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Discrete Random Variables and Expected Value — Answers

Answer sheet with worked solutions for the Year 12 probability worksheet.

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Success TutoringWorksheetsAnswer sheetYear 12 · Mathematics · Probability

Discrete Random Variables and Expected Value — Answers

Part A: Probability distributions

A random variable X assigns a number to each outcome of a random experiment. It is discrete if it takes separate (countable) values, such as the number of sixes in four rolls. Its probability distribution lists each value x with its probability P(X = x).

In a valid distribution, every probability is between 0 and 1 and the probabilities add to 1.

Worked example. X is the number shown on a spinner with P(X = 1) = 0.5, P(X = 2) = 0.3 and P(X = 3) = c. Then 0.5 + 0.3 + c = 1, so c = 0.2, and P(X > 1) = 0.3 + 0.2 = 0.5.

1.The random variable X has P(X = 0) = 0.1, P(X = 1) = 0.3, P(X = 2) = k and P(X = 3) = 0.2. Find k.

Answer: k = 0.4

  1. The probabilities must add to 1: 0.1 + 0.3 + k + 0.2 = 1
  2. 0.6 + k = 1, so k = 0.4
2.For the distribution in question 1, find P(X ≥ 2).

Answer: 0.6

  1. P(X ≥ 2) = P(X = 2) + P(X = 3) = 0.4 + 0.2 = 0.6
3.A table with outcomes 1, 2, 3 and probabilities 0.2, 0.5 and 0.4 is a valid probability distribution.

Answer: False

The probabilities add to 0.2 + 0.5 + 0.4 = 1.1. A valid distribution needs every probability between 0 and 1 and a total of exactly 1.

4.Which of these is a discrete random variable?
  1. AThe number of heads when a coin is tossed 10 times
  2. BThe time taken to run 100 m
  3. CThe mass of a randomly chosen apple
  4. DThe height of a randomly chosen student

Answer: The number of heads when a coin is tossed 10 times

The number of heads can only take the separate values 0, 1, 2, …, 10. Time, mass and height are measured, so they are continuous.

Part B: Expected value

The expected value (mean) of a discrete random variable is E(X) = μ = Σ x P(X = x): multiply each value by its probability and add. It is the long-run average value of X over many repeats.

A game is fair when the expected gain (expected winnings − cost to play) is 0.

Worked example. A spinner pays $0 with probability 0.5, $2 with probability 0.3 and $10 with probability 0.2. E(payout) = 0 × 0.5 + 2 × 0.3 + 10 × 0.2 = 0.6 + 2 = $2.60. If a spin costs $3, the expected gain is 2.60 − 3 = −$0.40.

5.Using the distribution in question 1 (P(X = 0) = 0.1, P(X = 1) = 0.3, P(X = 2) = 0.4, P(X = 3) = 0.2), find E(X).

Answer: E(X) = 1.7

  1. E(X) = Σ x P(X = x)
  2. = 0 × 0.1 + 1 × 0.3 + 2 × 0.4 + 3 × 0.2
  3. = 0 + 0.3 + 0.8 + 0.6 = 1.7
6.A fair six-sided die is rolled. Find the expected value of the number shown.

Answer: 3.5

  1. E(X) = (1 + 2 + 3 + 4 + 5 + 6) × = = 3.5
  2. The expected value does not have to be a value X can actually take.
7.It costs $5 to play a game. A fair die is rolled: a 6 wins $20, a 4 or a 5 wins $5, and any other number wins nothing. Find the expected winnings and decide whether the game is fair.

Answer: Expected winnings $5, so the game is fair

  1. E(winnings) = 20 × + 5 × + 0 × = + = = $5
  2. Expected gain = 5 − 5 = $0 per game.
  3. The expected gain is zero, so the game is fair.
8.A raffle sells 500 tickets at $4 each. There is one prize of $1000 and two prizes of $250. Find the expected gain for a person who buys one ticket.

Answer: −$1

  1. Expected prize = 1000 × + 250 × = 2 + 1 = $3
  2. Expected gain = 3 − 4 = −$1 per ticket (on average a ticket loses $1; the organisers keep $500 of the $2000 raised).
9.A game has an expected gain of −$0.50 per play for the player. Which statement is correct?
  1. AOver many games, a player loses about 50 cents per game on average
  2. BA player loses exactly 50 cents every game
  3. CThe game is fair
  4. DA player cannot win any money

Answer: Over many games, a player loses about 50 cents per game on average

Expected value is a long-run average. Individual games can win or lose, but on average the player loses 50 cents per game, so the game favours the organiser.

Part C: Variance and standard deviation

The variance measures spread about the mean: Var(X) = E(X[sup 2]) − [E(X)][sup 2], where E(X2) = Σ x2 P(X = x). The standard deviation is SD(X) = √Var(X), in the same units as X.

For a linear change Y = aX + b: E(aX + b) = aE(X) + b and Var(aX + b) = a[sup 2] Var(X), so SD(aX + b) = |a| SD(X).

Worked example. X takes values 1 and 3, each with probability 0.5. E(X) = 2, E(X2) = 0.5 × 1 + 0.5 × 9 = 5, so Var(X) = 5 − 22 = 1 and SD = 1.

10.For the distribution in question 1, find Var(X) and the standard deviation of X.

Answer: Var(X) = 0.81, SD = 0.9

  1. E(X2) = 02 × 0.1 + 12 × 0.3 + 22 × 0.4 + 32 × 0.2 = 0 + 0.3 + 1.6 + 1.8 = 3.7
  2. Var(X) = E(X2) − [E(X)]2 = 3.7 − 1.72 = 3.7 − 2.89 = 0.81
  3. SD = √0.81 = 0.9
11.A random variable has E(X) = 4 and Var(X) = 2.25. Find E(2X + 3), Var(2X + 3) and the standard deviation of 2X + 3.

Answer: E = 11, Var = 9, SD = 3

  1. E(2X + 3) = 2E(X) + 3 = 2 × 4 + 3 = 11
  2. Var(2X + 3) = 22 Var(X) = 4 × 2.25 = 9 (adding 3 shifts the values but does not change the spread)
  3. SD = √9 = 3
12.For constants a and b, Var(aX + b) = (answer space) × Var(X).

Answer: a2

Multiplying by a scales the spread by a, so the variance (a squared quantity) is scaled by a2; adding b has no effect on spread.

Part D: Building a distribution

To build a distribution, list every value X can take, find each probability (using a tree diagram, a table of outcomes or counting), and check that the probabilities add to 1.

Worked example. A fair die is rolled and X = 1 if a six is rolled, otherwise X = 0. Then P(X = 1) = , P(X = 0) = and E(X) = .

13.Two fair coins are tossed and X is the number of heads. Write the probability distribution of X and find E(X).

Answer: P(X = 0) = , P(X = 1) = , P(X = 2) = ; E(X) = 1

  1. Outcomes HH, HT, TH, TT are equally likely.
  2. P(X = 0) = (TT), P(X = 1) = = (HT, TH), P(X = 2) = (HH)
  3. E(X) = 0 × + 1 × + 2 × = 1
14.A bag holds 4 red and 2 blue counters. Two are drawn without replacement and X is the number of red counters drawn. Find the probability distribution of X and E(X).

Answer: P(X = 0) = , P(X = 1) = , P(X = 2) = ; E(X) =

  1. P(X = 0) = P(BB) = × = =
  2. P(X = 2) = P(RR) = × = =
  3. P(X = 1) = 1 − − = (check: P(RB) + P(BR) = + = = )
  4. E(X) = 0 × + 1 × + 2 × = = [frac 4/3]

Part E: Extended problem

The daily profit X (in dollars) of a market stall has this distribution: P(X = −50) = 0.1, P(X = 100) = 0.3, P(X = 200) = 0.4, P(X = 300) = 0.2.

(a) Find the probability that the stall makes a profit of at least $200 on a given day.

(b) Find the expected daily profit.

(c) Find the standard deviation of the daily profit, to the nearest cent.

(d) Find the expected total profit over 20 trading days.

(e) Next season the owner expects every daily profit to rise by 10%, but a new fee of $30 per day must be paid, so the new profit is Y = 1.1X − 30. Find E(Y) and the standard deviation of Y. Is the change worthwhile?

15.Answer parts (a) to (e), showing full working.[10 marks]

Model answer: (a) P(X ≥ 200) = 0.4 + 0.2 = 0.6. (b) E(X) = −50 × 0.1 + 100 × 0.3 + 200 × 0.4 + 300 × 0.2 = −5 + 30 + 80 + 60 = $165. (c) E(X2) = 2500 × 0.1 + 10 000 × 0.3 + 40 000 × 0.4 + 90 000 × 0.2 = 250 + 3000 + 16 000 + 18 000 = 37 250; Var(X) = 37 250 − 1652 = 37 250 − 27 225 = 10 025; SD = √10 025 ≈ $100.12. (d) 20 × 165 = $3300. (e) E(Y) = 1.1 × 165 − 30 = 181.50 − 30 = $151.50; SD(Y) = 1.1 × 100.12… ≈ $110.14. The expected daily profit falls from $165 to $151.50 and the profit becomes more variable, so on these figures the change is not worthwhile.

Success criteria:

  • (a) adds the correct probabilities to get 0.6 (1 mark)
  • (b) correct expected value of $165 with working (2 marks)
  • (c) finds E(X2) = 37 250 and Var(X) = 10 025, SD ≈ $100.12 (3 marks)
  • (d) $3300 (1 mark)
  • (e) E(Y) = $151.50 and SD(Y) ≈ $110.14 using the linear rules, with a conclusion based on the expected value (3 marks)
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