Use Expected Cost to Compare Options
In Comparing Two Random Variables by Mean and SD, you compared the centers and variability of two distributions. For a decision involving warranty plans or games, you may instead want to know which option has the lower expected cost. The key is to define cost consistently, include every relevant charge or payment, and compare the probability-weighted averages.
A cost can be fixed, such as a fee paid regardless of what happens, or variable, such as a repair bill paid only if a product breaks. Define a random variable for the total cost under each option. Then use the probability distribution and the expected-value formula from The Mean of a Discrete Random Variable: multiply each possible cost by its probability and add the products.
Before calculating, decide what counts as cost. For a warranty comparison, that might include the plan price and any repair bill the customer must pay. For a game, it might be the entry fee minus the prize received. Use the same time period and define costs from the same person’s perspective for both options. A plan’s premium is a cost to the customer, for example, not a benefit to the customer.
A useful shortcut applies when an option has a fixed charge plus an additional cost only when an event occurs. If the fixed charge is \(a\), the extra cost is \(b\) when event \(E\) occurs, and \(P(E)=p\), then the expected total cost is \(a+bp\). This shortcut agrees with calculating the weighted average from the full distribution. It helps separate a certain charge from a chance-dependent charge.
A Process for Comparing Expected Costs
Specify which options are being compared, whose costs are counted, and the time period.
List what payments are included and connect each possible outcome to its total cost.
Confirm that all possible outcomes are represented, each probability is between 0 and 1, and the probabilities sum to 1, as in Checking Whether a Probability Distribution Is Valid.
Multiply each cost by its probability and add. Include fixed fees and costs that occur only for particular outcomes.
State which option has the lower expected cost and by how much. Explain that the result is about modeled averages, not a guaranteed individual outcome.
If the options have the same fixed charge, that charge contributes equally to both expected costs. You can still include it in both calculations, or note that it cancels when comparing them. Do not cancel a charge that applies to only one option. Also, the expected-cost criterion does not describe how much costs vary; a person may care about the possibility of a particularly large bill as well as the average.
Worked Example: Comparing Two Warranty Plans
Worked Example: Comparing Two Warranty Plans
A customer is considering two ways to cover a fictional appliance for one year. Plan A costs \(\$78\) and covers any repair during the year. Under Plan B, the customer pays a \(\$25\) enrollment charge and remains responsible for a \(\$300\) repair bill if the appliance breaks. The model assigns a probability of \(0.1833\) to a breakdown during the year. Compare the customer’s expected total costs.
State. We will compare the expected one-year out-of-pocket cost to the customer under Plans A and B. The relevant costs are the plan charge and, under Plan B, the repair bill if a breakdown occurs.
Plan. Let \(C_A\) and \(C_B\) be the total one-year costs, in dollars, under Plans A and B. Plan A always costs \(\$78\). Under Plan B, the total is \(\$25\) without a breakdown and \(\$325\) with a breakdown. The two probabilities for Plan B are \(1-0.1833=0.8167\) and \(0.1833\); both are between 0 and 1 and sum to 1. The breakdown and no-breakdown outcomes cover the modeled possibilities. Both plans use the same customer perspective and one-year period, so their expected costs are comparable.
Do. Plan A has a single possible total cost:
For Plan B, calculate the probability-weighted average from its two possible total costs:
Check the result by separating the fixed enrollment charge from the repair bill that occurs only after a breakdown:
The two calculations agree. Plan B’s expected cost is \(\$79.99-\$78=\$1.99\) higher than Plan A’s.
Conclude. Under the stated one-year model, Plan A has the lower expected cost: \(\$78\), compared with \(\$79.99\) for Plan B, a difference of \(\$1.99\). This comparison favors Plan A by expected cost, but an individual customer’s actual cost under Plan B could be \(\$25\) or \(\$325\). The expected cost is not a promise about that customer’s bill.
Worked Example: Premiums and Repair Deductibles
Worked Example: Premiums and Repair Deductibles
Suppose two fictional warranty plans cover the same appliance for one year. The model gives the appliance a \(0.15\) probability of breaking during the year, so the probability of no breakdown is \(0.85\). Plan A costs \(\$45\) and leaves the customer with a \(\$350\) deductible if the appliance breaks. Plan B costs \(\$65\) and has a \(\$200\) deductible if it breaks. Compare the expected total costs to the customer.
Let \(C_A\) and \(C_B\) be the one-year total costs under Plans A and B. The possible costs for Plan A are \(\$45\) without a breakdown and \(\$395\) with one. For Plan B they are \(\$65\) and \(\$265\), respectively. The probabilities \(0.85\) and \(0.15\) are each between 0 and 1 and sum to 1. The outcomes include both possibilities in this model; both plans cover the same year and use the same breakdown probability and customer perspective.
Calculate Plan A’s expected cost. The plan charge is paid whether or not the appliance breaks; the deductible is paid only in the breakdown outcome.
As a check, use the fixed-charge shortcut: \(\$45+(\$350)(0.15)=\$45+\$52.50=\$97.50\). The two forms give the same result.
Calculate Plan B’s expected cost.
The shortcut gives \(\$65+(\$200)(0.15)=\$65+\$30=\$95.00\), which checks the calculation. Plan A’s expected cost is \(\$97.50-\$95.00=\$2.50\) greater.
Conclude. Under this model, Plan B has the lower expected one-year cost, \(\$95.00\) compared with \(\$97.50\) for Plan A. The difference is \(\$2.50\) in expected cost. This comparison does not mean Plan B costs less for every customer: if there is no breakdown, Plan A costs \(\$45\) and Plan B costs \(\$65\); if there is a breakdown, Plan B’s total cost is lower.
Worked Example: Comparing the Cost of Two Games
Worked Example: Comparing the Cost of Two Games
A player is choosing between two fictional games. Game A costs \(\$5\) to play and awards a \(\$12\) prize with probability \(0.40\); otherwise, there is no prize. Game B costs \(\$2\) and awards an \(\$8\) prize with probability \(0.25\); otherwise, there is no prize. Compare the player’s expected net cost, defined as the entry fee minus the prize received.
Let \(C_A\) and \(C_B\) be the player’s net costs per play. A negative net cost means the prize exceeds the entry fee, so the player has a net gain on that outcome. For Game A, the no-prize probability is \(1-0.40=0.60\); for Game B, it is \(1-0.25=0.75\). Each game’s two probabilities are between 0 and 1 and sum to 1, and each table below includes both possible prize outcomes.
| Game A outcome | Net cost \(C_A\) | Probability |
|---|---|---|
| No prize | \(\$5\) | 0.60 |
| \(\$12\) prize | \(-\$7\) | 0.40 |
The expected net cost of Game A is:
| Game B outcome | Net cost \(C_B\) | Probability |
|---|---|---|
| No prize | \(\$2\) | 0.75 |
| \(\$8\) prize | \(-\$6\) | 0.25 |
The expected net cost of Game B is:
As a check, expected net cost equals the entry fee minus the expected prize. For Game A, \(\$5-(\$12)(0.40)=\$5-\$4.80=\$0.20\). For Game B, \(\$2-(\$8)(0.25)=\$2-\$2=\$0.00\). These calculations match the distribution-based results.
Conclude. By expected net cost, Game B is the better choice in this model: its expected net cost is \(\$0.00\) per play, compared with \(\$0.20\) per play for Game A. This does not guarantee that a player will break even on Game B or win on either game. The comparison describes average net cost across many plays under the stated probabilities.
Common Mistakes and AP Exam Tips
- Leaving out a cost that applies only in one outcome. Include a repair bill, deductible, or other conditional charge in the total cost for the outcome where it occurs. A fixed fee alone is not the total cost if additional payments are possible.
- Using the wrong cost for an outcome. Add the charges that are actually paid in that outcome. For a game, define whether the variable is entry fee, prize, net gain, or net cost; do not mix those quantities.
- Comparing different time periods or perspectives. Compare costs for the same period and from the same person’s perspective. Do not compare one plan’s annual customer cost with another plan’s monthly premium.
- Calling expected cost a guaranteed bill. Expected cost is a probability-weighted average, not necessarily a cost that can occur. State the possible individual costs when that distinction matters.
- Claiming the lower expected cost must be the best option for everyone. A person may also care about how variable costs are, the chance of a large bill, or other features not included in the model. Make a conclusion about expected cost rather than claiming the comparison settles every preference.
- Rounding too early. Keep enough digits during the calculation and round the final comparison consistently. If a probability is supplied as a rounded decimal, make clear that the expected cost is based on that model value.
A strong written conclusion identifies the options, reports both expected costs with units and the relevant time period, says which is lower, and gives the difference when useful. It then connects the result to the decision in context. For example: “Under the stated one-year model, Plan B has an expected customer cost of \(\$95.00\), which is \(\$2.50\) less than Plan A’s \(\$97.50\). This compares modeled averages, not every customer’s actual bill.”
Key Takeaway
To compare options by expected cost, define a total-cost random variable for each one, include fixed and outcome-dependent charges, and calculate the probability-weighted average. Choose the option with the lower expected cost only if the decision criterion is modeled average cost; explain what the result does and does not establish.
Check Your Understanding
Use the fictional one-year warranty options below. A breakdown has probability \(0.10\), and no breakdown has probability \(0.90\).
| Option | Cost with no breakdown | Cost with breakdown |
|---|---|---|
| Plan P | \(\$40\) | \(\$340\) |
| Plan Q | \(\$70\) | \(\$220\) |
- Define a random variable for the total one-year customer cost under each plan.
- Check that the two outcome probabilities form a valid distribution.
- Calculate the expected cost of each plan using the probability-weighted outcomes.
- Check each expected cost by separating its no-breakdown charge from the additional cost of a breakdown.
- Write a conclusion comparing the expected costs in context, and explain why it does not guarantee which plan will cost less for one customer.