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Probability of Combined Events · Form 4

Probability of Combined Events: Worked Examples (KBAT)

Real-situation problems where the technique is hidden: reading a survey with overlap, a two-draw prize game without replacement, and two independent traffic lights solved by the complement. Trains students to choose the right tool.

Written by the spmmath.com.my editorial team.

Worked example 1

In a canteen survey of 200 students, 90 buy nasi lemak, 70 buy mee goreng, and 25 buy both. A student is chosen at random.

Find the probability that the student buys neither dish.

  1. Students buying at least one dish = 90 + 70 − 25 = 135 (remove the 25 counted in both).
  2. Students buying neither = 200 − 135 = 65.
  3. P(neither) = 65/200.
  4. Simplify: 65/200 = 13/40.

Worked example 2

A prize box holds 12 tokens: 4 give a drink voucher and 8 are blank. Aina draws one token and keeps it, then her friend draws one.

Find the probability that exactly one of the two draws is a voucher.

  1. Voucher tokens = 4, blank = 8, total = 12.
  2. Path Voucher then Blank: 4/12 × 8/11 = 32/132.
  3. Path Blank then Voucher: 8/12 × 4/11 = 32/132.
  4. Exactly one voucher = 32/132 + 32/132 = 64/132 = 16/33.

Worked example 3

On a school morning a student passes two independent traffic lights. The light at Junction A is green with probability 0.6, and the light at Junction B is green with probability 0.5.

She reaches class on time if at least one light is green. Find the probability that she is on time.

  1. P(A red) = 1 − 0.6 = 0.4; P(B red) = 1 − 0.5 = 0.5.
  2. The lights are independent, so P(both red) = 0.4 × 0.5 = 0.2.
  3. 'On time' means not both red, use the complement.
  4. P(on time) = 1 − 0.2 = 0.8.

Worked example 4

In a group of 120 tourists, 75 visited the beach, 50 visited the museum and 20 visited both places. A tourist is chosen at random.

Find the probability that the tourist visited exactly one of the two places.

  1. Visited beach only = 75 − 20 = 55.
  2. Visited museum only = 50 − 20 = 30.
  3. Visited exactly one place = 55 + 30 = 85.
  4. P(exactly one) = 85/120 = 17/24.

Worked example 5

A box contains 5 good light bulbs and 3 defective ones. Two bulbs are chosen at random without replacement.

Find the probability that at least one of the chosen bulbs is defective.

  1. Use the complement: P(at least one defective) = 1 − P(none defective).
  2. P(both good) = 5/8 × 4/7 = 20/56 = 5/14.
  3. P(at least one defective) = 1 − 5/14.
  4. P(at least one defective) = 9/14.

Worked example 6

A student answers two multiple-choice questions purely by guessing. The first question has 4 options and the second has 5 options, each with one correct answer.

Find the probability that the student gets exactly one of the two questions correct.

  1. P(correct Q1) = 1/4, P(wrong Q1) = 3/4; P(correct Q2) = 1/5, P(wrong Q2) = 4/5.
  2. Exactly one correct = (correct Q1 and wrong Q2) or (wrong Q1 and correct Q2).
  3. = (1/4 × 4/5) + (3/4 × 1/5) = 4/20 + 3/20.
  4. P(exactly one correct) = 7/20.

Worked example 7

A factory has two machines. Machine A produces 60% of the day's items, of which 5% are defective.

Machine B produces the remaining 40% of the day's items, of which 8% are defective. An item is chosen at random from the day's production.

Find the probability that it is defective.

  1. Model as a tree: first branch = which machine made it, second branch = defective or not
  2. P(from A and defective) = 0.6 × 0.05 = 0.03
  3. P(from B and defective) = 0.4 × 0.08 = 0.032
  4. P(defective) = P(from A and defective) + P(from B and defective) = 0.03 + 0.032 = 0.062

Worked example 8

A security system has three independent sensors protecting a store. Sensor 1 detects an intrusion with probability 0.9, Sensor 2 with probability 0.8, and Sensor 3 with probability 0.7.

Find the probability that exactly two of the three sensors detect an intrusion.

  1. Exactly two detecting means one of the three sensors fails to detect; consider each case
  2. Sensors 1 and 2 detect, Sensor 3 fails: 0.9 × 0.8 × 0.3 = 0.216
  3. Sensors 1 and 3 detect, Sensor 2 fails: 0.9 × 0.2 × 0.7 = 0.126
  4. Sensors 2 and 3 detect, Sensor 1 fails: 0.1 × 0.8 × 0.7 = 0.056
  5. P(exactly two detect) = 0.216 + 0.126 + 0.056 = 0.398

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

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Frequently asked questions

What do KBAT questions on combined events usually look like?

KBAT questions wrap combined-event probability inside a real-life scenario, like a game, a quality-control check, or a school activity, where you must first figure out which events are involved and how they relate (independent, mutually exclusive, or with/without replacement) before you can even choose a method.

What does the examiner reward most in a probability KBAT answer?

Marks go to correctly identifying the type of events and setting up an organised approach, a tree diagram, list, or systematic calculation, before computing. An answer that jumps straight to a formula without showing this reasoning, even if the final number is right, often loses method marks.

What trips students up most in KBAT probability questions?

Students often misread the scenario and apply the wrong rule, for example, treating a "without replacement" situation as independent, or missing that an event described in words is actually mutually exclusive. Read the scenario twice and identify the event relationship before writing any working.

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