Final answer:
The probability that a radio functions for more than 10 years is e^(-10/8), and the conditional probability, given it has lasted 2 years, for it to last at least another 7 years is e^(-7/8), with the rates obtained using the provided mean of 8 years. None of the options provided in the question are correct.
Step-by-step explanation:
If we denote the lifetime of the radio as X, which is an exponential random variable, the expected value (mean) E(X) is given as 8 years. The rate parameter λ of the exponential distribution can be calculated using the formula λ = 1/mean, so λ = 1/8.
For part (a), the probability that a radio will be functioning 10 years from now is P(X > 10), which we calculate using the exponential distribution formula P(X > x) = e-λx. Plugging in the values λ = 1/8 and x = 10, we get:
P(X > 10) = e-(1/8)10 ≈ 0.1353, not in the options provided so none of the options are correct.
For part (b), the conditional probability that the radio lasts at least an additional 7 years given that it has already lasted 2 years is P(X > 9 | X > 2). We use the memoryless property of the exponential distribution to calculate this, which gives us P(X > 7) because the additional 2 years do not impact the remaining time. Thus:
P(X > 9 | X > 2) = P(X > 7) = e-(1/8)7 ≈ 0.4724, not in the options provided so none of the options are correct.
It seems that there was a possible miscalculation in the original question options, as they do not match the calculations based on the provided average value of 8 years for an exponential distribution.