Probability

Tree Diagrams

Year 10 · Year 11

✓Construct tree diagrams to represent sequences of events.
✓Calculate probabilities of combined events using tree diagrams for independent events.
✓Calculate probabilities of combined events using tree diagrams for dependent events (without replacement).
✓Understand the difference between independent and dependent events in the context of tree diagrams.

Key concepts

What is a Tree Diagram?

A tree diagram is a visual tool used in probability to show all possible outcomes of a sequence of events. Each 'branch' represents a possible outcome, and the probability of that outcome is written on the branch. The sum of probabilities on branches stemming from a single point must always be 1. The final outcomes are listed at the end of the 'leaves' of the tree.

Independent Events

Two events are independent if the outcome of the first event does not affect the probability of the second event. For example, flipping a coin and then rolling a die are independent events. When drawing a tree diagram for independent events, the probabilities on the second set of branches (and subsequent sets) remain the same, regardless of the outcome of the first event. To find the probability of a sequence of independent events, you multiply the probabilities along the relevant branches.

P(A and B) = P(A) × P(B)

Dependent Events (Without Replacement)

Two events are dependent if the outcome of the first event *does* affect the probability of the second event. A common scenario for dependent events is 'without replacement', where an item is selected and not returned to the group before the next selection is made. This changes the total number of items and potentially the number of specific items available for the second selection. When drawing a tree diagram for dependent events, the probabilities on the second set of branches (and subsequent sets) change based on the outcome of the preceding event. To find the probability of a sequence of dependent events, you multiply the probabilities along the branches, using the updated probabilities for each subsequent event.

P(A and B) = P(A) × P(B|A) (where P(B|A) is the probability of B occurring given that A has already occurred)

Key facts to remember

1Tree diagrams are visual representations of sequences of events and their probabilities.
2Probabilities are written on the branches, and outcomes are listed at the end of the branches.
3The sum of probabilities on branches originating from a single point must always equal 1.
4To find the probability of a sequence of events, multiply the probabilities along the relevant path of branches.
5For independent events, the probability of the second event is not affected by the first event.
6For dependent events (e.g., 'without replacement'), the probabilities for subsequent events change based on the outcomes of previous events.
7To find the probability of 'either/or' outcomes (e.g., at least one, or one of each), sum the probabilities of all the individual sequences that satisfy the condition.

Worked examples

Example 1

A fair coin is flipped and a fair six-sided die is rolled. a) Draw a tree diagram to represent all possible outcomes. b) Calculate the probability of getting a Head and an even number. c) Calculate the probability of getting a Tail and a number greater than 4.

Ia) First Event (Coin Flip): The possible outcomes are Head (H) or Tail (T). P(H) = 1/2, P(T) = 1/2.

IISecond Event (Die Roll): From each coin outcome, the possible outcomes are 1, 2, 3, 4, 5, 6. P(each number) = 1/6.

IIIThe tree diagram would show two main branches for the coin, and from each of those, six sub-branches for the die. The final outcomes would be (H,1), (H,2), ..., (H,6), (T,1), ..., (T,6).

IVb) To find the probability of a Head and an even number, we identify the paths: (H,2), (H,4), (H,6).

VP(Head) = 1/2.

VIP(Even number) = P(2) + P(4) + P(6) = 1/6 + 1/6 + 1/6 = 3/6 = 1/2.

VIISince these are independent events, P(Head and Even) = P(Head) × P(Even).

VIIIP(Head and Even) = (1/2) × (1/2) = 1/4.

9c) To find the probability of a Tail and a number greater than 4, we identify the paths: (T,5), (T,6).

10P(Tail) = 1/2.

11P(Number > 4) = P(5) + P(6) = 1/6 + 1/6 = 2/6 = 1/3.

12Since these are independent events, P(Tail and Number > 4) = P(Tail) × P(Number > 4).

13P(Tail and Number > 4) = (1/2) × (1/3) = 1/6.

Answer

a) [Tree diagram description: First branches H (1/2) and T (1/2). From H, branches 1,2,3,4,5,6 (1/6 each). From T, branches 1,2,3,4,5,6 (1/6 each).] b) 1/4 c) 1/6

Ensure all branches are labelled with outcomes and probabilities. The sum of probabilities for each set of branches must be 1.

Example 2

A bag contains 5 red counters and 3 blue counters. Two counters are chosen at random without replacement. a) Draw a tree diagram to represent all possible outcomes. b) Calculate the probability that both counters are red. c) Calculate the probability that one counter is red and one is blue.

Ia) Total counters in the bag = 5 (red) + 3 (blue) = 8 counters.

IIFirst Draw:

IIIP(Red first) = 5/8.

IVP(Blue first) = 3/8.

VSecond Draw (after Red first): If a red counter was chosen first, there are now 4 red counters and 3 blue counters left, making a total of 7 counters.

VIP(Red second | Red first) = 4/7.

VIIP(Blue second | Red first) = 3/7.

VIIISecond Draw (after Blue first): If a blue counter was chosen first, there are now 5 red counters and 2 blue counters left, making a total of 7 counters.

9P(Red second | Blue first) = 5/7.

10P(Blue second | Blue first) = 2/7.

11The tree diagram would show branches for the first draw (R, B) and then conditional branches for the second draw from each of those.

12b) To calculate the probability that both counters are red, we follow the path Red then Red (RR).

13P(Red first and Red second) = P(Red first) × P(Red second | Red first).

14P(RR) = (5/8) × (4/7) = 20/56 = 5/14.

15c) To calculate the probability that one counter is red and one is blue, we consider two possible paths: Red then Blue (RB) OR Blue then Red (BR).

16P(Red first and Blue second) = P(Red first) × P(Blue second | Red first).

17P(RB) = (5/8) × (3/7) = 15/56.

18P(Blue first and Red second) = P(Blue first) × P(Red second | Blue first).

19P(BR) = (3/8) × (5/7) = 15/56.

20P(one red and one blue) = P(RB) + P(BR).

21P(one red and one blue) = 15/56 + 15/56 = 30/56 = 15/28.

Answer

a) [Tree diagram description: First branches R (5/8) and B (3/8). From R, branches R (4/7) and B (3/7). From B, branches R (5/7) and B (2/7).] b) 5/14 c) 15/28

Remember to adjust the total number of items and the number of specific items for the second draw when events are 'without replacement'.

Common mistakes

✗Not adjusting probabilities for dependent events (e.g., forgetting to reduce the total number of items and the number of specific items when drawing 'without replacement').
✗Adding probabilities along branches instead of multiplying them to find the probability of a sequence of events.
✗Forgetting to sum the probabilities of all relevant paths when the question asks for a combined outcome (e.g., 'one of each colour' involves two paths).
✗Incorrectly simplifying fractions or making arithmetic errors, especially when adding or multiplying fractions.
✗Not clearly labelling the branches of the tree diagram with both the outcome and its probability.

Exam tips

★Always draw the tree diagram clearly and label all branches with both the outcome (e.g., 'Red') and its probability (e.g., '5/8').
★Double-check that the probabilities on branches stemming from a single point add up to 1; this helps catch errors early.
★Read the question carefully to determine if events are independent ('with replacement') or dependent ('without replacement') as this dictates how probabilities change.
★Show all working steps, especially when multiplying and adding probabilities, as this can earn partial marks even if the final answer is incorrect.

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