Answer:
The average yearly replacement cost is approximately $32,211,500
Explanation:
Exponential Distribution
P(X < x) = 1 - e -x/
![P(X < x) = 1 - e^{(-x)/(\mu)}](https://img.qammunity.org/2022/formulas/mathematics/college/qv9wm6c6ahjnrhmb1j5jo60rpi1epfdkfy.png)
For the CPU, μ = 3 years
The probability that a CPU fails in less than 1 year is therefore;
P(C) =
![P(X < 1) = 1 - e^{(-1)/(3)} \approx 0.2835](https://img.qammunity.org/2022/formulas/mathematics/college/epi62kc32m3um2fpmvlo6zxyw16s52ojjb.png)
For the RAM, μ = 2 years
The probability that a RAM fails in less than 1 year is therefore;
P(R) =
![P(X < 1) = 1 - e^{(-1)/(2)} \approx 0.3935](https://img.qammunity.org/2022/formulas/mathematics/college/xckt38xctwhz8gqq5qgodz7zn0srrjchfh.png)
For the hard drive, μ = 5 years
The probability that a hard drive fails in less than 1 year is therefore;
P(H) =
![P(X < 1) = 1 - e^{(-1)/(5)} \approx 0.1813](https://img.qammunity.org/2022/formulas/mathematics/college/qepj294hbavxr7a41i033vphat1pa26wzn.png)
The probability that either a CPU, or a RAM or a hard drive fails in 1 year so that a replacement is required, P(C ∪ R ∪ H), is given as follows;
P(C ∪ R ∪ H) = P(C) + P(R) + P(H) - P(C∩R) - P(C∩H) - P(R∩H) + P(C∩R∩H)
∴ P(C ∪ R ∪ H) = 0.2835 + 0.3935 + 0.1813 - (0.2835 × 0.3935) - (0.2835 × 0.1813) -(0.3935 × 0.1813) + (0.2835 × 0.3935 × 0.1813) ≈ 0.64423
∴ P(C ∪ R ∪ H) ≈ 0.64423
There is a 0.64423 chance that a laptop will require replacement
For 100,000 laptops sold a year and a replacement cost of $500 per laptop, we have;
The average yearly replacement cost, 'Cost' is given as follows;
Cost = 500 × 100,000 × 0.64423 = 32,211,500
The average yearly replacement cost, Cost ≈ $32,211,500