What Is the Distribution of the First Failure Time for Two Independent Machines?

• topgun08
However, since this is a proof it would be better to state my assumptions explicitly, but this is what I arrived at.
topgun08
Question:
Two faulty machines, M1 and M2, are repeatedly run synchronously in parallel (i.e., both machines execute
one run, then both execute a second run, and so on). On each run, M1 fails with probability p1 and M2 with
probability p2, all failure events being independent. Let the random variables X1, X2 denote the number of
runs until the ﬁrst failure of M1, M2 respectively; thus X1, X2 have geometric distributions with parameters
p1, p2 respectively.
Let X denote the number of runs until the ﬁrst failure of either machine. Show that X also has a geometric
distribution, with parameter p1 + p2 − p1p2

Attempt at an answer:
X1 has a geometric distribution of (1-p1)^i-1 * p1
X2 has a geometric distribution of (1-p2)^i-1 * p2

I'm confused an don't know how to proceed. Any help is appreciated.

topgun08 said:
Question:
Two faulty machines, M1 and M2, are repeatedly run synchronously in parallel (i.e., both machines execute
one run, then both execute a second run, and so on). On each run, M1 fails with probability p1 and M2 with
probability p2, all failure events being independent. Let the random variables X1, X2 denote the number of
runs until the ﬁrst failure of M1, M2 respectively; thus X1, X2 have geometric distributions with parameters
p1, p2 respectively.
Let X denote the number of runs until the ﬁrst failure of either machine. Show that X also has a geometric
distribution, with parameter p1 + p2 − p1p2

Attempt at an answer:
X1 has a geometric distribution of (1-p1)^i-1 * p1
X2 has a geometric distribution of (1-p2)^i-1 * p2

I'm confused an don't know how to proceed. Any help is appreciated.

Hey topgun08.

I'll start with one hint: in terms of failures, consider the probability distribution P(X1 OR X2)

After much deliberation here's what I arrived at.

Consider both machines as biased coins such that Tails means the machine fails and Heads means it runs. Thus for Machine1 and Machine 2,
Pr[T1] = p1 and Pr[H1] = 1-p1
Pr[T2] = p2 and Pr

= 1-p2 So the running of both machines can be considered as flipping both these biased coins together, giving us the below table: Pr[H1H2] = (1-p1)(1-p2) Pr[H1T2] = (1-p1)(p2) = p2-p1p2 Pr[T1H2] = (p1)(1-p2) = p1-p1p2 Pr[T1T2] = (p1)(p2) = p1p2 And the questions asks for when either machine fails, so adding the probabilities that contain a tails together: Pr[H1T2] + Pr[T1H2] + Pr[T1T2] = p2-p1p2 + p1-p1p2 + p1p2 = p2 + p1 - p1p2 Which is equal to what the question gave us. Please let me know if my logic is sound or if there's a better way to solve this, and thank you for your help!

topgun08 said:
After much deliberation here's what I arrived at.

Consider both machines as biased coins such that Tails means the machine fails and Heads means it runs. Thus for Machine1 and Machine 2,
Pr[T1] = p1 and Pr[H1] = 1-p1
Pr[T2] = p2 and Pr

That looks pretty good, and the logic looks very sound. Since events are independent you can apply product rule and since the events are disjoint you can simply add them together so under these two assumptions the algebra should work. If the events were not independent or the events were not disjoint this would not work, but since this is not the case it should be ok.

The geometric distribution is a probability distribution that describes the number of trials needed to achieve a success in a sequence of independent trials. In this scenario, we have two machines, M1 and M2, that are repeatedly run synchronously in parallel and have a probability of failing on each run.

We can see that X1 and X2 have geometric distributions with parameters p1 and p2, respectively. This means that the probability of M1 failing on the first run is p1, the probability of failing on the second run is (1-p1)*p1, and so on. Similarly, the probability of M2 failing on the first run is p2, the probability of failing on the second run is (1-p2)*p2, and so on.

Now, let's consider the random variable X, which represents the number of runs until the first failure of either machine. This means that X can take on values of 1, 2, 3, and so on. The probability of X being equal to i, where i is any positive integer, is equal to the probability that both M1 and M2 do not fail on the first i-1 runs, but fail on the ith run. This can be written as:

P(X=i) = P(M1 does not fail on first i-1 runs) * P(M2 does not fail on first i-1 runs) * P(M1 fails on ith run) * P(M2 fails on ith run)

Since the failure events of M1 and M2 are independent, we can multiply the individual probabilities to get:

P(X=i) = (1-p1)^(i-1) * (1-p2)^(i-1) * p1 * p2

This is the same as the geometric distribution with parameter p1 + p2 - p1p2, which is the probability of both M1 and M2 failing on the same run. Therefore, we can conclude that X also has a geometric distribution with parameter p1 + p2 - p1p2.

1. What is the definition of geometric distribution?

The geometric distribution is a probability distribution that models the number of trials needed to achieve a success in a series of independent Bernoulli trials with a constant probability of success.

2. How is geometric distribution different from binomial distribution?

While both distributions involve a series of independent trials with a fixed probability of success, the binomial distribution models the number of successes in a fixed number of trials, while the geometric distribution models the number of trials needed to achieve a single success.

3. What is the formula for calculating the probability of success in geometric distribution?

The probability of success in geometric distribution is calculated using the formula P(x) = (1-p)^(x-1) * p, where x is the number of trials and p is the probability of success.

4. What is the expected value of geometric distribution?

The expected value of geometric distribution is equal to 1/p, where p is the probability of success. This means that on average, it takes 1/p trials to achieve a success.

5. In what real-life situations can geometric distribution be applied?

Geometric distribution can be applied in situations where there are repeated independent trials with a constant probability of success, such as in gambling, sports, and quality control processes.

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