Continuous probability distributions and Bayes' theorem

Question

I came across the following question while reading a book on continuous probability distributions.

$Q.$ Suppose that the life length of a radio tube can be modelled as a continuous random variable $X$ with density $f(x) = \frac {100}{x^2} \space;\space x > 100$. What is the probability that a tube will last more than $200$ hours if it is known that it is still functioning after $150$ hours of service?

Here it is easy to see that $$F(t)=\int_{100}^tf(x)dx=\int_{100}^t\frac {100}{x^2}dx=1-\frac{100}t$$

Therefore $P($life of bulb ends by time $t)=1-\frac{100}t$.

In the question we need to calculate $P(t>200|t>150)$. I am confused about the expression I have to use here.

Answer 1

Using the definition of conditional probability, we have $$ P(t > 200 \mid t > 150) = \frac{P(t > 150, t > 200)}{P(t> 150)} $$ Now $\{t > 200\} \cap \{t > 150\} = \{t > 200\}$, hence $$ P(t > 200 \mid t > 150) = \frac{P(t > 200)}{P(t> 150)} = \frac{\frac 12}{\frac 23} = \frac 34. $$