Year 10 Extended · Unit 4 Sets, Venn diagrams and conditional probability · Lesson 5 of 6

What does it change to know that something has happened?

Knowing that one event has happened changes the sample space. This lesson reads conditional probabilities from a table, a Venn diagram and a tree, and asks what happens when the condition is turned round.

Statistics and ProbabilityNo calculator

Warm-up

Three quick ones from earlier work. Try each in your head or on paper before you open the answer.

  1. From lesson 4: A and B are independent with P(A) = 0.2 and P(B) = 0.5. Find P(A ∩ B).
    Answer0.1. 0.2 × 0.5.
  2. From Year 10, Standard: a bag has 3 red and 2 blue counters. One is taken and not put back. How many counters are left?
    Answer4. One counter has been removed, so 5 − 1 = 4 remain.
  3. From Year 7: write 18 out of 40 as a decimal.
    Answer0.45. 18 ÷ 40.
Words used in this lesson
conditional probability
the probability of an event when another has happened. For example: the chance a bus is late given that it is raining.
given
the event we know has happened, written after the bar |. For example: in P(B | A) the given event is A.
reduced sample space
the outcomes that remain once the given event is known. For example: only the walkers, when we know the student walks.
multiplication rule
P(A ∩ B) = P(A) × P(B | A). For example: multiplying along a route of a tree.
without replacement
the item taken is not put back, so the next probabilities change. For example: taking two cards from a pack.
two-way table
a table of counts for two categories. For example: walk or not against glasses or not.

Same students, different chance

A school in Kuala Lumpur surveys 200 students. Group A is the students who play a team sport. Group B is the students who do more than one hour of homework on a typical night. The survey teacher says: “Pick any student at random and there is a certain chance that they do more than an hour. Now pick a student who plays a team sport.”

Predict first: will the chance of more than an hour of homework be the same for a team sport player as for a student chosen from everyone, or different? Then test it in the figure, and look at what happens when the question is turned round: the chance of team sport for a student who does more than an hour.

P(B | A)
48/100count over the group
0.48value
0.4without the condition

P(B | A): the group given is A, which has 100 students, and 48 of them are in B. So P(B | A) = 48/100 = 0.48, against P(B) = 0.4 for all 200.

Show me the steps
  1. P(B | A), the opening state: the group given is A, with 48 + 52 = 100 students, and 48 of them are in B. P(B | A) = 48/100 = 0.48, which is more than P(B) = 80/200 = 0.4.
  2. Slide to P(B | not A): the group is not A, with 32 + 68 = 100 students, and 32 are in B. P(B | not A) = 32/100 = 0.32, less than 0.4. Being in A raises the chance of B.
  3. Slide to P(A | B): the group is now B, with 48 + 32 = 80 students, and 48 are in A. P(A | B) = 48/80 = 0.6. The same 48 students, but a different group to divide by, so a different answer from P(B | A).
  4. Slide to P(A | not B): the group is not B, with 52 + 68 = 120 students, and 52 are in A. P(A | not B) = 52/120 ≈ 0.43, less than P(A) = 100/200 = 0.5.

Conditional probability. P(B | A), read “the probability of B given A”, is the chance of B when we know A has happened. The sample space shrinks to A, so P(B | A) = P(A ∩ B) ÷ P(A), or from counts, n(A ∩ B) ÷ n(A).

Multiplication rule. P(A ∩ B) = P(A) × P(B | A). On a tree diagram the second-stage branches are these conditional probabilities, so without replacement they change with the first pick.

P(B | A) and P(A | B) usually differ, because the groups we divide by are different. A and B are independent exactly when P(B | A) = P(B).

P(A | B) is not P(B | A). In a group of 100 students, 10 are in the choir (C) and 40 are in Year 10 (Y), and 8 are in both. P(C | Y) = 8/40 = 0.2, but P(Y | C) = 8/10 = 0.8. The same 8 students, divided by a different group.

Worked examples

1. Of 120 students, 40 walk to school, and 18 of the walkers wear glasses. Find P(glasses | walks).

  1. The group given is the walkers: 40 students, not 120.
  2. 18 of them wear glasses: 18 ÷ 40 = 0.45.

2. A box holds 6 green and 4 purple counters. Two are taken without replacement. Find the probability that both are purple.

  1. First counter purple: P = 4/10.
  2. Given that, 3 purple remain among 9: P(second purple | first purple) = 3/9.
  3. P(both purple) = 4/10 × 3/9 = 12/90 = 2/15.

3. n(U) = 60, n(A) = 24, n(B) = 30 and n(A ∩ B) = 9. Find P(A | B) and P(B | A).

  1. On a Venn diagram the regions are A only 24 − 9 = 15, both 9, B only 30 − 9 = 21 and neither 60 − 45 = 15. Given B, only the B circle (21 + 9 = 30) counts; given A, only the A circle (15 + 9 = 24).
  2. Given B, the space is B with 30 elements, and 9 are in A: P(A | B) = 9/30 = 3/10.
  3. Given A, the space is A with 24 elements: P(B | A) = 9/24 = 3/8.

Your go. The first step is given. P(rain) = 0.3. If it rains the chance a bus is late is 0.5, and if it does not rain the chance is 0.1. Find P(late).

  1. Rain and late: P(rain ∩ late) = 0.3 × 0.5 = 0.15.
  2. Draw the tree: rain 0.3 and no rain 0.7, then late or on time on each branch. Find the no-rain-and-late route, and add the two routes that end in late.
Check the answer

P(no rain ∩ late) = 0.7 × 0.1 = 0.07, so P(late) = 0.15 + 0.07 = 0.22.

Check before you go on

1. P(A) = 0.4 and P(A ∩ B) = 0.3. What is P(B | A)?

Your turn

Start at Level 1 and go as far as you can. Type negative numbers with a minus sign, for example -7. A wrong answer gets a reason, not just a cross: read it, then try again.

Level 1 · Getting started

2. Of 150 hikers, 60 carry a camera, and 21 of the camera carriers also carry binoculars. A camera carrier is chosen at random. What is the probability, as a decimal, that this person carries binoculars?

3. A bag has 6 red, 3 blue and 2 green counters. One is taken and not replaced; it is red. What is the probability that the next counter is blue? Give a decimal.

Level 2 · Building

4. P(A) = 0.8 and P(B | A) = 0.45. Find P(A ∩ B).

5. P(A ∩ B) = 0.15 and P(B) = 0.6. Find P(A | B).

Level 3 · Stretch

6. n(U) = 200, n(A) = 100, n(B) = 40 and n(A ∩ B) = 12. A student outside B is chosen at random. Find P(A | B′) as a decimal.

7. A bag has 4 red counters and n blue counters. Two are taken without replacement, and the probability that both are red is 1/6. Find n.

Something to think about

In the table, P(B | A) is 0.48 and P(B) is 0.4. Show that P(B) is a weighted average of P(B | A) and P(B | not A), and say what the weights are. What would have to be true of the table for P(B | A) to equal P(B)?

One way to think about it

P(B) = P(A) × P(B | A) + P(not A) × P(B | not A) = 0.5 × 0.48 + 0.5 × 0.32 = 0.4, with weights equal to the shares of A and not A. P(B | A) = P(B) only if P(B | not A) = P(B) as well, because the average cannot sit away from both. That is the table where the share in B is the same inside every row, which is independence.

Exit ticket

8. P(A) = 0.15, P(B) = 0.45 and P(A ∩ B) = 0.09. Which pair of values is correct?