Expected Value (Discrete Variable)

\( E(X)=\sum x\,P(X=x) \)
Probability GCSE
Question 8 of 20
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Lottery: £50(0.02), £10(0.18), £0(0.8). Find E(X).

Tips: use ^ for powers, sqrt() for roots, and type pi for π.
Hint (H)
Multiply and sum.

Explanation

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Statement

The expected value of a discrete random variable represents the long-term average outcome if the random experiment is repeated many times. It is calculated as the sum of each possible outcome multiplied by its probability:

\[ E(X) = \sum x \, P(X=x) \]

Where:

  • \(E(X)\) = expected value (the mean of the distribution)
  • \(x\) = each possible value the random variable can take
  • \(P(X=x)\) = probability of \(X\) being equal to \(x\)

Why it works

  • The expected value is a weighted average of outcomes, weighted by their probabilities.
  • If you repeated the experiment a very large number of times, the mean result would approach \(E(X)\).
  • It generalises the idea of an average to situations where outcomes are not equally likely.

Recipe (Steps)

  1. List all possible values of the random variable \(x\).
  2. Write down the probability of each value, \(P(X=x)\).
  3. Multiply each outcome \(x\) by its probability.
  4. Add up all these products to get \(E(X)\).

Spotting it

You need this formula when the question gives a probability distribution table and asks for the expected value, or when you must compute the mean of a random variable.

Common pairings

  • Variance and standard deviation of a discrete variable.
  • Binomial distribution: expected value = \(np\).
  • Games of chance: expected winnings or losses.

Mini examples

  1. Given: A fair die: \(x = 1,2,3,4,5,6\), each with probability \(1/6\). Find: \(E(X)\). Answer: 3.5.
  2. Given: A biased coin: \(P(\text{Head})=0.7, P(\text{Tail})=0.3\). Let \(X=1\) for head, \(0\) for tail. Find: \(E(X)\). Answer: 0.7.

Pitfalls

  • Probabilities must sum to 1. Always check this before calculating.
  • Expected value may not be one of the possible outcomes (e.g. die mean = 3.5).
  • Mixing up frequency with probability. Use probabilities in the formula.

Exam Strategy

  • Set out your working in a table with columns for \(x\), \(P(X=x)\), and \(xP(X=x)\).
  • Double-check probabilities sum to 1.
  • Interpret your answer in context: it is an average, not a guaranteed result.

Worked Micro Example

A spinner has outcomes 1, 2, 3 with probabilities 0.2, 0.5, 0.3. Find \(E(X)\).

Compute products: \(1\times 0.2 = 0.2\), \(2\times 0.5 = 1.0\), \(3\times 0.3 = 0.9\).

Add: \(0.2 + 1.0 + 0.9 = 2.1\).

So, \(E(X)=2.1\).

Summary

The expected value formula provides the mean outcome of a discrete random variable. By multiplying each possible outcome by its probability and summing, we obtain the theoretical long-run average. This underpins much of probability, from dice and coins to real-world statistics like average demand or insurance payouts.

Worked examples

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  1. A fair die is rolled. Find E(X).
    1. \( x=1..6 \)
    2. \( P=1/6 each \)
    3. \( E(X)=(1+2+3+4+5+6)/6=3.5 \)
    Answer: 3.5
  2. \( X takes values 0,1 with P(X=1)=0.7, P(X=0)=0.3. Find E(X). \)
    1. \( E=0*0.3+1*0.7=0.7 \)
    Answer: 0.7
  3. X has distribution: 1(0.2), 2(0.5), 3(0.3). Find E(X).
    1. \( 1*0.2=0.2 \)
    2. \( 2*0.5=1 \)
    3. \( 3*0.3=0.9 \)
    4. \( Sum=2.1 \)
    Answer: 2.1
  4. X: 2(0.25), 4(0.5), 6(0.25). Find E(X).
    1. \( 2*0.25=0.5 \)
    2. \( 4*0.5=2 \)
    3. \( 6*0.25=1.5 \)
    4. \( Sum=4 \)
    Answer: 4
  5. Biased die: outcomes 1..6 with probs (0.1,0.2,0.2,0.2,0.2,0.1). Find E(X).
    1. \( Compute products and sum=3.5 \)
    Answer: 3.5
  6. X: 0(0.1), 1(0.2), 2(0.4), 3(0.3). Find E(X).
    1. \( 0*0.1=0 \)
    2. \( 1*0.2=0.2 \)
    3. \( 2*0.4=0.8 \)
    4. \( 3*0.3=0.9 \)
    5. \( Sum=1.9 \)
    Answer: 1.9
  7. Lottery ticket pays £10 with prob 0.1, £0 with prob 0.9. Find E(X).
    1. \( 10*0.1=1 \)
    2. \( 0*0.9=0 \)
    3. \( Sum=1 \)
    Answer: 1
  8. Game: win 5 with prob 0.4, win 2 with prob 0.4, lose 3 with prob 0.2. Find E(X).
    1. \( 5*0.4=2 \)
    2. \( 2*0.4=0.8 \)
    3. \( -3*0.2=-0.6 \)
    4. \( Sum=2.2 \)
    Answer: 2.2
  9. X: -1(0.5), 1(0.5). Find E(X).
    1. \( -1*0.5=-0.5 \)
    2. \( 1*0.5=0.5 \)
    3. \( Sum=0 \)
    Answer: 0
  10. X: 1(0.1), 2(0.2), 3(0.3), 4(0.4). Find E(X).
    1. \( 1*0.1=0.1 \)
    2. \( 2*0.2=0.4 \)
    3. \( 3*0.3=0.9 \)
    4. \( 4*0.4=1.6 \)
    5. \( Sum=3.0 \)
    Answer: 3.0