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An organization has N employecs where N is a large numbcr. Each employcc has one of three possible job classificatious Ac changes classifications (independently) ac...

Question

An organization has N employecs where N is a large numbcr. Each employcc has one of three possible job classificatious Ac changes classifications (independently) according Markov chain with transition probabilities0.7 0.20.1 0.What percentage of employccs are in each classification?

An organization has N employecs where N is a large numbcr. Each employcc has one of three possible job classificatious Ac changes classifications (independently) according Markov chain with transition probabilities 0.7 0.2 0.1 0. What percentage of employccs are in each classification?



Answers

A company employs Mr. Withrow, Mr. Church, Ms. David, Ms. Henry, Mr. Fielding, Mr. Smithson, Ms. Penny, and Mr. Butler and plans to select a group of three of these employees to receive advanced training. If the three employees who will receive advanced training are selected at random, what is the probability that Mr. Church, Ms. Henry, and Mr. Butler will be selected for advanced training?

All right? So here we are given the transition metrics, which is this, right? And uh let uh we have to find out the expected time when it takes to reach from uh To reach ST three. Right? So, we will state that uh let X be the expected time uh to reach ST three uh from state one. Right? So, let this will be uh To reach that three prom state one. Right? And why will be the expected time to reach the two? Uh tourists take three prom state too. Right. So, let uh so then we get from the state. Let United below here. So, let this will be X. And this will be one hub one plus X Plus one hour. 6 times one plus Y plus one or three Times Uh one. Right? So this will be X. And this will be one or two plus X. Or two plus one or six plus Y. Were six plus one or three. Right? And this will be X minus X or two. And this will be one or two plus 1/6 plus one or three plus. Why? You were six? Right? Now, this will be 0.5 X. And this will be one plus why? You are six? And this will be one hub X. And this will be one plus Y over sex. Right? Now this will be X. And this will be two plus to why we're sex, right? And from here, this X will be two plus Y over three. And X will be this this will be let This will be uh Why? And uh why were three Will be X. Men is too right? And why will be three X. Menos six? Right? This y will be three X. Men is sex right now. Uh Prom stirred to let perhaps state to Why will be they write it here, Bram state too. Let uh Let Why will be This will be one or four times one plus X plus one or two times one plus Y plus one or four times one. Right? Now here, why will be won or poor Plus X. Where poor plus one or 2 plus? Why were two plus one? We're all right now why will be one plus? This will be one plus X. Were poor plus Why? Or two? And here X were poor will be why men as Y were two men as one. And the X were poor will be Why over 2 -1. So now uh this will be uh let this will be 0.25 x. And this will be 0.5 Y -1. Right? And while this is why? Because y is three X. Menos sex right now. Here uh let this is 0125 X. And this will be 0.5 times three X minus six and minus one. Right? I have just put it this value instead of Y. Right? And this is 0.25 x. and this is 1.5 x minus three minus one. And this will be 0125 X. And this will be minus. Uh This will be just this is Menace here. Right? Let this is man is here. This will be minus 1.5 X. And this will be Menace pour, right? And Menace, 1.5 X will be Menace pour. And from here, X will be poor or 1.25 R x will be 3.2. Right? So excess here. And this is 3.2 Now, uh the expected time to reach 3rd 3 Prime State one is 3.2. Right?

Alright. So here uh in part uh the transition matrix will be let the transition the resins Ishan metrics. This will be uh one manners, Alpa and vita. Alpa. one Menace Beat. All right. And this will be A And this is here a 11. This is a 12. This is a to one and this is a 22 right now. In part be here. Uh ah In part we here uh The second matrix, uh this will be a little late. In part B is beta and alpha. Uh This is uh metrics beat a Menace. Alpha, beta plus alpha beta. Right? And this is alpha beta plus alpha. Manus alphabet to write. And this will be better, alpa. Right. So hence the stationary distribution is the stationary the stationary distribution is one. We're all populous beta times, metrics are beta and alpha. Right.

Hi there. So for this problem and we start with finding H. Is the event that the uh corn land and t. Is the event of uh corn land land too. So we know the corn can eat the land heads or tails? Probably the T. Is he called to one minus the probability. Okay. So which is the call to 1.0 point six? Because one minus 0.4? 0.6. Mhm. Um let's we know there are 40 independent cost tosses. But let's uh start with it's small number. As an example of, let's say we have four tosses. Um Now we can divide into three pairs, three pairs of consecutive tosses. Uh if we have five um tosses, we can divide into four different consecutive pairs of tosses. So for every end number of tosses we have We have N -1 consecutive pair. All right. So, We know for 40 tosses, It's like 40 tosses. Mhm. We have 39 pips. Now let's see what the CCP has two placeholders. Let's see how many cases it can take on. So, case number one the pair can take up is both of the coin tosses or land ahead. The second case is the first lens had the second length tell The 3rd case in the first lens. Tell the second length head, The 4th cases, both tosses and then tell What's the probably the first one happening. It is equal to 0.4 Time 0.4. You called to 0.16. The second case could be 0.47 0.6. You called to 0.24. The third case would be 0.670 point 40.24. The fourth case could be 0.670 point six which is equal to zero for four 36. Now we know there's a task over. It's a shame over here is gonna come is different from the one preceding it. So these are the two the only two cases which have a task over because it went from head to tail and from tail to head. So we know the probability of change over. Yeah. Is he called to probably your head tail plus the probability of tells him. Yeah. Which we found each of them. We go to zero control form. So it's open 24. Let's be open 24 equals there for going 48. So we know the probability probability of a changeover happened in a pair. We know that the expected value would be the number of uh trials sounded, the probability of that event. Um, we know the number of trials 39. The probability is 0.48. So the expected value will be 39 times 0.30 48. We should work out to be 18 0.72 18.72. That will be our answer. I hope

Hi, I'm David. And I'm here to have you to answer your question. So let me bring up your caution here. Now in discussion we have a company employs a certain number of the employees. So the first person to be with the withdraw with the church. David, Henry feeling smithson, Penny and Bella. So we have 123 4567 and eight people here. And dan. Uh They wouldn't choose select a group of three people among the people. And then if the three employees will, who will receive advance channing are selected at random. The probability done limited church with Henry immutable uh will be selected for advanced training. Now energy find this probability the first time we did you find the doctor number uh possible. It's just fine. A group of three. Now we will have to leave eight people. And then we want to select only three. So will be the combination under eight three. And this one is equal to the in Victoria. And you're running by three Victoria. Eight monastery Victoria. And if we compare this one, we have to choose three. And then we can go to the 56 total number of the combinations. And now we see that the one we want to be the meter judgment Henry interpreter must be in the group. Therefore the probability. Do you have the mid uh church? Yeah. And meet uh Henry and the middle uh Bella uh on the list on the group Which we called you one even by 56. And that will be the answer


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