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Year 11

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Conditional probability in a tree diagram

I can calculate a conditional probability from a tree diagram.

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New

Year 11

Higher

Conditional probability in a tree diagram

I can calculate a conditional probability from a tree diagram.

Download all resources

Share activities with pupils

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Lesson details

Key learning points

A tree diagram can be used to calculate conditional probabilities
A tree diagram can be used to calculate conditional probabilities involving sets of events
Tree diagrams allow us to identify which events are mutually exclusive
Tree diagrams can be used to identify which events are independent

Keywords

Probability - The probability that an event will occur is the proportion of times the event is expected to happen in a suitably large experiment.
Probability tree - Each branch of a probability tree shows a possible outcome from an event or from a stage of a trial, along with the probability of that outcome happening.
Independent events - Event A is independent of event B if the probability of event A occurring is not affected by whether or not event B occurs.
Mutually exclusive - Mutually exclusive events have no outcomes in common.

Common misconception

When pupils construct probability trees for scenarios where two marbles are removed from a bag, they may mistakenly subtract 2 from the denominators of the probabilities instead of 1.

Even though two marbles are removed from the bag at once, you can think of it at choosing one marble and then immediately choosing a second marble. At the end of the process, two marbles have been removed.

Some of the examples, from this lesson, could be enacted with objects (e.g. removing two marbles from a bag).

Teacher tip

Licence

This content is © Oak National Academy Limited (2024), licensed on Open Government Licence version 3.0 except where otherwise stated. See Oak's terms & conditions (Collection 2).

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Starter quiz

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6 Questions

Q1.

Event A and event B are mutually exclusive events. The P(A and B) is .

Correct Answer: 0, zero

Q2.

From the frequency Venn diagram, P(A | B) = . Write your answer as a decimal.

Correct Answer: 0.3

Q3.

From the frequency Venn diagram, P(A | B') = . Write your answer as a decimal.

Correct Answer: 0.3

Q4.

The diagram shows the number of outcomes for two events A and B. The events A and B are __________ events.

exhaustive

Correct answer: independent

mutually exclusive

Q5.

Here is an incomplete frequency Venn diagram. P(C) = $$1\over3$$, P(D') = $$1\over6$$, P(C ∩ D) = $$1 \over 4$$ and P(C | D) = $$3 \over 10$$. The value of $$x$$ is .

Correct Answer: 15

Q6.

Here is an incomplete frequency Venn diagram. P(C) = $$1\over3$$, P(D') = $$1\over6$$, P(C ∩ D) = $$1 \over 4$$ and P(C | D) = $$3 \over 10$$. The value of $$y$$ is .

Correct Answer: 5

Exit quiz

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6 Questions

Q1.

Two events A and B are independent. This means that...

P(A ∩ B) = P(A)

P(A ∪ B) = P(A)

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

Correct answer: P(A | B) = P(A)

Correct answer: P(A | B') = P(A)

Q2.

Sofia has to catch a bus and then a train to get to school. When the bus is late, Sofia is more likely to miss the train and so be late for school. Match each value to its probability.

Correct Answer:$$x$$,0.2

0.2

Correct Answer:$$y$$,0.1

0.1

Correct Answer:$$z$$,0.7

0.7

Q3.

Sofia has to catch a bus and then a train to get to school. When the bus is late, Sofia is more likely to miss the train and so be late for school. P(Sofia is late for school) = .

Correct Answer: 0.14

Q4.

Here is a general probability tree. Which of these probabilities give the probability of the highlighted events?

Correct answer: P(A' ∩ C)

P(A' ∪ C)

P(A') × P(C | A)

Correct answer: P(A') × P(C | A')

P(A') × P(A' | C)

Q5.

One marble is taken at random from a bag of 7 purple and 5 green marbles. The marble is not replaced. A second marble is then taken. The tree diagram shows these two events. Find the value of $$x$$.

$$7 \over 12$$

$$5 \over 12$$

$$7 \over 11$$

Correct answer: $$6 \over 11$$

$$5 \over 11$$

Q6.

One marble is taken at random from a bag of purple and green marbles. The marble is not replaced. A second marble is then taken. The tree diagram shows the two events. Find P(both marbles are green).

$$\frac{25}{132}$$

$$\frac{25}{144}$$

$$\frac{103}{132}$$

Correct answer: $$\frac{5}{33}$$

$$\frac{4}{33}$$