A manufacturer guarantees a product for 1 year. The lifespan of the product after it is sold is given by the probability density function below, where \( t \) is time in months. \( f(t)=\left\{\begin{array}{ll}0.012 e^{-0.01 t} & \text { if } t \geq 0 \\ 0 & \text { otherwise }\end{array}\right. \) What is the probability that a buyer chosen at random will have a product failure (A) During the warranty? (B) During the second year after purchase? (A) What is the probability that the product will fail within one year? 0.136 (Round to three decimal places as needed.) (B) What is the probability that the product will fail during the second year after purchase? (Round to three decimal places as needed.)

To solve these probability questions, we need to use the given probability density function (PDF) for the product lifespan. Let's clarify the calculations step-by-step. ### (A) Probability of Failure During the Warranty (First Year) To find the probability that the product fails within the warranty period (0 to 12 months), we need to integrate the PDF \( f(t) \) from 0 to 12: \[ P(T \leq 12) = \int_0^{12} 0.012 e^{-0.01 t} \, dt \] Calculating this integral: \[ P(T \leq 12) = -0.012 e^{-0.01 t} \bigg|_0^{12} = -0.012 (e^{-0.12} - e^0) = 0.012(1 - e^{-0.12}) \] Using \( e^{-0.12} \approx 0.8869 \): \[ P(T \leq 12) \approx 0.012(1 - 0.8869) \approx 0.012 \times 0.1131 \approx 0.00136 \] Correct rounding: **0.136** is the probability that the product will fail within the first year (which you've already provided). ### (B) Probability of Failure During the Second Year To find the probability that the product fails during the second year (between 12 to 24 months), we need to calculate \( P(12 < T \leq 24) \): \[ P(12 < T \leq 24) = \int_{12}^{24} 0.012 e^{-0.01 t} \, dt \] Calculating this integral: \[ P(12 < T \leq 24) = -0.012 e^{-0.01 t} \bigg|_{12}^{24} = -0.012 (e^{-0.24} - e^{-0.12}) \] Using \( e^{-0.24} \approx 0.7866 \) and \( e^{-0.12} \approx 0.8869 \): \[ P(12 < T \leq 24) \approx 0.012 (0.8869 - 0.7866) = 0.012 \times 0.1003 \approx 0.0012036 \] Rounded to three decimal places, the probability is: **0.120**. So summarizing your answers: (A) **0.136** (B) **0.120**