UniGeniusGCSE Maths · Edexcel 1MA1

Worksheets · Foundation and Higher

Independent and dependent event trees

8 exam-style questions, grades 4 to 7. Worked solutions and the marks are on the last page.

  1. Question 1Non-calculator · 4 marks

    The probability that it rains on any day is 0.3. Assume the weather on Monday and on Tuesday is independent.

    (a) Work out the probability that it rains on both days. (2)

    (b) Work out the probability that it rains on neither day. (2)

  2. Question 2Non-calculator · 2 marks

    (a) A bag contains 2 blue counters and 5 white counters. A counter is taken at random, replaced, and then another counter is taken at random. Work out the probability that both counters are blue. Give a fraction in its simplest form. (2)

  3. Question 3Calculator · 2 marks

    (a) The probability that bus A is late is 0.2. The probability that bus B is late is 0.35. Assume the buses being late are independent events. Work out the probability that neither bus is late. (2)

  4. Question 4Non-calculator · 5 marks

    A bag contains 4 red and 6 blue counters. Two counters are drawn at random without replacement.

    (a) Find the probability that the colours differ. (3)

    (b) How would the probability differ if the first counter were replaced? Find the new probability. (2)

  5. Question 5Non-calculator · 4 marks

    A bag contains 5 red counters and 3 blue counters. Two counters are taken at random, one after the other, without replacement.

    (a) Work out the probability that both counters are the same colour. (4)

  6. Question 6Non-calculator · 3 marks

    The probability that Sam's bus is late is 0.2. The probability that his train is late is 0.1. The two events are independent.

    (a) Work out the probability that exactly one of them is late. (3)

  7. Question 7Non-calculator · 3 marks

    (a) A box contains 5 green tokens and 4 yellow tokens. Three tokens are chosen at random without replacement. Work out the probability that all three are the same colour. Give a fraction in its simplest form. (3)

  8. Question 8Non-calculator · 3 marks

    A box contains 4 green balls and 6 yellow balls. Three balls are taken at random without replacement.

    (a) Work out the probability that at least one of the balls is green. (3)

Worked solutions and marks

Question 1

(a) 0.09

  1. Independent events: multiply. 0.3×0.3=0.090.3 \times 0.3 = 0.09.
  • M1 0.3×0.30.3 \times 0.3.
  • A1 Correct answer: 0.09.

(b) 0.49

  1. P(no rain) =0.7= 0.7 each day: 0.7×0.7=0.490.7 \times 0.7 = 0.49.
  • M1 0.7×0.70.7 \times 0.7.
  • A1 Correct answer: 0.49.

Question 2

(a) 4/49

  1. 2/7×2/72/7\times 2/7
  2. Replacement leaves P(blue) = 2/7 on each draw.
  3. Multiply independent probabilities: 2/7 ×\times 2/7 = 4/49.
  • P1 Establishing 2/7×2/72/7\times 2/7 or an equivalent valid method.
  • A1 Correct answer: 4/49

Question 3

(a) 0.52

  1. 0.8×0.650.8\times 0.65
  2. Use complements: P(A not late) = 0.8 and P(B not late) = 0.65.
  3. For independent events multiply: 0.8 ×\times 0.65 = 0.52.
  • P1 Establishing 0.8×0.650.8\times 0.65 or an equivalent valid method.
  • A1 Correct answer: 0.52

Question 4

(a) 8/158/15

  1. Find the probability of red then blue.
    4/10×6/94/10\times 6/9
  2. Add the separate blue-then-red route.
    4/10×6/9+6/10×4/94/10\times 6/9+6/10\times 4/9
  3. Therefore 8/158/15.
  • M1 Find the probability of red then blue.
  • M1 Add the separate blue-then-red route.
  • A1 Correct answer: 8/158/15

(b) 12/2512/25

  1. With replacement both second-draw denominators remain 10.
    4/10×6/10+6/10×4/104/10\times 6/10+6/10\times 4/10
  2. Therefore 12/2512/25.
  • M1 With replacement both second-draw denominators remain 10.
  • A1 Correct answer: 12/2512/25

Question 5

(a) 1328\frac{13}{28}

  1. Both red.
    58×47=2056\tfrac{5}{8} \times \tfrac{4}{7} = \tfrac{20}{56}
  2. Both blue.
    38×27=656\tfrac{3}{8} \times \tfrac{2}{7} = \tfrac{6}{56}
  3. Add the two ways.
    2056+656=2656=1328\tfrac{20}{56} + \tfrac{6}{56} = \tfrac{26}{56} = \tfrac{13}{28}
  • P1 58×47\frac{5}{8} \times \frac{4}{7} for both red.
  • P1 38×27\frac{3}{8} \times \frac{2}{7} for both blue.
  • P1 Adding the two products.
  • A1 2656\frac{26}{56} or equivalent.

Question 6

(a) 0.26

  1. Bus late, train on time.
    0.2×0.9=0.180.2 \times 0.9 = 0.18
  2. Bus on time, train late.
    0.8×0.1=0.080.8 \times 0.1 = 0.08
  3. 0.18+0.08=0.260.18 + 0.08 = 0.26
  • P1 One correct product, 0.2×0.90.2 \times 0.9 or 0.8×0.10.8 \times 0.1.
  • P1 Both products added.
  • A1 Correct answer: 0.26.

Question 7

(a) 1/6

  1. 5/9×4/8×3/75/9\times 4/8\times 3/7
  2. 4/9×3/8×2/74/9\times 3/8\times 2/7
  3. The disjoint possibilities are three green or three yellow.
  4. P(GGG) = 5/9 ×\times 4/8 ×\times 3/7 = 60/504. P(YYY) = 4/9 ×\times 3/8 ×\times 2/7 = 24/504.
  5. Add the paths: 84/504 = 1/6.
  • P1 Establishing 5/9×4/8×3/75/9\times 4/8\times 3/7 or an equivalent valid method.
  • P1 Establishing 4/9×3/8×2/74/9\times 3/8\times 2/7 or an equivalent valid method.
  • A1 Correct answer: 1/6

Question 8

(a) 56\frac{5}{6}

  1. "At least one green" is everything except "no green", so use the complement.
  2. All three yellow.
    610×59×48=120720=16\tfrac{6}{10} \times \tfrac{5}{9} \times \tfrac{4}{8} = \tfrac{120}{720} = \tfrac{1}{6}
  3. 1−16=561 - \tfrac{1}{6} = \tfrac{5}{6}
  • P1 Using 1−P(no green)1 - P(\text{no green}) (or listing all seven green-containing cases).
  • P1 610×59×48\frac{6}{10} \times \frac{5}{9} \times \frac{4}{8}.
  • A1 56\frac{5}{6} or equivalent.

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Independent and dependent event trees

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