Question

Suppose that the life length (in hours) of a certain radio tube is a continuous random variable X with pdf f(x).

f(x) = 100/x², x > 100, and 0 elsewhere

a. What is the probability that a tube will last less than 200 hours if it is known that the tube is still functioning after 150 hours of service?

Answer 1

The probability that a radio tube will last less than 200 hours given it is still functioning after 150 hours is found through conditional probability, involving integrals of the given pdf over different intervals and then dividing them appropriately.

Step-by-step explanation:

The student asks for the probability that a radio tube will last less than 200 hours given it has already lasted 150 hours. This is a conditional probability scenario which can be addressed by considering the pdf of the radio tube's life, which is f(x) = 100/x² for x > 100. To find the conditional probability P(X < 200 | X > 150), we integrate the pdf from 150 to 200 and then divide by the integral from 150 to infinity because we are conditioning on the tube having lasted over 150 hours already. The calculation involves the following steps:

1. Calculate the probability the tube lasts more than 150 hours, P(X > 150), by integrating the pdf from 150 to infinity.
2. Calculate the probability the tube lasts less than 200 hours and more than 150 hours, P(150 < X < 200), by integrating the pdf from 150 to 200.
3. Divide the result of step 2 by the result of step 1 to get the conditional probability.

The integrals can be solved analytically using basic calculus techniques. The solution to these integrals yield the individual probabilities required for the final conditional probability calculation.