Poisson process, probability of lifetime

Question

Beginning at time t = O. we start using bulbs. one at a time, to illuminate a room. Bulbs are replaced immediately upon failure. Each new bulb is selected independently by an equally likely choice between a type-A bulb and a type-B bulb. The lifetime, X , of any particular bulb of a particular type is a random variable:

for type-A Bulbs:$$ f(x) = e^{-x},x>=0$$

for type-B Bulbs:$$ f(x) = 3*e^{-3*x},>=0$$

Find the probability that there are no bulb failures before time t.

For solving this problem i have used the total probability theorem: Let D be the event of no Bulb failures on time t.

$$ P(D) = P(D|A)*P(A) + P(D|B)*P(B)$$

$$ P(D) = 1/2 * \int_0^t e^{-x} + 1/2 * \int_0^t 3*e^{-3*x}$$

But i have got the wrong answer: $1 - 1/2 * e^{-t} - 1/2 * e^{-3t}$

The correct answer is: $1/2 * e^{-t} + 1/2 * e^{-3*t}$

Can you tell me, where i have mad a mistake? Thank you

Answer 1

You should not do the integration. Please note that the pdf functions accept time $t$ as input and give the probability of a bulb not failed to the time $t$. So, by plugging in $t$ to each function, you have the probabilities required to use in the total probability theorem.

Answer 2

Simply put, what you calculated was the probability of at least one failure up to time $t$, rather than the probability of no failures up to time $t$.

Answer 3

Your answer is almost correct, but you've misunderstood the problem. The problem is the event of no Bulb failures on time $t$, which means that the time that either $A$ or $B$ went wrong is after $t$, so the probability should be $P(T>t)=0.5\times P(T>t|A)+0.5\times P(T>t|B)$. What you have calculated is the probabilty that either $A$ or $B$ went wrong before $t$, so the correct answer is to use $1$ to minus your answer.