Suppose that the probability that a particular computer chip fails after \( \mathrm{t}=\mathrm{a} \) hours of operation is \( 0.00004 \int_{\mathrm{a}}^{\infty} e^{-0.00004 \mathrm{t}} \mathrm{dt} \). a. Find the probability that the computer chip fails after \( 13,000 \mathrm{hr} \) of operation (that is, the chip lasts at least \( 13,000 \mathrm{hr} \) ). b. Of the chips that are still in operation after \( 13,000 \mathrm{hr} \), what fraction of these will operate for at least another \( 13,000 \mathrm{hr} \) ? c. Evaluate \( 0.00004 \int_{0}^{\infty} e^{-0.00004 \mathrm{t}} \mathrm{dt} \) and interpret its meaning.

**a. Probability the chip fails after 13,000 hours** The probability that the chip fails after \( t = a \) hours is given by \[ P(T \ge a) = 0.00004 \int_{a}^{\infty} e^{-0.00004\,t}\,dt. \] For \( a = 13000 \), we have \[ P(T \ge 13000) = 0.00004 \int_{13000}^{\infty} e^{-0.00004\,t}\,dt. \] Evaluate the integral: \[ \int_{13000}^{\infty} e^{-0.00004\,t}\,dt = \left[-\frac{1}{0.00004} e^{-0.00004\,t}\right]_{13000}^{\infty} = \frac{1}{0.00004}\, e^{-0.00004 \times 13000}. \] Multiplying by \(0.00004\), we obtain \[ P(T \ge 13000) = 0.00004 \cdot \frac{1}{0.00004}\, e^{-0.00004 \times 13000} = e^{-0.00004 \times 13000}. \] Since \[ 0.00004 \times 13000 = 0.52, \] the probability is \[ P(T \ge 13000) = e^{-0.52}. \] --- **b. Fraction operating for at least another 13,000 hours given survival to 13,000 hours** We are asked to compute the conditional probability that a chip will operate for at least another 13,000 hours given that it has already operated for 13,000 hours. This is given by \[ P(T \ge 26000 \mid T \ge 13000) = \frac{P(T \ge 26000)}{P(T \ge 13000)}. \] We already found \[ P(T \ge 13000) = e^{-0.00004 \times 13000} = e^{-0.52}. \] Similarly, \[ P(T \ge 26000) = e^{-0.00004 \times 26000} = e^{-1.04}. \] Thus, \[ P(T \ge 26000 \mid T \ge 13000) = \frac{e^{-1.04}}{e^{-0.52}} = e^{-1.04 + 0.52} = e^{-0.52}. \] --- **c. Evaluation of \( 0.00004 \int_{0}^{\infty} e^{-0.00004\,t}\,dt \) and its interpretation** First, evaluate the integral: \[ \int_{0}^{\infty} e^{-0.00004\,t}\, dt = \left[-\frac{1}{0.00004} e^{-0.00004\,t}\right]_{0}^{\infty} = \frac{1}{0.00004}\left( e^{-0.00004 \times 0} - \lim_{t\to\infty} e^{-0.00004\,t} \right). \] Since \[ e^{-0.00004 \times 0} = 1 \quad \text{and} \quad \lim_{t\to\infty} e^{-0.00004\,t} = 0, \] we have \[ \int_{0}^{\infty} e^{-0.00004\,t}\, dt = \frac{1}{0.00004}. \] Multiplying by \(0.00004\): \[ 0.00004 \int_{0}^{\infty} e^{-0.00004\,t}\, dt = 0.00004 \cdot \frac{1}{0.00004} = 1. \] **Interpretation:** The result \(1\) confirms that the total probability of the chip failing at some time (i.e., over its entire lifetime) is \(1\). This means the probability distribution is properly normalized; the chip is certain to eventually fail.