Suppose a dealership’s has a fraction x of good cars and 1 - x of bad cars. Your willingness to pay is $10,000 for a good car and $4,000 for a bad car. You cannot tell whether a given car is good or bad, but you know the value of x. The dealership requires at least $6,000 to sell you a good car. If you offer the expected value of a randomly chosen car (based on x), for what values of x will the dealership be willing to sell you a good car?

The value of x is ____ % (enter 20, not 0.2). Also round you answers to nearest 2 decimals (example: 20.333 = 20.33 and 20.335 = 20.34)

Question

Suppose a dealership’s has a fraction x of good cars and 1 - x of bad cars. Your willingness to pay is $10,000 for a good car and $4,000 for a bad car. You cannot tell whether a given car is good or bad, but you know the value of x. The dealership requires at least $6,000 to sell you a good car. If you offer the expected value of a randomly chosen car (based on x), for what values of x will the dealership be willing to sell you a good car?

The value of x is ____ % (enter 20, not 0.2).  Also round you answers to nearest 2 decimals (example: 20.333 = 20.33 and 20.335 = 20.34)

BlackTom AI · Accepted Answer

We start by identifying the key values involved. A good car is worth 10,000 to me and a bad car is worth 4,000 to me. The fraction of good cars is x, so the expected value (EV) of a randomly chosen car before knowing its type is EV = x*(10,000) + (1 - x)*(4,000).

Compute the EV: EV = 10,000x + 4,000 - 4,000x = 6,000x + 4,000.

The dealership will sell a good car if the EV is at least 6,000, since that is their minimum acceptable price for a good car.

Set up the inequality: 6,000x + 4,000 ≥ 6,000.

Subtract 4,000 from both sides: 6,000x ≥ 2,000.

Divide both sides by 6,000: x ≥ 2,000 / 6,000 = 1/3 ≈ 0.3333.

Convert to a percentage by multiplying by 100: x ≥ 33.33% (rounded to two decimals as requested).

Interpretation: The dealership will be willing to sell you a good car when the probability that a random car is good is at least about one-third. If x is below 33.33%, the expected value falls short of the 6,000 threshold; if x is at or above 33.33%, the expected value meets or exceeds the threshold and a deal is possible.

If you manipulate the numbers, you can see why: the cheaper bad cars pull the EV down unless x is high enough to compensate, and 6,000x + 4,000 crosses 6,000 precisely at x = 1/3.

COMM_V 295 105 106 2025W1 Participation/Math Quiz #3

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