91Ó°ÊÓ

The probability density function (pdf) of an exponential distribution is given by: $$ f(x) = \lambda e^{-{ \lambda x}}, \quad x \geq 0 $$ And the cumulative distribution function (cdf) is given by: \( F(x) = P(X \leq x) = 1 - e^{ - \lambda x } \) where \(X\) is the random variable representing the repair time, and \(x\) is the specific repair time.

We have to find \(P(X > 2)\), which can also be written as \(1 - P(X \leq 2)\). Using the cdf formula: \(P(X > 2) = 1 - F(2)\) Replace the values of \(\lambda\) and \(x\) in the cdf formula: \(P(X > 2) = 1 - (1 - e^{ - \frac{1}{2} \cdot 2 })\) Now, calculate the probability: \(P(X > 2) = 1 - (1 - e^{ -1 })\) \(P(X > 2) = e^{ -1 } \approx 0.368\) Therefore, the probability that a repair time exceeds 2 hours is approximately \(0.368\).

We are asked to find the conditional probability of a repair taking at least 10 hours, given that it exceeds 9 hours. Using the conditional probability formula: \(P(X \geq 10 \mid X > 9) = \frac{P(X \geq 10 \cap X > 9)}{P(X > 9)} = \frac{P(X \geq 10)}{P(X > 9)}\) Since \(X \geq 10\) implies \(X > 9\), we can write it as: \(P(X \geq 10 \mid X > 9) = \frac{P(X \geq 10)}{P(X > 9)}\) Now, we can find the probabilities using the same approach as in Step 2: \(P(X \geq 10) = 1 - F(10) = e^{-\frac{1}{2} \cdot 10} \approx 6.74 \times 10^{-5}\) \(P(X > 9) = 1 - F(9) = e^{-\frac{1}{2} \cdot 9} \approx 1.87 \times 10^{-4}\) Now we can find the conditional probability: \(P(X \geq 10 \mid X > 9) = \frac{6.74 \times 10^{-5}}{1.87 \times 10^{-4}} \approx 0.360\) Therefore, the conditional probability that a repair takes at least 10 hours, given its duration exceeds 9 hours, is approximately \(0.360\).