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Cambridge IGCSE0580

Probability of combined events

Mathematics 0580 Chapter Notes

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Probability of combined events
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In this chapter

  1. 1Sample space diagrams
  2. 2Venn diagrams with two sets
  3. 3Tree diagrams, with replacement
  4. 4Without replacement
  5. 5Several ways to reach one outcome

1. Sample space diagrams

A sample space diagram is a table that lists every possible outcome of two events, such as two dice. Each cell is one outcome.

For fair dice every cell is equally likely. Count the cells that match the event, then divide by the total number of cells.

An event that cannot happen has probability 0. An event that must happen has probability 1.

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

Examiner insight

“Very few attempted a probability space diagram which would have simplified the problem and helped identify the required outcomes.”

Worked example 13 marks

Two fair 4-sided dice have faces numbered 1 to 4. The dice are thrown, and the numbers are multiplied to give a total. The table shows all the possible results. Find the probability the total is

(i) 12

(ii) 5

(iii) a number that is less than 20.

×1234
11234
22468
336912
4481216
  1. 1

    (i) 12 appears twice, so P = 2/16 = 1/8 [1]

  2. 2

    (ii) 5 does not appear: P = 0 [1]

  3. 3

    (iii) all 16 totals are less than 20: P = 1 [1]

2. Venn diagrams with two sets

In a Venn diagram with two sets, fill the overlap first. Then take the overlap away from each set total to find the part of each circle that is in that set only.

The numbers outside the circles go in the box. The numbers in all the regions add up to the total in the group.

The probability of a region is its number divided by the total. Extended papers may write the probability of being in both sets as P(A ∩ B), and the probability of being in A or B or both as P(A ∪ B).

Examiner insight

“Most candidates put 15 and 12 in the E and S circle and left the intersection blank – not recognising that some of the students who speak English also speak Spanish and vice versa.” “Understanding the notation n(EUS) was challenging for all candidates except the most able. Few candidates recognised this was the number of students in the circles (10 + 5 + 7 = 22) with most confusing it for the number of students in the intersection (5).”

Worked example 14 marks

In a class of 30 students, 12 play football (F) and 9 play tennis (T). 4 students play both. A student is chosen at random. Find the probability that the student plays

(i) both sports

(ii) football or tennis or both

(iii) neither sport.

  1. 1

    overlap = 4; football only = 12 - 4 = 8; tennis only = 9 - 4 = 5 [1]

  2. 2

    (i) both: P(F ∩ T) = 4/30 = 2/15 [1]

  3. 3

    (ii) football or tennis or both: P(F ∪ T) = (8 + 4 + 5)/30 = 17/30 [1]

  4. 4

    (iii) neither = 30 - 17 = 13, so P(neither) = 13/30 [1]

3. Tree diagrams, with replacement

A tree diagram shows two or more events in turn. Write the probability by the side of each branch and the outcome at the end of the branch.

The probabilities on each set of branches add up to 1. When an item is replaced, the second set of branches is the same as the first.

Multiply along the branches for "and". Add the outcomes you want for "or".

P(A then B) = P(A) × P(B)

Examiner insight

“Some candidates also put different probabilities on the second set of branches, despite the counters being replaced after the colour noted.” “Of those who did attempt this part, the majority did not use multiplication along the branches but added instead.”

Worked example 13 marks

Marwa has these 10 numbered tiles: 1, 2, 2, 3, 3, 4, 4, 4, 4, 4. Marwa picks one of the tiles at random. She replaces the tile and then picks a second tile at random. Find the probability that the two tiles have the same even number.

  1. 1

    both 2: 2/10 × 2/10 = 4/100 [1]

  2. 2

    both 4: 5/10 × 5/10 = 25/100 [1]

  3. 3

    add the two outcomes: 4/100 + 25/100 = 29/100 [1]

4. Without replacement

If the first item is not replaced, there is one fewer item for the second pick. The denominator goes down by 1 on the second branches.

The number of the colour that was picked also goes down by 1.

The probabilities on each set of branches still add up to 1. Use this to check.

Examiner insight

“Only the weaker candidates struggled to make any progress with some attempting the question with replacement.” “Some candidates incorrectly combined probabilities, multiplying three or more fractions or adding where multiplication was appropriate (and vice versa).”

Worked example 14 marks

Juan has a bag containing 2 red counters, 3 green counters and 5 blue counters. Juan picks two counters from the bag at random, without replacement. Find the probability that the two counters are different colours.

  1. 1

    method: P(different) = 1 - P(same colour), with 10 counters then 9

  2. 2

    P(RR) = 2/10 × 1/9 = 2/90 [1]

  3. 3

    P(GG) = 3/10 × 2/9 = 6/90 and P(BB) = 5/10 × 4/9 = 20/90 [1]

  4. 4

    P(same) = (2 + 6 + 20)/90 = 28/90 [1]

  5. 5

    P(different) = 1 - 28/90 = 62/90 = 31/45 [1]

5. Several ways to reach one outcome

One outcome can often happen in more than one order. Red then blue and blue then red are different branches. Find each one and add them.

For three picks, count every order. Two 4s and a 3 can come as 4, 4, 3 or 4, 3, 4 or 3, 4, 4.

When there are many ways, it can be quicker to find the probability of the outcomes you do not want and take it from 1.

Examiner insight

“Finding the probability of only one of the two combinations, almost always red then blue, was the most common error.” “A common error was to find the probability of at least one yellow counter being picked, including the case of yellow then yellow.”

Worked example 13 marks

From the 10 numbered tiles (1, 2, 2, 3, 3, 4, 4, 4, 4, 4), Marwa now picks three tiles at random without replacement. Find the probability that the sum of the numbers on the three tiles is 11.

  1. 1

    the only way to make 11 is 4, 4, 3, in any order [1]

  2. 2

    one order: 5/10 × 4/9 × 2/8 = 40/720 = 1/18 [1]

  3. 3

    three orders: 3 × 1/18 = 1/6 [1]

Recap

  • Probability = successful outcomes ÷ total outcomes.
  • A sample space diagram lists every outcome. Count cells.
  • Venn diagram: fill the overlap first. The regions add up to the total.
  • Tree diagram: multiply along branches, add between branches.
  • With replacement, the second branches repeat the first.
  • Extended: without replacement, the denominator goes down by 1.
  • Extended: add every order that gives the outcome, or use 1 minus the rest.
  • Extended: P(A ∩ B) is in both sets. P(A ∪ B) is in A or B or both.

End-of-chapter exercise

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

  1. Do not use a calculator. Two fair 6-sided dice are thrown and the scores are added. Find the probability that the total is 7.2 marks
  2. Do not use a calculator. Spinner A has the numbers 1, 2 and 3. Spinner B has the numbers 1, 2, 3 and 4. Both spinners are fair. They are spun and the numbers are multiplied. Find the probability that the product is even.3 marks
  3. In a group of 40 students, 18 study French and 15 study German. 6 students study both languages. A student is chosen at random. Find the probability that the student studies neither language.3 marks
  4. Do not use a calculator. A bag contains 3 red counters and 7 blue counters. Ravi picks a counter at random, replaces it, then picks a second counter at random. Find the probability that both counters are red.2 marks
  5. A fair coin is tossed and a fair 6-sided dice is thrown. Find the probability of getting a head and a number greater than 4.2 marks
  6. A survey of 50 people finds that 30 own a bike and 20 own a scooter. 8 people own both. One person is chosen at random. Find the probability that the person owns exactly one of the two.3 marks
  7. Do not use a calculator. A bag contains 4 green sweets and 6 yellow sweets. Two sweets are taken at random without replacement. Find the probability that both are green.2 marks
  8. A box holds 4 red pens, 3 white pens and 2 black pens. Two pens are taken at random without replacement. Find the probability that the two pens are the same colour.4 marks
  9. Do not use a calculator. A choir has 24 singers. 14 can read music (M) and 9 can play the piano (P). 5 singers can do neither. (a) Find P(M ∩ P) for a singer chosen at random. (b) Two different singers are chosen at random. Find the probability that exactly one of them can both read music and play the piano.5 marks
  10. A box holds 9 batteries. 4 are new, 3 are part used and 2 are dead. Three batteries are taken at random, without replacement, to go in a torch. Find the probability that at least two of the three batteries are of the same kind.5 marks

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