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Probability of Combined Events · Form 4

Probability of Combined Events: Worked Examples (Medium)

Combines techniques over two or three steps: the addition rule with overlap, independent events by multiplication, and dependent draws without replacement. For students ready to link ideas.

Written by the spmmath.com.my editorial team.

Worked example 1

A number is chosen at random from the whole numbers 1 to 20. Find the probability that it is a multiple of 3 or a multiple of 5.

  1. Multiples of 3 from 1–20: {3, 6, 9, 12, 15, 18} → 6 numbers.
  2. Multiples of 5 from 1–20: {5, 10, 15, 20} → 4 numbers.
  3. Common to both (multiple of 15): {15} → 1 number, counted twice.
  4. P(3 or 5) = P(3) + P(5) − P(both) = 6/20 + 4/20 − 1/20 = 9/20.

Worked example 2

A game spinner lands on 'Win' with probability 0.3. It is spun twice, and the results are independent.

Find the probability of winning exactly once.

  1. P(Win) = 0.3, so P(Lose) = 1 − 0.3 = 0.7.
  2. Winning exactly once happens two ways: (Win, Lose) or (Lose, Win).
  3. P = (0.3 × 0.7) + (0.7 × 0.3) = 0.21 + 0.21.
  4. = 0.42.

Worked example 3

A box has 5 red and 3 yellow pens. Two pens are taken one after another without replacement.

Find the probability that both are red.

  1. Total pens = 5 + 3 = 8, so P(first red) = 5/8.
  2. One red is now gone: 4 red left out of 7 pens, so P(second red) = 4/7.
  3. Dependent events, multiply along the branch: 5/8 × 4/7 = 20/56.
  4. Simplify: 20/56 = 5/14.

Worked example 4

A number is chosen at random from the integers 1 to 30. Find the probability that the number is a multiple of 4 or a multiple of 6.

  1. Multiples of 4: 4, 8, 12, 16, 20, 24, 28 → 7 numbers.
  2. Multiples of 6: 6, 12, 18, 24, 30 → 5 numbers.
  3. Multiples of both 4 and 6 (i.e. of 12): 12, 24 → 2 numbers.
  4. P(4 or 6) = (7 + 5 − 2) ÷ 30 = 10/30 = 1/3.

Worked example 5

A fair coin is tossed and a fair die is rolled. Find the probability of obtaining a head on the coin and a number greater than 4 on the die.

  1. The two events are independent.
  2. P(head) = 1/2.
  3. P(number greater than 4) = P(5 or 6) = 2/6 = 1/3.
  4. P(head and >4) = 1/2 × 1/3 = 1/6.

Worked example 6

A box contains 4 green balls and 6 blue balls. Two balls are drawn one after another without replacement.

Find the probability that both balls are the same colour.

  1. Total balls = 10.
  2. P(both green) = 4/10 × 3/9 = 12/90.
  3. P(both blue) = 6/10 × 5/9 = 30/90.
  4. P(same colour) = 12/90 + 30/90 = 42/90 = 7/15.

Worked example 7

A factory has two independent alarm systems protecting a warehouse. Alarm A works when triggered with probability 0.9, and Alarm B works when triggered with probability 0.85.

Find the probability that at least one alarm works when triggered.

  1. P(at least one works) = 1 − P(both fail)
  2. P(A fails) = 1 − 0.9 = 0.1; P(B fails) = 1 − 0.85 = 0.15
  3. P(both fail) = 0.1 × 0.15 = 0.015
  4. P(at least one works) = 1 − 0.015 = 0.985

Worked example 8

A drawer has 9 black socks and 3 white socks. Two socks are taken out at random, one after another, without replacement.

Find the probability that the two socks are of different colours.

  1. Different colours means Black-then-White, or White-then-Black
  2. P(B then W) = (9/12) × (3/11) = 27/132
  3. P(W then B) = (3/12) × (9/11) = 27/132
  4. P(different colours) = 27/132 + 27/132 = 54/132 = 9/22

Source:DSKP KSSM Mathematics Form 4 and 5 (Versi English)

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Frequently asked questions

What kind of questions come up at the medium level for combined events?

Medium questions usually involve tree diagrams for two events without replacement, for example, drawing two balls or two cards one after another from the same box without putting the first one back. You need to adjust the probabilities on the second branch because the total number of items has changed.

How should I set out a probability tree diagram to get full marks?

Label every branch with its probability, and make sure branches from the same point add up to 1. Write the combined probability for each full path by multiplying along the branches, then add the paths that match the event asked.

Showing this working clearly is what earns method marks, not just the final number.

What's the most common error in "without replacement" probability questions?

The most common error is using the same probability for both draws, forgetting that removing an item changes the total and possibly the count of that type. Always recheck the denominator (and numerator, if the same type was drawn) before writing the second branch's probability.

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