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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.

What is the answer given by the way ??
ans is .181

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/e2/5 -   1/e3/5  (according to the given function for P(x >= t) = e-t/5)

=  0.1215

But we are given it works fine till 2 years .

So  ,   P(x >= 2 ) = 1 / e0.4

= 0.6703

Therefore , P((2 <= x <= 3) | (x >= 2) )   =    P(2 <= x <= 3) ∩ (x >=2 )) / P(x >= 2 )

=    0.1215 / 0.6703

=    0.181

Hence the required probability should be 0.181

thanks ..i got it.
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