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Cambridge IGCSE Mathematics — 0580 Extended

Topic 8.3: Probability — Combined Events

Two experiments at once — two coins, a coin and a die, two dice — need a sample space: a complete list (or grid) of equally likely outcomes. Count the matching cells, then divide by the total. Extended papers also distinguish with replacement from without replacement.

Listing a sample space

Write every equally likely outcome once. Two coins give four, not three: HT and TH are different.

Two fair coins are flipped. Find \(\mathrm{P}(\text{two heads})\) and \(\mathrm{P}(\text{one of each})\).

Four coin-pair outcomes HH, HT, TH and TT
\(\mathrm{P}(\text{HH}) = 1/4\). One of each is HT or TH, so \(2/4 = 1/2\).

A fair coin and a fair die are used together. Find \(\mathrm{P}(\text{heads and even})\).

Twelve outcomes of a coin and a die, with H2, H4 and H6 as heads and even
Twelve equally likely pairs. H2, H4 and H6 are the three favourable, so \(3/12 = 1/4\).

Two dice

A \(6 \times 6\) grid has 36 equally likely pairs. The outcomes are the pairs, not the sums. “A total of 7” is six cells, not one.

Two fair dice are rolled. Find \(\mathrm{P}(\text{sum} = 7)\).

6 by 6 sample space with the six pairs that sum to 7 highlighted, giving 6/36 = 1/6
Six highlighted pairs. \(\mathrm{P}(\text{sum} = 7) = 6/36 = 1/6\).

AND and OR

AND means both: count the overlap of the two lists. OR means either: join the lists, but do not count a cell twice. Adding two probabilities that share outcomes double-counts the overlap.

On two dice, contrast doubles with an even sum.

Two-dice grid with doubles in amber and other even sums in blue, showing that adding the two probabilities would double-count doubles
Doubles sit inside the even-sum cells. \(\mathrm{P}(\text{double or even sum})\) is just \(\mathrm{P}(\text{even sum})\), not the sum of the two probabilities.

With replacement or without

If the first counter is put back, the bag is unchanged and the second probabilities stay the same. If it is not put back, the denominator (and often the numerator) changes. Trees in the next topic make this automatic; the bag picture is the reason.

Two bags: with replacement still 3 red and 2 blue; without replacement, one red removed so 2 red and 2 blue remain
With replacement, \(\mathrm{P}(\text{2nd red}) = 3/5\). Without, after a red is taken, it is \(2/4\).

Paper 2 (non-calculator)

Keep 36 in the denominator until the last line, then cancel: \(6/36 = 1/6\). Do not write \(0.166\ldots\) on Paper 2.

Try this

Two fair dice are rolled. Find \(\mathrm{P}(\text{both the same})\).

Show answer
Answer

  1. Doubles: (1,1), (2,2), …, (6,6) — six pairs out of 36

    \[ \mathrm{P}(\text{both the same}) = \dfrac{6}{36} = \dfrac{1}{6} \]

Exam Traps

  • HT and TH are different outcomes. Collapsing them into “one head and one tail” before you count will make \(\mathrm{P}(\text{HH})\) look like \(1/3\).

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