Cambridge Lower Secondary CheckpointStage 9

Statistics and Probability (9Sp)

Mathematics Stage 9 Chapter Notes

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Statistics and Probability - Probability
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In this chapter

  1. 1Mutually Exclusive Events
  2. 2Experimental Probability & Relative Frequency
  3. 3Calculating Expected Frequency
  4. 4Probability of Independent Events
  5. 5Using Tree Diagrams for Combined Events

1. Mutually Exclusive Events

In probability, events are 'mutually exclusive' if they cannot happen at the same time. For example, when you roll a standard six-sided dice once, you can get a 3 or a 4, but you can't get both on the same roll. The outcomes 'rolling a 3' and 'rolling a 4' are mutually exclusive. A crucial rule is that for a set of mutually exclusive events that covers all possible outcomes, their probabilities must add up to 1. This is often used to find a missing probability.

For mutually exclusive events A and B: P(A or B) = P(A) + P(B)

For a complete set of mutually exclusive events: ΣP(all outcomes) = 1

Key term

Mutually Exclusive: Events that cannot happen at the same time.

Examiner insight

Examiners expect you to show that the sum of probabilities equals 1 when solving for an unknown probability in a complete set of outcomes. Always write the equation (e.g., P(A) + P(B) + ... = 1) to secure method marks.

Common pitfall

Assuming events are mutually exclusive when they are not. For example, in a deck of cards, 'drawing a King' and 'drawing a Heart' are not mutually exclusive because you can draw the King of Hearts.

Worked example 12 marks

A spinner has only blue, red and yellow sectors. The probability of the spinner landing on a particular colour is shown in the table. What is the value of x?

Colourblueredyellow
Probability3/104/10x
  1. 1

    The events 'landing on blue', 'landing on red', and 'landing on yellow' are mutually exclusive and cover all possible outcomes.

  2. 2

    Therefore, the sum of their probabilities must be 1.

  3. 3

    P(blue) + P(red) + P(yellow) = 1

  4. 4

    3/10 + 4/10 + x = 1

  5. 5

    7/10 + x = 1

  6. 6

    To find x, subtract 7/10 from 1: x = 1 - 7/10

  7. 7

    x = 10/10 - 7/10 = 3/10

Worked example 23 marks

The train Jana takes to work on Monday will either be early, on time, late or cancelled. The table shows the probability of each of these events. What is the value of x?

StatusEarlyOn timeLateCancelled
Probabilityx0.65x0.05
  1. 1

    The four outcomes are mutually exclusive and cover all possibilities, so their probabilities sum to 1.

  2. 2

    P(Early) + P(On time) + P(Late) + P(Cancelled) = 1

  3. 3

    x + 0.65 + x + 0.05 = 1

  4. 4

    Combine the 'x' terms and the number terms: 2x + 0.70 = 1

  5. 5

    Subtract 0.70 from both sides: 2x = 1 - 0.70

  6. 6

    2x = 0.30

  7. 7

    Divide by 2: x = 0.15

Recap

  • Mutually exclusive events cannot both happen in the same trial.
  • The probability of either of two mutually exclusive events occurring is the sum of their individual probabilities.
  • If a set of events are mutually exclusive and exhaustive (cover all possibilities), their probabilities add up to 1.
  • This 'sum to 1' rule is key to finding unknown probabilities.

Quick check

  1. The probability of a biased dice landing on 6 is 0.3. What is the probability of it not landing on 6?1 mark
  2. A bag contains only red, green and blue counters. P(red) = 0.2 and P(green) = 0.5. What is P(blue)?1 mark

2. Experimental Probability & Relative Frequency

While theoretical probability tells us what should happen in theory (e.g., P(heads) = 0.5), experimental probability is about what actually happens when you perform an experiment. We find this by calculating the 'relative frequency'. This is the ratio of the number of times an event occurs to the total number of trials. A key idea is that as you conduct more and more trials, the relative frequency usually gets closer and closer to the true theoretical probability.

Relative Frequency = (Number of times an event occurs) / (Total number of trials)

Key term

Relative Frequency: An estimate of probability based on the results of an experiment or observation.

Common pitfall

Confusing theoretical probability (what should happen based on equal likelihoods) with experimental probability (what actually happened in a specific number of trials).

Fun fact

The 18th-century mathematician Buffon had a coin tossed 4040 times, resulting in 2048 heads. His calculated relative frequency was 2048/4040 ≈ 0.5069, incredibly close to the theoretical 0.5.

Worked example 12 marks

A dice is rolled 50 times and lands on a six 8 times. What is the relative frequency of obtaining a six?

  1. 1

    Identify the number of times the desired event occurred: 8 times.

  2. 2

    Identify the total number of trials: 50 rolls.

  3. 3

    Use the formula: Relative Frequency = (Number of successful outcomes) / (Total number of trials).

  4. 4

    Relative Frequency = 8 / 50.

  5. 5

    This can be simplified to 4/25 or expressed as a decimal, 0.16.

Worked example 24 marks

A spinner is spun 200 times. The table shows the relative frequency of it landing on orange after 50, 100, 150 and 200 spins.

Number of spins50100150200
Relative frequency of orange0.220.420.440.46

a) How many times did the spinner land on orange after 150 spins?b) What is the best estimate of the theoretical probability of landing on orange?

  1. 1

    a) Use the formula: Relative Frequency = Number of successes / Number of trials.

  2. 2

    Rearrange to find the number of successes: Number of successes = Relative Frequency × Number of trials.

  3. 3

    At 150 spins, the relative frequency is 0.44. So, Number of orange = 0.44 × 150 = 66 times.

  4. 4

    b) The best estimate for the theoretical probability is the relative frequency calculated from the largest number of trials.

  5. 5

    After 200 spins, the relative frequency is 0.46. This is the most reliable estimate we have from the data provided.

Recap

  • Experimental probability is based on the results of an actual experiment.
  • Relative frequency is the measure of experimental probability.
  • The formula for relative frequency is successes divided by total trials.
  • More trials generally lead to a more reliable estimate of the theoretical probability.
  • Relative frequency can be given as a fraction, decimal or percentage.

Quick check

  1. A drawing pin is dropped 100 times. It lands point up 65 times. What is the relative frequency of it landing point up?1 mark

3. Calculating Expected Frequency

Expected frequency is a prediction. It's the number of times you would expect an event to happen over a certain number of trials. It's a bridge between theoretical probability and real-world outcomes. If you know the probability of an event, you can predict how often it will occur. For example, if you know a biased coin lands on heads with a probability of 0.6, you can expect it to land on heads about 60 times in 100 throws. This is a powerful tool for making predictions based on probability.

Expected Frequency = P(event) × Number of trials

Key term

Expected Frequency: The number of times an outcome is predicted to occur in a set number of trials.

Examiner insight

Clearly show the multiplication step (probability × number of trials) to secure method marks, even if you make a calculation error.

Common pitfall

Giving the answer as a fraction or probability. Expected frequency is a count of how many times something is expected to happen, so it should be a number (which can be a decimal).

Worked example 12 marks

The probability that Xisco wins a game of chess is 0.4. He plays 65 games. How many games would he be expected to win?

  1. 1

    Identify the probability of the event: P(win) = 0.4.

  2. 2

    Identify the number of trials: 65 games.

  3. 3

    Use the formula: Expected Frequency = P(event) × Number of trials.

  4. 4

    Expected wins = 0.4 × 65.

  5. 5

    0.4 × 65 = 26. So, Xisco is expected to win 26 games.

Worked example 23 marks

A biased six-sided dice is rolled 300 times. The probability of it landing on a 4 is 0.15. How many times is it expected to land on a number other than 4?

  1. 1

    First, find the probability of the event 'not landing on a 4'. The outcomes 'landing on 4' and 'not landing on 4' are mutually exclusive and cover all possibilities.

  2. 2

    P(not 4) = 1 - P(4) = 1 - 0.15 = 0.85.

  3. 3

    Now, identify the number of trials: 300 rolls.

  4. 4

    Use the formula for expected frequency: Expected Frequency = P(not 4) × Number of trials.

  5. 5

    Expected number of non-4s = 0.85 × 300.

  6. 6

    0.85 × 300 = 255. The dice is expected to land on a number other than 4 on 255 rolls.

Recap

  • Expected frequency is a calculation of how many times you predict an outcome will happen.
  • The formula is the probability of the event multiplied by the number of trials.
  • It connects theoretical probability to a predicted number of occurrences.
  • The answer represents a number of times, not a probability.
  • Expected frequency might not be a whole number, and that is acceptable.

Quick check

  1. The probability of rain on any given day in April is 0.2. How many rainy days would you expect in the 30 days of April?2 marks

4. Probability of Independent Events

Two events are 'independent' if the outcome of one has absolutely no effect on the outcome of the other. Classic examples include flipping a coin and then rolling a dice, or drawing a card from a deck, replacing it, and then drawing another. To find the probability of two independent events BOTH happening (e.g., getting a 'Head' AND rolling a '6'), you multiply their individual probabilities. This is often called the 'AND' rule.

For independent events A and B: P(A and B) = P(A) × P(B)

Key term

Independent Events: Events where the outcome of one has no effect on the probability of the outcome of the other.

Common pitfall

Adding probabilities instead of multiplying for independent events. Remember: 'AND' means multiply, 'OR' (for mutually exclusive events) means add.

Worked example 13 marks

A fair dice is thrown and a fair coin is spun. Find the probability of getting a number greater than 4 and a tail.

  1. 1

    These are independent events. We need to find the probability of each event separately.

  2. 2

    Event A: Getting a number greater than 4 on a dice. The numbers are 5 and 6. So there are 2 successful outcomes out of 6. P(A) = 2/6 = 1/3.

  3. 3

    Event B: Getting a tail on a coin. There is 1 successful outcome out of 2. P(B) = 1/2.

  4. 4

    To find the probability of both happening, we multiply the probabilities: P(A and B) = P(A) × P(B).

  5. 5

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

Worked example 23 marks

The probability that a bus is on time is 0.8. The probability that a train is on time is 0.9. These events are independent. What is the probability that both the bus and the train are not on time?

  1. 1

    First, find the probability of each event NOT happening.

  2. 2

    P(bus is not on time) = 1 - P(bus is on time) = 1 - 0.8 = 0.2.

  3. 3

    P(train is not on time) = 1 - P(train is on time) = 1 - 0.9 = 0.1.

  4. 4

    Since the original events are independent, their complements are also independent.

  5. 5

    Use the multiplication rule for independent events: P(bus not on time AND train not on time) = P(bus not on time) × P(train not on time).

  6. 6

    Probability = 0.2 × 0.1 = 0.02.

Recap

  • Independent events do not affect each other's outcomes.
  • To find the probability of event A AND event B happening, you multiply their probabilities.
  • Remember the 'AND' rule: P(A and B) = P(A) × P(B).
  • Always check if events are independent before multiplying.
  • Common examples are coin flips, dice rolls, or events 'with replacement'.

Quick check

  1. P(A) = 0.5 and P(B) = 0.2. If A and B are independent, what is P(A and B)?1 mark

5. Using Tree Diagrams for Combined Events

Tree diagrams are a brilliant way to map out all possible outcomes and probabilities for a sequence of events. Each set of branches represents an event, and the probabilities are written on the branches. There are two golden rules for tree diagrams:

  1. MULTIPLY along the branches to find the probability of a single combined outcome (e.g., the path for 'Red then Blue').
  2. ADD the probabilities of the final outcomes if there is more than one way to achieve the desired result (e.g., to find the probability of getting 'one of each colour', you would find P(Red then Blue) and P(Blue then Red) and add them together).

Multiply along branches for 'AND' (e.g., P(A then B))

Add down final outcomes for 'OR' (e.g., P(path 1 or path 2))

Key term

Tree Diagram: A diagram used to represent the probabilities of a sequence of events.

Examiner insight

Always label your branches clearly. When a question asks for a probability, examiners look for the multiplication along the branches and, if necessary, the addition of the final outcomes. Show both steps.

Common pitfall

Forgetting to adjust the probabilities on the second set of branches for 'without replacement' scenarios. The total number of items and the number of the chosen item both decrease by one.

Worked example 14 marks

A bag has four red balls and one green ball. A ball is taken out at random and then replaced. A second ball is then taken out. Draw a tree diagram and use it to find the probability of getting two balls of the same colour.

  1. 1

    First, establish the initial probabilities: P(Red) = 4/5, P(Green) = 1/5. Since the ball is replaced, these probabilities remain the same for the second draw.

  2. 2

    Draw the first set of branches for the first ball: one branch for Red (4/5) and one for Green (1/5).

  3. 3

    From the end of each first branch, draw the second set of branches. From the 'Red' branch, draw two more: Red (4/5) and Green (1/5). Do the same from the 'Green' branch.

  4. 4

    To find the probability of getting two balls of the same colour, we need P(Red and Red) OR P(Green and Green).

  5. 5

    Multiply along the branches: P(Red and Red) = (4/5) × (4/5) = 16/25.

  6. 6

    Multiply along the branches: P(Green and Green) = (1/5) × (1/5) = 1/25.

  7. 7

    Add the final probabilities: P(same colour) = P(Red and Red) + P(Green and Green) = 16/25 + 1/25 = 17/25.

Worked example 24 marks

A bag contains 5 red and 3 blue sweets. A sweet is picked and eaten, then a second is picked and eaten. Find the probability that the two sweets picked are of different colours.

  1. 1

    This is 'without replacement'. The probabilities for the second pick will change. Total sweets = 8.

  2. 2

    Initial probabilities: P(Red) = 5/8, P(Blue) = 3/8.

  3. 3

    Draw the tree diagram. First branches are R (5/8) and B (3/8).

  4. 4

    Second branches (after picking a Red): There are now 7 sweets left (4 Red, 3 Blue). So P(R after R) = 4/7, P(B after R) = 3/7.

  5. 5

    Second branches (after picking a Blue): There are now 7 sweets left (5 Red, 2 Blue). So P(R after B) = 5/7, P(B after B) = 2/7.

  6. 6

    We want 'different colours'. This can be Red then Blue (RB) or Blue then Red (BR).

  7. 7

    P(RB) = P(R on first) × P(B on second) = (5/8) × (3/7) = 15/56.

  8. 8

    P(BR) = P(B on first) × P(R on second) = (3/8) × (5/7) = 15/56.

  9. 9

    Add the results: P(different colours) = P(RB) + P(BR) = 15/56 + 15/56 = 30/56. This simplifies to 15/28.

Recap

  • Tree diagrams show outcomes for multiple events.
  • Label each branch with the event and its probability.
  • Probabilities from a single point on a tree diagram must sum to 1.
  • Multiply probabilities along a path to find the probability of that sequence.
  • Add the probabilities of final paths to find the probability of an overall event.
  • For 'without replacement' problems, remember to adjust the probabilities for the second event.

Quick check

  1. On a tree diagram, the path for 'Win then Win' has probabilities 0.7 and 0.7. What is the probability of winning both games?1 mark

End-of-chapter exercise

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

  1. A biased six-sided dice has the following probabilities of landing on each number: | Number | 1 | 2 | 3 | 4 | 5 | 6 | |-------------|------|------|------|-----|------|-----| | Probability | 0.2 | 0.25 | 0.15 | x | 0.15 | 0.1 | Find the value of x.2 marks
  2. A spinner is tested by spinning it 500 times. It lands on the red section 120 times. What is the relative frequency of the spinner landing on red?2 marks
  3. The probability that a seed from a packet will germinate is 0.85. If a packet contains 240 seeds, how many seeds are expected to germinate?2 marks
  4. A bag contains only red, yellow and green badges. The probability a randomly chosen badge is red is 0.22. The probability it is yellow is twice the probability it is red. What is the probability the badge is green?4 marks
  5. The probability of a phone battery lasting a full day is 0.9. The probability of having a good network signal is 0.7. Assuming these events are independent, what is the probability that the battery lasts a full day but the network signal is not good?3 marks
  6. A bag contains 6 red counters and 4 blue counters. A counter is taken at random, its colour is noted, and it is not replaced. A second counter is then taken. Find the probability that the two counters are different colours.4 marks
  7. At a junction, the traffic light is red for 50% of the time, green for 45% of the time, and amber for 5% of the time. What is the probability that on two consecutive, independent occasions, the light is the same colour?4 marks
  8. A biased six-sided dice is thrown. The probability of throwing a 6 is 1/12. The probability of throwing a 2 is three times the probability of throwing a 6. The probability of throwing a 1 is 1/6. The probabilities of throwing a 3, 4, or 5 are all equal. Find the probability of throwing a number greater than 3.5 marks
  9. There are 100 students in a year group. 60 study French, 35 study German, and 15 study both. A student is chosen at random. Find the probability that the student studies neither French nor German.4 marks
  10. A bag contains 3 red balls and 'n' blue balls. Two balls are taken from the bag without replacement. The probability of taking two red balls is 1/15. Show that n^2 + 5n - 36 = 0, and hence find the number of blue balls in the bag.6 marks

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