There are 4 choices for each digit.
    4 × 4 = 16 codes.

23. Question 23 · Product rule · OCR 11.02a · Higher only

    A menu has 3 starters, 5 mains and 4 desserts. How many three-course meals are possible?

    Answer:

    Use the product rule for counting.
    3 × 5 × 4 = 60 meals.

24. Question 24 · Product rule · OCR 11.02a · Higher only

    How many ways can 4 different books be arranged on a shelf?

    Answer:

    4 choices for the first place, then 3, 2 and 1.
    4 × 3 × 2 × 1 = 24.

25. Question 25 · Sample spaces · OCR 11.02a

    Two coins are flipped. List the sample space.

    Answer:

    Each coin is H or T.
    HH, HT, TH, TT.

26. Question 26 · Sample spaces · OCR 11.02a

    Two fair dice are rolled and the scores added. Find P(total of 7).

    Answer:

    There are 36 outcomes; 6 give 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1).
    636 = 16.

27. Question 27 · Listing · OCR 11.02a

    Ann, Ben and Cal line up in a queue. List all the orders.

    Answer:

    3 × 2 × 1 = 6 orders.
    ABC, ACB, BAC, BCA, CAB, CBA.

28. Question 28 · Counting · OCR 11.02a

    A café offers 3 soups and 4 sandwiches. How many soup-and-sandwich choices are possible?

    Answer:

    Each of 3 soups can be paired with 4 sandwiches.
    3 × 4 = 12 choices.

29. Question 29 · Venn diagrams · OCR 11.02c

    In a class of 30, 18 play football, 12 play tennis and 5 play both. How many play neither?

    Answer:

    Football or tennis = 18 + 12 − 5 = 25.
    30 − 25 = 5.

30. Question 30 · Venn diagrams · OCR 11.02c

    Using the class above (30 pupils, 18 football, 12 tennis, 5 both), find P(tennis only).

    Answer:

    Tennis only = 12 − 5 = 7.
    730.

31. Question 31 · Two-way tables · OCR 11.02c · Higher only

    In a two-way table, 12 of the 30 Year 10s and 18 of the 50 Year 11s chose art. A pupil who chose art is picked. Find P(Year 10).

    Answer:

    30 chose art in total.
    1230 = 25.

32. Question 32 · Sets · OCR 11.02c

    In a class of 30, 12 study art, 10 study music and 4 study both. How many study art or music?

    Answer:

    Adding 12 and 10 counts the 4 students studying both twice.
    Subtract the overlap: 12 + 10 − 4 = 18.

33. Question 33 · Sets · OCR 11.02c

    A set A has 8 members and B has 6; they share 3. How many are in A union B?

    Answer:

    Add the set sizes and subtract the 3 counted twice.
    8 + 6 − 3 = 11.

34. Question 34 · Conditional · OCR 11.02c · Higher only

    A group has 12 girls and 8 boys. Five girls wear glasses. A girl is chosen at random. Find P(glasses).

    Answer:

    The condition restricts the sample to the 12 girls.
    Five of those 12 wear glasses, so P = 512.

35. Question 35 · Conditional · OCR 11.02c · Higher only

    Of 40 students, 24 take music; 6 of the music students also take drama. Find P(drama given music).

    Answer:

    The condition restricts the sample to the 24 music students.
    Six also take drama, so P = 624 = 14.

36. Question 36 · Frequency trees · OCR 11.02d

    What is a frequency tree used for?

    Answer:

    To sort a number of people or items into groups at each stage.
    It shows frequencies, not probabilities.

37. Question 37 · Sum to 1 · OCR 11.02e

    P(rain) = 0.35. Find P(not rain).

    Answer:

    P(not A) = 1 − P(A).
    1 − 0.35 = 0.65.

38. Question 38 · Sum to 1 · OCR 11.02e

    A spinner lands on red, blue or green. P(red) = 0.2 and P(blue) = 0.45. Find P(green).

    Answer:

    Probabilities sum to 1.
    1 − 0.65 = 0.35.

39. Question 39 · Sum to 1 · OCR 11.02e

    A spinner has P(A) = x, P(B) = 2x and P(C) = 0.4. Find x.

    Answer:

    x + 2x + 0.4 = 1, so 3x = 0.6.
    x = 0.2.

40. Question 40 · Mutually exclusive · OCR 11.02e

    What are mutually exclusive events?

    Answer:

    Events that cannot happen at the same time.
    P(A or B) = P(A) + P(B).

41. Question 41 · Mutually exclusive · OCR 11.02e

    P(red) = 0.3 and P(blue) = 0.25 for one spin. Find P(red or blue).

    Answer:

    They are mutually exclusive.
    0.3 + 0.25 = 0.55.

42. Question 42 · Sum to 1 · OCR 11.02e

    P(win) = 38 and draws are impossible. Find P(lose).

    Answer:

    1 − 38.
    58.

43. Question 43 · Mutual exclusivity · OCR 11.02e

    On a fair die, find P(1 or 2).

    Answer:

    A roll cannot be both 1 and 2, so add their probabilities.
    16 + 16 = 26 = 13.

44. Question 44 · Independent events · OCR 11.02f

    What are independent events?

    Answer:

    One event does not affect the probability of the other.
    P(A and B) = P(A) × P(B).

45. Question 45 · Independent events · OCR 11.02f

    A coin is flipped and a dice is rolled. Find P(head and a 6).

    Answer:

    12 × 16.
    112.

46. Question 46 · Tree diagrams · OCR 11.02f

    P(pass driving test) = 0.7. Two friends take it independently. Find P(both pass).

    Answer:

    0.7 × 0.7.
    0.49.

47. Question 47 · Tree diagrams · OCR 11.02f

    P(pass) = 0.7 for each of two independent friends. Find P(exactly one passes).

    Answer:

    P(pass, fail) + P(fail, pass) = 0.7 × 0.3 + 0.3 × 0.7.
    0.21 + 0.21 = 0.42.

48. Question 48 · Tree diagrams · OCR 11.02f

    Why do branches from one point on a tree diagram add up to 1?

    Answer:

    They show every possible outcome at that stage.
    The outcomes are mutually exclusive and exhaustive.

49. Question 49 · Without replacement · OCR 11.02f

    A bag has 5 red and 3 blue balls. Two are taken without replacement. Find P(both red).

    Answer:

    58 × 47.
    2056 = 514.

50. Question 50 · Without replacement · OCR 11.02f

    A box has 4 milk and 6 dark chocolates. Two are eaten. Find P(both dark).

    Answer:

    610 × 59.
    3090 = 13.

More OCR GCSE Maths topics

• OCR 1 Number Operations and Integers
• OCR 2 Fractions, Decimals and Percentages
• OCR 3 Indices and Surds
• OCR 4 Approximation and Estimation
• OCR 5 Ratio, Proportion and Rates Of Change
• OCR 6 Algebra
• OCR 7 Graphs of Equations and Functions
• OCR 8 Basic Geometry
• OCR 9 Congruence and Similarity
• OCR 10 Mensuration
• OCR 11 Probability
• OCR 12 Statistics

Other Maths courses: AQA GCSE Maths, Edexcel GCSE Maths.