A car is new at the beginning of a calendar year. The time, in years, before the car experiences its first failure is exponentially distributed with mean 2. Calculate the probability that the car experiences its first failure in the last quarter of some calendar year.
Step 1:Let x=time in years before car experience first failure.
$$\sum \limits_{n=0}^{\infty} P(0.75+n<x<1+n)$$ n=some year
Step 2: Since x exponential the cumulative function is $F(x)=1-$e^-(x/$\theta$)
Step 3. $F(1)-F(0.75)=0.081$
the correct answer is 0.205,can someone help me get on the right track to solve this problem.
Your series summation results in
$$\frac{e^{-0.75/\theta}-e^{-1/\theta}}{1-e^{-1/\theta}}$$
or
$$\frac{F(1)-F(0,75)}{F(1)} = 0.205 \, .$$