We have three light bulbs with lifetimes T1,T2,T3 distributedaccording to Exponential(λ1), Exponential(λ2), Exponential(λ3). Inother word, for example bulb #1 will break at a random time T1,where the distribution of this time T1 is Exponential(λ1). Thethree bulbs break independently of each other. The three lightbulbs are arranged in series, one after the other, along acircuit—this means that as soon as one or more light bulbs fail,the circuit will break. Let T be the lifetime of the circuit—thatis, the time until the circuit breaks.

(a) What is the CDF of T, the lifetime of the circuit?

(b) Next, suppose that we only check on the circuit once everysecond (assume the times T1,T2,T3,T are measured in seconds). Let Sbe the first time we check the circuit and see that it’s broken. Forexample, if the circuit breaks after 3.55 seconds, we will onlyobserve this when 4 seconds have passed, and so S = 4. Calculatethe PMF of S.

(c) Finally, suppose that instead of checking on the circuitevery second, we instead do the following: after each second, werandomly decide whether to check on the circuit or not. Withprobability p we check, and with probability 1−p we do not check.This decision is made independently at each time. Now let N be thenumber of times we check and see the circuit working. For example,if the circuit breaks at time 3.55, and our choices were to checkat time 1 second, not to check at times 2 or 3 or 4, and to checkat time 5, then N = 1, since the circuit was broken the 2nd time wechecked. What is the PMF of N? (Hint: start by finding the joint PMFof N and S. It’s fine if your answer is in summation form.)