If a Tl carrier system slips and loses track of where it is, it tries to resynchronize using the

step 1 This is a binomial distribution problems. You can see I've gone ahead and written out the binomi

If a Tl carrier system slips and loses track of where it is,...

Key Concepts

Frame Synchronization

Frame synchronization is the process in digital communication systems of aligning the receiver’s timing to the beginning of data frames. It ensures that the receiver correctly identifies the boundaries of each frame using specific marker bits or sequences, which is essential for interpreting the transmitted data accurately.

Geometric Distribution and Expected Value

The geometric distribution describes the number of independent, identically distributed trials needed to achieve the first success in a series of Bernoulli experiments. Its expected value, which is the average number of trials needed, is the reciprocal of the success probability. This concept is crucial when determining how many frames need to be inspected on average to correctly reestablish synchronization.

Error Probability and System Reliability

Error probability in communication systems refers to the likelihood that a synchronization event or another communication process will fail or be incorrect. Specifying a very low error probability, such as 0.001, provides a measure of confidence in the synchronization process and directly influences the calculation of the average number of frames that must be inspected to meet the reliability standard.

Transcript

00:01 This is a binomial distribution problems.

00:03 You can see i've gone ahead and written out the binomial distribution here.

00:06 Just like with any binomial problem, we need to find n, our number of trials total.

00:12 We need to find r, the number of successes that we want.

00:16 We need to find p and q, the probability of success on any given trial, and q the probability of failure on any given trial.

00:23 We know that we are doing a run of 100 items, 100 items.

00:27 That would represent our n.

00:29 N is 100.

00:34 We are told that the probability of a defective part is 0 .001.

00:39 If the probability of a defective part is 0 .001, that is going to be either our p or our q.

00:46 The question is, what are we being asked about? if the question we're being asked about is about defective parts, then that's our probability of success, right? because that's the probability of defectiveness, and we care about defective, so we'd consider it a success.

01:02 If we're being asked about the probability of non -defective parts, then 0 .001 would be our q because that's not the probability of what we care about.

01:11 If we look at our question, we are being asked about defective.

01:15 We're being asked what the probability is that no more than two are defective.

01:19 So defective is what we're interested in, which means the probability of a part being defected would be our p, our probability of success, quote, unquote.

01:28 So p would be .001.

01:33 We know that 1 minus p would give us q.

01:38 So if we take 1 minus .001, we get our q, which would be .999, which is, we could have figured that anyways probably, but you know, if you need to use a calculator, so be it.

01:53 R will not be the same every time because we are asked the probability that no more than 2.

01:59 So what they're saying is, r cannot be more than two, cannot be higher than two.

02:08 So two is okay, but we also could have anything lower, so we could, instead of having two defective parts, we could just have one, or we could actually have zero defective parts.

02:20 All three of those are possible r values.

02:23 So what we're going to have to do is we're going to have to do three separate binomial distributions and then add up all those problems.

02:29 Probabilities at the end, meaning i want to know the probability of zero defective parts, i want to know the probability of one defective part, and i want to know the probability of two defective parts.

02:44 Then we can be done with this problem.

02:46 Then we can solve this problem.

02:48 So we'll start with zero defective parts.

02:51 We have 100 trials for n, choose zero successes.

02:58 Then we're multiplying that by our probability of success, which is .001, and we want to take that to the zero power, because we don't want any defective parts.

03:11 Remember, defective parts are actually successes for the way this problem was worded.

03:14 We don't want any defective parts in this case, r is zero.

03:18 But we do have to multiply that by the probability of failure, which is 0 .999, and that will be taken to the 100 minus 0 .0 .0 .0 .000...

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