Question

A company estimates that 0.5% of their products will fail after the original warranty period but within 2 years of the purchase, with a replacement cost of $200.

If they offer a 2 year extended warranty for $11, what is the company's expected value of each warranty sold?

Answer 1

The company's expected value for each extended warranty sold is $9.945, computed by considering the 0.5% probability of a product failure with an associated $200 replacement cost.

Step-by-step explanation:

To calculate the company's expected value for each extended warranty sold, we need to consider both the probability of a product failing and the cost to the company when a product does fail. There is a 0.5% or 0.005 probability that a product will fail after the original warranty period but within 2 years, which would incur a $200 replacement cost. Conversely, there is a 99.5% or 0.995 probability that the product won't fail, and the company would retain the full $11 from the sale of the extended warranty.

The expected value (EV) calculation is as follows:

Expected value for failure: 0.005 (probability of failure) × -$200 (cost) = -$1

Expected value for no failure: 0.995 (probability of no failure) × $11 (revenue) = $10.945

Add both expected values to get the total expected value for each warranty sold:

EV = -$1 + $10.945 = $9.945

Therefore, the company's expected value of each extended warranty sold is $9.945.

Answer 2

The company is expected to make
`\$ 1` on every extended warranty sold.

Step-by-step explanation:

Let
`X` denote whether a product with the extended warranty requires replacement in that two years. (
`X\!` would be a random variable.) Assume that
`X = 1` means that the product requires replacement, and
`X = 0` otherwise.

Assume that this product requires replacement in that two years. That is:
`X = 1`. The company would then bear a cost of
`200 * 1 =200` for replacing this product (since
`X = 1\!`, that cost would be the same as
`200 = 200 * 1 = 200\, X`.)

On the other hand, assume
`X = 0` (that is: this product does not require replacement in that two years.) The company would not need to pay for replacing this product. Since
`X = 0\!`, the expression
`200\, X` would still represent the cost for the company for this warranty.

Either way,
`200\, X` would denote the replacement cost that the company would bear for this product. However, given that the company would charge
`\$ 11` for the extended warranty, the net revenue of the company on this warranty would be
`((-200) \, X + 11)`. (An earning of
`\$11 \!` minus a spending of
`200\, X \!`.)

The question states that
`0.5\%` of the products would need replacement in this period. In other words, the expected value of
`X` would be
`\mathbb{E}(X) = 0.005`.

The expected revenue of the company on this warranty would be:

`\mathbb{E}(200\, X + 11)`.

Apply the linearity of expected values to find the value of
`\mathbb{E}(200\, X + 11)`:

`\begin{aligned}& \mathbb{E}((-200)\, X + 11) \\ &= (-200)\, \mathbb{E}(X) + 11 \\ &= (-200) * 0.005 + 11 = 1\end{aligned}`.

Hence, the company is expected to make
`\$ 1` on every extended warranty that it sold.