Question

Suppose that the life in hours of a certain part of radio tube is an r.v X with p.d.f given by 

\(f(x) = \begin{cases}100/x^2,\,\,\,x≥100 \\0,\,\,\,\, otherwise\end{cases}\)

(i) What is the probability that all of three Such tubes in a given radio set will have to be replaced in the first 150 hours? 

(ii) What is the probability that none of the three tubes will be replaced?

Answer 1

(i) A tube in the radio set will have to be replaced during the first 150 hours if its life is less than 150 hours. Hence the required probability is

The probability that all three of the original tubes will have to be replaced during the first 150 hours is (1/3)³ = 1/27

(ii) The probability that a tube is not replaced is given by P(X > 150) 

= 1 – P(X ≤ 150) 

= 1 – 1/3 

= 2/3 

Hence the probability that none of the three tubes will be replaced during the 150 hours of operation is (2/3)³ = 8/27