We have three light bulbs with lifetimes T1, T2, T3 distributed according to Exponential(1), Exponential(2), Exponential(3). In other word, for example bulb #1 will break at a random time T1, where the dis- tribution of this time T1 is Exponential(1). The three bulbs break independently of each other. The three light bulbs are arranged in series, one after the other, along a circuitthis means that as soon as one or more light bulbs fail, the circuit will break. Let T be the lifetime of the circuitthat is, 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 every second (assume the times T1, T2, T3, T are measured in seconds). Let S be the first time we check the circuit and see that it's broken. For example, if the circuit breaks after 3.55 seconds, we will only observe this when 4 seconds have passed, and so S = 4. Calculate the PMF of S.

(c)Finally, suppose that instead of checking on the circuit every second, we instead do the following: after each second, we randomly decide whether to check on the circuit or not. With probability p we check, and with probability 1 p we do not check. This decision is made independently at each time. Now let N be the number of times we check and see the circuit working. For example, if the circuit breaks at time 3.55, and our choices were to check at time 1 second, not to check at times 2 or 3 or 4, and to check at time 5, then N = 1, since the circuit was broken the 2nd time we checked.

What is the PMF of N? (Hint: start by finding the joint PMF of N and S. It's fine if your answer is in summation form.)