The lifetime of two brands of bulbs $X$ and $Y$ are exponentially distributed with the mean life of $100$ hours. Bulb $X$ is switched on $15$ hours after Bulb $Y$ has been switched on. The probability that bulb $X$ fails before Bulb $Y$ is? The options are: $(A). > \frac{15}{100} (B). \frac{1}{2} (C).\frac{85}{100} (D). 0$ (See Linked Question)
As mentioned in the answers, the options only make sense if Y survives the first 15 hours.
However, if the options were not given, then would the following solution be correct? $$P(Y<15)=F_Y(15)= 1-e^{-\lambda x}=1-e^{0.01x}\approx0.1392$$
P(Y fails before X) $$=P(Y<15) + P(Y>15)P(Y fails before X)$$
P(Y fails before X) = 1/2 by memoryless property if it survives to 15: $$=0.1392+(0.8607)(0.5)=0.56955$$
The probability we want is the complement of Y fails before X: $$=1-0.56955=0.43045$$
If you apply the difference of two exponential random variables, then also you get the same answer.
