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Cambridge IGCSE0607

Probability of combined events

International Mathematics 0607 Chapter Notes

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Probability of combined events
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In this chapter

  1. 1Sample space diagrams
  2. 2Tree diagrams and multiplying along branches
  3. 3Adding the paths for 'or'
  4. 4Venn diagrams and probability
  5. 5Without replacement
  6. 6Same colour, different colours and 'at least one'
  7. 7Choosing from a Venn diagram without replacement

1. Sample space diagrams

A sample space diagram is a table that lists all the outcomes of two events. Each cell is one outcome, and the outcomes are equally likely when the coin, spinner or dice is fair.

Count the cells that fit the event, then divide by the total number of outcomes. A probability can never be greater than 1.

P(event) = (number of favourable outcomes)/(total number of outcomes)

Examiner insight

“Some gave answers greater than 1, which showed a fundamental misunderstanding of probability.”

Worked example 12 marks

A fair coin has sides heads (H) and tails (T). A fair 4-sided spinner is numbered 1, 2, 3 and 4. The coin and the spinner are both spun. The table lists all the outcomes. Find the probability that the outcome is a head and an odd number.

1234
HH1H2H3H4
TT1T2T3T4
  1. 1

    Head and odd number: H1 and H3, so 2 outcomes out of 8 [1]

  2. 2

    2/8 (= 1/4) [1]

2. Tree diagrams and multiplying along branches

On a tree diagram the outcomes are written at the end of the branches and the probabilities by the side of the branches. The two branches from one point cover every possibility, so their probabilities add to 1.

To find the probability of a path, multiply the probabilities along its branches. For independent events, P(A and B) = P(A) × P(B).

With replacement, the probabilities on the second pair of branches are the same as on the first.

P(A and B) = P(A) × P(B)

P(not A) = 1 - P(A)

Examiner insight

“Many incorrectly used addition to combine the probabilities showing they did not understand that, in this situation, the probabilities should be multiplied.” “Candidates did not check that each pair of branches added to 1.”

Worked example 12 marks

In a café, the probability that a customer buys a cup of tea is 1/4. 2 customers in the café are chosen at random. Work out the probability that the 2 customers each buy a cup of tea.

  1. 1

    Tea and tea: 1/4 × 1/4 [1]

  2. 2

    = 1/16 [1]

3. Adding the paths for 'or'

Some events can happen in more than one way. Find the probability of each path, then add them.

The paths must be outcomes that cannot happen together. These are mutually exclusive events, and for them P(A or B) = P(A) + P(B).

So multiply along a path, and add between paths.

P(A or B) = P(A) + P(B)

Worked example 13 marks

The probability that a seed grows is 0.9. Two seeds are planted. Find the probability that exactly one seed grows.

  1. 1

    Grows then does not grow: 0.9 × 0.1 = 0.09 [1]

  2. 2

    Does not grow then grows: 0.1 × 0.9 = 0.09 [1]

  3. 3

    Add the two paths: 0.09 + 0.09 = 0.18 [1]

4. Venn diagrams and probability

In a Venn diagram each region holds a number of items. The probability of a region is its number divided by the total number of items.

In the Extended notation, P(A ∩ B) is the probability of being in both sets and P(A ∪ B) is the probability of being in A or B or both.

Fill in every region first, including the items outside both sets.

P(region) = (number in the region)/(total number)

P(A ∩ B) is both, P(A ∪ B) is either or both

Examiner insight

“The most common errors were to misplace the 1 or to omit all or some of the elements outside the two sets.” “Most shaded the correct area although a few confused intersection and union.”

Worked example 11 mark

There are 24 students in a class. G = {students who study German}. S = {students who study Spanish}. The table gives the number of students in each region of the Venn diagram. One of the 24 students is chosen at random. Find the probability that this student studies German or Spanish but not both.

RegionNumber of students
German only6
German and Spanish9
Spanish only8
Neither1
  1. 1

    German only 6 and Spanish only 8, so 6 + 8 = 14 students

  2. 2

    14/24 [1]

5. Without replacement

If an item is not replaced, the next pick is from a smaller group. The second fraction has a denominator one less than the first.

If the first pick is of one type, the number of that type also drops by 1 for the second pick. The probabilities on the second pair of branches are not the same as on the first.

Still multiply along the path.

7/10 × 6/9 (first red of 7 in 10, then red of 6 in 9)

Examiner insight

“Many candidates, however, used both denominators as 50, not appreciating that the question involved probabilities without replacement.” “The majority of candidates treated this as a question involving probability with replacement.”

Worked example 12 marks

A bag contains 10 discs. 7 discs are red and 3 discs are blue. A disc is picked at random from the bag and not replaced. A second disc is then picked at random from the bag. Find the probability that both discs are red.

  1. 1

    7/10 × 6/9 [1]

  2. 2

    = 42/90 (= 7/15) [1]

6. Same colour, different colours and 'at least one'

Write down every path that fits the event. Two different colours can happen in two orders, so both orders are added.

'At least one' can be found as 1 minus the probability of none.

Check that you have not left out a path.

P(same colour) = P(both blue) + P(both red)

P(at least one) = 1 - P(none)

Examiner insight

“Some candidates omitted the probability of 2 S tiles being chosen.” “Of those who did make an error, the most common was the usual mistake of omitting the reverse pairs.”

Worked example 13 marks

A bag contains 4 blue balls and 3 red balls. Rita takes 2 balls from the bag at random without replacement. Work out the probability that the 2 balls are the same colour.

  1. 1

    Both blue: 4/7 × 3/6 = 12/42 [1]

  2. 2

    Both red: 3/7 × 2/6 = 6/42 [1]

  3. 3

    12/42 + 6/42 = 18/42 (= 3/7) [1]

7. Choosing from a Venn diagram without replacement

A Venn diagram gives the numbers in a group. Picking two items from a region without replacement is then a without replacement tree.

Use the number in the region as the first denominator, and one less as the second. Pick out the two orders when the two items are of different kinds, and add them.

Examiner insight

“This proved the most difficult part of the question, with many candidates not dealing with the two ways of achieving the outcome, and so giving an answer which was half that required.”

Worked example 13 marks

There are 24 students in a class. G = {students who study German}. S = {students who study Spanish}. The table gives the number of students in each region of the Venn diagram. Two of the students who study German are chosen at random. Find the probability that one student studies Spanish and one student does not study Spanish.

RegionNumber of students
German only6
German and Spanish9
Spanish only8
Neither1
  1. 1

    15 students study German: 9 also study Spanish and 6 do not

  2. 2

    Spanish then not: 9/15 × 6/14 and not then Spanish: 6/15 × 9/14 [1]

  3. 3

    Two orders, so 2 × 54/210 [1]

  4. 4

    = 108/210 = 18/35 [1]

Recap

  • Sample space: favourable outcomes ÷ total outcomes.
  • Tree diagram: branches from one point add to 1.
  • Multiply along a path. Add between paths.
  • Venn diagram: number in the region ÷ total number.
  • Extended: without replacement, the second denominator is one less.
  • Extended: different colours come in two orders. At least one = 1 minus none.

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. Do not use a calculator. A fair coin is tossed and a fair 4-sided spinner numbered 1 to 4 is spun. Find the probability of a head and a 3.2 marks
  2. Do not use a calculator. Two fair 4-sided spinners, each numbered 1 to 4, are spun and the scores are added. Find the probability that the total is 5.3 marks
  3. The probability that a bus is late on any day is 0.2. Find the probability that it is late on both Monday and Tuesday.2 marks
  4. A biased coin lands on heads with probability 0.3. The coin is thrown twice. Find the probability that it lands on heads exactly once.3 marks
  5. Do not use a calculator. A bag has 3 red and 7 blue counters. A counter is taken, then replaced, and a second counter is taken. Find the probability that both counters are the same colour.3 marks
  6. Do not use a calculator. In a class of 30 students, 12 study art, 15 study music and 5 study both. One student is chosen at random. Find the probability that the student studies art or music or both.3 marks
  7. Do not use a calculator. A bag has 5 red and 3 green sweets. Two sweets are taken at random without replacement. Find the probability that both are red.2 marks
  8. Do not use a calculator. A drawer holds 5 black socks and 3 white socks. Two socks are taken at random without replacement. Find the probability that at least one sock is white.3 marks
  9. The 8 letters of the word PARALLEL are each written on a separate card. Three cards are taken at random without replacement. Find the probability that the three cards can be arranged to spell the word ALL. Give your answer as a fraction in its simplest form.4 marks
  10. A year group has 30 students. 18 study physics, 15 study chemistry and 4 study neither subject. Two of the students who study physics are chosen at random. Find the probability that exactly one of them also studies chemistry.5 marks

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