Decision Probability

Grade 11 · statistics · 100 practice problems · read aloud

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Decision Probability

What Is It? 🤔

Decision Probability uses probability to make informed choices between different options. It combines the likelihood of various outcomes with their consequences (like costs or rewards) to find the best decision.

Why it's useful: It helps you make rational, data-driven decisions in uncertain situations, from business strategies to everyday life.

How to Solve Problems

  1. List Options & Outcomes: Identify your possible choices and the uncertain events that could follow.
  2. Assign Probabilities: Determine the probability for each outcome.
  3. Assign Values: Attach a value (cost, profit, score) to each outcome.
  4. Calculate Expected Value (EV): For each option: EV = Σ(Probability × Value).
  5. Compare & Decide: Choose the option with the highest expected value (or lowest, if dealing with costs).

Worked Examples

Example 1: The Game Show
You can choose Door A (50% chance to win $100) or Door B (30% chance to win $250). Which is better?

  1. Option A EV: (0.50 × $100) + (0.50 × $0) = $50
  2. Option B EV: (0.30 × $250) + (0.70 × $0) = $75
  3. Decision: Choose Door B (higher EV).

Example 2: The Commute
You can take the Highway (70% chance of a 30-min trip, 30% chance of a 60-min traffic jam) or the Streets (100% chance of a 45-min trip). Which minimizes travel time?

  1. Highway EV: (0.70 × 30) + (0.30 × 60) = 21 + 18 = 39 minutes
  2. Streets EV: (1.00 × 45) = 45 minutes
  3. Decision: Take the Highway (lower expected time).

Common Mistakes ⚠️

Ignoring All Outcomes: Every possible outcome must be included in your EV calculation, even the ones with $0 or negative value.

Confusing Probability & Value: A high probability doesn't guarantee a good outcome if the value is low. Always multiply them.

Probability Sum Check: The probabilities for all outcomes of a single choice must add up to 1 (or 100%).

Tips & Tricks

EV = "Average Long-Run Result": Think of the Expected Value as the average result you'd get if you could make the same decision hundreds of times.

Organize with a Table: Create a simple table with columns for Choice, Outcome, Probability, and Value to stay organized.

Decision Rule: For profits/rewards, pick the highest EV. For costs/losses, pick the lowest EV.

How to Practice

  • Start with simple 2-choice scenarios (like the examples above).
  • Create your own real-world problems (e.g., "Should I buy a lottery ticket or save the money?").
  • Practice drawing simple decision trees to visualize the options and outcomes.
  • Always double-check that your probabilities for each scenario sum to 1.

Practice problems

6 of the 100, worked through step by step — try them before opening the answer.

1 P(X ≥ 3) = ? where X ~ Poisson(λ = 2.5)

Hint: For a Poisson distribution, the probability of at least k events equals 1 minus the cumulative probability of k-1 events. Consider using the complement rule and Poisson probability mass function.

Show the answer

Answer: 0.4562

  1. Recall the Poisson probability formula: P(X = k) = (λ^k * e^(-λ)) / k! where λ = 2.5, e ≈ 2.71828.
  2. P(X ≥ 3) = 1 − P(X ≤ 2) because it’s easier to compute the complement for small values.
  3. Compute P(X = 0): λ^0 = 1, 0! = 1 P(X = 0) = (1 * e^(-2.5)) / 1 = e^(-2.5) Numerically: e^(-2.5) ≈ 0.082085
  4. Compute P(X = 1): λ^1 = 2.5, 1! = 1 P(X = 1) = (2.5 * e^(-2.5)) / 1 = 2.5 * 0.082085 ≈ 0.205212
  5. Compute P(X = 2): λ^2 = 6.25, 2! = 2 P(X = 2) = (6.25 * e^(-2.5)) / 2 = (6.25 * 0.082085) / 2 First: 6.25 * 0.082085 ≈ 0.513031 Divide by 2: ≈ 0.256516
  6. Sum P(X ≤ 2): P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.082085 + 0.205212 + 0.256516 = 0.543813
  7. P(X ≥ 3) = 1 − 0.543813 = 0.456187
  8. Round to 4 decimal places: 0.4562 Final answer: 0.4562

We are given: X ~ Poisson(λ = 2.5) We want: P(X ≥ 3)

2 P(X ≥ 2) where X ~ Binomial(n=5, p=0.4) = ?

Hint: For a binomial distribution, the probability of at least k successes equals 1 minus the cumulative probability of k-1 successes. Consider using the complement rule to simplify calculation.

Show the answer

Answer: 0.66304

  1. Recall that for a binomial distribution, P(X = k) = C(n, k) * p^k * (1-p)^(n-k) where C(n, k) = n! / (k! (n-k)!) Here n = 5, p = 0.4, 1-p = 0.6.
  2. Instead of computing P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) directly, we can use the complement rule: P(X ≥ 2) = 1 - P(X < 2) = 1 - [P(X = 0) + P(X = 1)].
  3. Compute P(X = 0): C(5, 0) = 1 P(X = 0) = 1 * (0.4)^0 * (0.6)^5 = 1 * 1 * (0.6)^5 0.6^5 = 0.6 * 0.6 = 0.36 0.36 * 0.36 = 0.1296 0.1296 * 0.6 = 0.07776 So P(X = 0) = 0.07776.
  4. Compute P(X = 1): C(5, 1) = 5 P(X = 1) = 5 * (0.4)^1 * (0.6)^4 0.4^1 = 0.4 0.6^4 = 0.6^2 * 0.6^2 = 0.36 * 0.36 = 0.1296 So P(X = 1) = 5 * 0.4 * 0.1296 First 5 * 0.4 = 2.0 2.0 * 0.1296 = 0.2592.
  5. Sum P(X = 0) + P(X = 1): 0.07776 + 0.2592 = 0.33696.
  6. Apply complement rule: P(X ≥ 2) = 1 - 0.33696 = 0.66304. Final answer: 0.66304

We are given: X ~ Binomial(n = 5, p = 0.4) We want: P(X ≥ 2)

3 P(X ≥ 4) where X ~ Binomial(n=10, p=0.35) = ?

Hint: For a binomial distribution, the probability of at least k successes equals 1 minus the cumulative probability of k-1 successes. Use the binomial cumulative distribution function.

Show the answer

Answer: 0.4862

  1. We need P(X ≥ 4) where X ~ Binomial(n=10, p=0.35)
  2. P(X ≥ 4) = 1 - P(X ≤ 3)
  3. Calculate P(X ≤ 3) using the binomial cumulative distribution function
  4. P(X = 0) = C(10,0) * (0.35)^0 * (0.65)^10 = 1 * 1 * 0.0135 = 0.0135
  5. P(X = 1) = C(10,1) * (0.35)^1 * (0.65)^9 = 10 * 0.35 * 0.0207 = 0.0725
  6. P(X = 2) = C(10,2) * (0.35)^2 * (0.65)^8 = 45 * 0.1225 * 0.0319 = 0.1757
  7. P(X = 3) = C(10,3) * (0.35)^3 * (0.65)^7 = 120 * 0.0429 * 0.0490 = 0.2522
  8. P(X ≤ 3) = 0.0135 + 0.0725 + 0.1757 + 0.2522 = 0.5138
  9. P(X ≥ 4) = 1 - 0.5138 = 0.4862

The answer is 0.4862.

4 Emma's investment: gain $1500 with P=0.45, lose $500 with P=0.55. Expected value = ?

Hint: Calculate the weighted average of all possible outcomes by multiplying each outcome by its probability and summing the results.

Show the answer

Answer: 400

  1. Identify the outcomes and their probabilities Gain $1500 with probability 0.45 Lose $500 with probability 0.55
  2. Calculate expected value using formula E(X) = Σ[x × P(x)] E(X) = (1500 × 0.45) + (-500 × 0.55)
  3. Perform the multiplications 1500 × 0.45 = 675 -500 × 0.55 = -275
  4. Sum the results 675 + (-275) = 400
  5. Interpret the result The expected value is $400, meaning Emma can expect an average gain of $400 per investment decision.

The answer is 400.

5 A triangular region is bounded by the parabola y = 4 - x² and the x-axis. Calculate the exact area of this region using integration methods.

Hint: Consider finding the points where the curve intersects the horizontal axis to determine the limits of integration. The area under a curve can be found by evaluating a definite integral.

Show the answer

Answer: 32/3

  1. Find the x-intercepts of the parabola by setting y = 0: 4 - x² = 0 → x² = 4 → x = -2 and x = 2.
  2. The area under the curve from x = -2 to x = 2 is given by the definite integral: ∫ from -2 to 2 of (4 - x²) dx.
  3. Find the antiderivative: The antiderivative of 4 is 4x, and the antiderivative of x² is x³/3. So the antiderivative of (4 - x²) is 4x - x³/3.
  4. Evaluate from -2 to 2: [4(2) - (2)³/3] - [4(-2) - (-2)³/3] = [8 - 8/3] - [-8 - (-8/3)] = [8 - 8/3] - [-8 + 8/3].
  5. Simplify: (24/3 - 8/3) - (-24/3 + 8/3) = (16/3) - (-16/3) = 16/3 + 16/3 = 32/3. The exact area is 32/3.

6 A triangular region is bounded by the parabola y = 4 - x² and the x-axis. A rectangle is inscribed within this region such that its base lies along the x-axis and its top corners touch the parabola. What is the maximum possible area of such a rectangle?

Hint: Consider how the width and height of the rectangle relate to the parabola's equation. The area can be expressed as a function of the x-coordinate of the corner point.

Show the answer

Answer: 6.158

  1. Let the rectangle have corners at (-a, 0), (a, 0), (a, 4 - a²), and (-a, 4 - a²), where a > 0.
  2. The width of the rectangle is 2a, and the height is 4 - a².
  3. The area function is A(a) = (2a)(4 - a²) = 8a - 2a³.
  4. To maximize the area, find the derivative: A'(a) = 8 - 6a².
  5. Set derivative equal to zero: 8 - 6a² = 0
  6. Solve for a: 6a² = 8, a² = 8/6 = 4/3, a = 2/sqrt(3)
  7. Calculate maximum area: A = 8(2/sqrt(3)) - 2(2/sqrt(3))³ = 16/sqrt(3) - 2(8/(3sqrt(3))) = 16/sqrt(3) - 16/(3sqrt(3)) = (48 - 16)/(3sqrt(3)) = 32/(3sqrt(3)) ≈ 6.158 The maximum possible area is approximately 6.158 square units.
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