You purchase a certain product. The manual states that the lifetime T of the
product, defined as the amount of time (in years) the product works properly
until it breaks down, satisfies

P(T >= t) = e^{(-t/5) ,for all t>=0}
you purchase the product and use it for two years without any problems. The
probability that it breaks down in the third year is _________ (upto three
decimal places).
plz solve it.

Basically the probability being asked is conditional probability :

P((2 <= x <= 3) | (x >= 2) ) meaning that it will break somewhere in the third year but we are given that the product is well till 2 years .The numerator term of probability can be written as :

P(2 <= x <= 3) ∩ (x >=2 )) = P(2 <= x <= 3 )

= P(x >= 2) - P(x >= 3)

= 1/e^{2/5} - 1/e^{3/5} (according to the given function for P(x >= t) = e^{-}^{t/5})

In this question we can't take it as a random variable and think it as a cdf and find P(2<=x<=3) =F(3)-F(2) and den minus it from 1 .?? Why can't we do like this sir