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Cambridge IGCSE Statistics revision notes

Probability distributions Revision Notes

Cambridge IGCSE revision notes for Probability distributions.

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Probability distributions

CambridgeIGCSEStatisticsProbability distributions

Complete revision guide 1 of 2

Determining and Displaying Probability Distributions for Simple Outcomes

This note explains how to identify all possible outcomes in straightforward probability scenarios, calculate their probabilities ensuring they sum to one, and present these probabilities clearly in a probability distribution table.

Syllabus source

Section 3, Topic 8.1, page 15, objective 8.1.1

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What you should be able to do

  • ✓Understand how to list all possible outcomes in a simple probability experiment.
  • ✓Calculate the probability of each outcome correctly.
  • ✓Verify that the total probability of all outcomes equals 1.
  • ✓Organise outcomes and their probabilities into a clear probability distribution table.
1

Identifying All Possible Outcomes

In any simple probability scenario, the first step is to list every possible outcome that can occur. This is essential because probabilities are assigned to these outcomes, and none should be omitted. For example, when rolling a fair six-sided die, the possible outcomes are the numbers 1, 2, 3, 4, 5, and 6. Each outcome is mutually exclusive, meaning only one can occur at a time, and collectively exhaustive, meaning one of them must occur.

  • List outcomes carefully to avoid missing any.
  • Ensure outcomes are mutually exclusive and collectively exhaustive.
2

Calculating Probabilities for Each Outcome

Once all outcomes are identified, calculate the probability of each. The probability of an outcome is the chance that it will occur, expressed as a number between 0 and 1. For equally likely outcomes, the probability is 1 divided by the number of outcomes. For example, with a fair six-sided die, each outcome has probability 1/6 ≈ 0.1667. If outcomes are not equally likely, probabilities must be assigned based on the context or given data, ensuring they are valid (between 0 and 1).

  • Probability of an outcome = (Number of favourable outcomes) / (Total number of possible outcomes) for equally likely cases.
  • Probabilities must be between 0 and 1 inclusive.
3

Ensuring the Sum of Probabilities Equals One

A fundamental property of probability distributions is that the sum of the probabilities of all possible outcomes must be exactly 1. This reflects the certainty that one of the possible outcomes will occur. After calculating individual probabilities, add them together to check this condition. If the sum is not 1, re-examine the calculations or the list of outcomes for errors.

  • Sum all probabilities: P(outcome 1) + P(outcome 2) +... + P(outcome n) = 1.
  • If sum ≠ 1, check for missing outcomes or calculation mistakes.
4

Presenting the Probability Distribution in Tabular Form

To communicate the probabilities clearly, organise the outcomes and their probabilities into a probability distribution table. This table has two columns: one for the outcomes and one for their corresponding probabilities. This format makes it easier to interpret and use the probabilities for further calculations or analysis. For example, the distribution for a fair six-sided die is:

| Outcome | Probability |\n|---------|-------------|\n| 1 | 1/6 |\n| 2 | 1/6 |\n| 3 | 1/6 |\n| 4 | 1/6 |\n| 5 | 1/6 |\n| 6 | 1/6 |

  • Use a two-column table: outcomes and probabilities.
  • Ensure probabilities are expressed as fractions, decimals, or percentages consistently.
  • Check that the table covers all outcomes and sums to 1.
5

Worked Example: Tossing a Fair Coin Twice

Consider tossing a fair coin twice. The possible outcomes are: HH (both heads), HT (head then tail), TH (tail then head), and TT (both tails). Each outcome is equally likely with probability 1/4.

Step 1: List outcomes: HH, HT, TH, TT.\nStep 2: Calculate probabilities: each is 1/4 = 0.25.\nStep 3: Sum probabilities: 0.25 + 0.25 + 0.25 + 0.25 = 1.\nStep 4: Present in a table:

| Outcome | Probability |\n|---------|-------------|\n| HH | 0.25 |\n| HT | 0.25 |\n| TH | 0.25 |\n| TT | 0.25 |

  • List all outcomes explicitly.
  • Calculate and verify probabilities.
  • Display in a clear table format.

At-a-glance summary

Comparison table

StepDetailExampleExam Use
Identify outcomesList all possible mutually exclusive and exhaustive outcomesRolling a die: outcomes 1 to 6Always list all outcomes to avoid missing any marks
Calculate probabilitiesAssign probability to each outcome, ensuring values between 0 and 1Each die face has probability 1/6Show correct fractions or decimals for full marks
Check sumSum all probabilities to confirm they equal 1Sum of 6 outcomes: 6 × 1/6 = 1Write sum explicitly to demonstrate understanding
Present distributionCreate a table with outcomes and their probabilitiesTable showing die faces and probabilitiesUse clear tables to organise answers and gain presentation marks

Active recall

Quick check

1. What must the sum of all probabilities in a distribution be?

The sum must be exactly 1, representing certainty that one outcome will occur.

2. If a fair six-sided die is rolled, what is the probability of rolling a 4?

The probability is 1/6, since all six outcomes are equally likely.

3. Why is it important to list all possible outcomes?

To ensure the probability distribution is complete and the total probability sums to 1.

4. How should probabilities be presented in a probability distribution?

In a table with two columns: one for outcomes and one for their corresponding probabilities.

Complete revision guide 2 of 2

Calculating Expected Value from a Probability Distribution and Solving Expectation Problems

This note explains how to compute the expected value of a discrete random variable using its probability distribution and apply this concept to solve problems involving expectation. It covers the calculation method, interpretation, and practical examples to develop a clear understanding of expected value as a measure of long-run average outcome.

Syllabus source

Section 3, Topic 8.1, page 15, objective 8.1.2

Open source document ↗

What you should be able to do

  • ✓Understand the concept of expected value as a weighted average of possible outcomes.
  • ✓Calculate expected value from a given probability distribution.
  • ✓Interpret the expected value in the context of the problem.
  • ✓Apply expected value calculations to solve practical problems involving chance and decision-making.
1

What is Expected Value?

Expected value is a fundamental concept in probability that represents the average or mean outcome of a random variable if an experiment is repeated many times. It is not necessarily a value the variable can take but rather a theoretical long-run average weighted by the probabilities of each outcome.

For example, if you roll a fair six-sided die many times, the expected value of the number shown is the average you would expect over a large number of rolls. This helps in predicting outcomes and making decisions under uncertainty.

  • Expected value is a weighted average of all possible outcomes.
  • Weights are the probabilities associated with each outcome.
  • It provides a measure of central tendency for a probability distribution.
2

Calculating Expected Value from a Probability Distribution

To calculate the expected value (denoted as E or μ) of a discrete random variable, follow these steps:

1. List all possible values the variable can take.

2. Identify the probability associated with each value; probabilities must sum to 1 and be between 0 and 1 inclusive. 3. Multiply each value by its corresponding probability to find the weighted value. 4. Sum all these weighted values to find the expected value.

  • Formula: E = Σ (x × P(x)) where x is a value and P(x) its probability.
  • Ensure probabilities sum to 1 before calculating.
  • Expected value may not be one of the possible outcomes.
3

Worked Example: Expected Value of a Dice Roll

Consider a fair six-sided die with outcomes 1, 2, 3, 4, 5, and 6, each with probability 1/6.

Calculate the expected value: E = (1 × 1/6) + (2 × 1/6) + (3 × 1/6) + (4 × 1/6) + (5 × 1/6) + (6 × 1/6)

= (1 + 2 + 3 + 4 + 5 + 6) × 1/6 = 21 × 1/6 = 3.5. This means that over many rolls, the average outcome tends towards 3.5, even though 3.5 is not a possible roll value.

  • Multiply each outcome by its probability.
  • Sum all products to get expected value.
  • Interpret result as long-run average outcome.
4

Applying Expected Value to Solve Problems

Expected value helps in decision-making and predictions where outcomes have different probabilities and values. For example, in games involving money or points, expected value can indicate the average gain or loss per play.

Example: A game costs £2 to play. You roll a fair die; if you get a 6, you win £10; otherwise, you win nothing. Calculate the expected net gain.

Step 1: List outcomes and probabilities: Win £10 with probability 1/6; win £0 with probability 5/6. Step 2: Calculate expected winnings: E = (10 × 1/6) + (0 × 5/6) = 10/6 ≈ £1.67. Step 3: Calculate expected net gain: Expected winnings - cost = £1.67 - £2 = -£0.33. This means on average, you lose 33p per game in the long run.

  • Use expected value to assess fairness or profitability.
  • Subtract costs to find expected net gain or loss.
  • Interpret results in context to guide decisions.
5

Key Points to Remember

Expected value is a theoretical average, not a guaranteed outcome.

Probabilities must be valid (between 0 and 1, summing to 1).

Expected value can be used to compare different scenarios or strategies in uncertain situations.

  • Check probabilities before calculating.
  • Interpret expected value carefully in context.
  • Use expected value to inform decisions involving risk.

At-a-glance summary

Comparison table

StepDetailExampleExam Use
Identify outcomesList all possible values of the variableOutcomes of rolling a die: 1,2,3,4,5,6Ensure all outcomes are included for full marks
Assign probabilitiesFind probability for each outcomeEach die face has probability 1/6Probabilities must sum to 1 to avoid errors
Calculate weighted valuesMultiply each value by its probability1×1/6, 2×1/6,..., 6×1/6Show all multiplications clearly
Sum weighted valuesAdd all products to find expected valueSum = 3.5 for a fair dieWrite final answer with correct units or context
Interpret resultExplain what expected value means in contextAverage roll value over many trials is 3.5Link answer to problem scenario for full marks

Active recall

Quick check

1. What is the expected value of a fair coin toss where Heads = 1 and Tails = 0?

E = (1 × 0.5) + (0 × 0.5) = 0.5; the expected value is 0.5, meaning on average half a head per toss over many tosses.

2. If a random variable X takes values 2, 4, and 6 with probabilities 0.2, 0.5, and 0.3 respectively, what is E(X)?

E = (2 × 0.2) + (4 × 0.5) + (6 × 0.3) = 0.4 + 2 + 1.8 = 4.2.

3. Why might the expected value not be a possible outcome?

Because expected value is a weighted average of outcomes, it can be a value between possible outcomes rather than one of them.

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