when this problem were given a scenario where Michelle is flipping coins and she ‘s keeping track of her streaks of heads. We ‘re told that we start with X of nothing is zero, and then x of em for normality is greater than or equal to. One is adequate to the size of the mottle. One party were given that she makes seven flips, her 1st 7 flips and we have heads, tales, three heads and details and then heads. So the room this works is that on the foremost flip, she has heads, so her mottle is of size one. But then she gets a tales which breaks the mottle. So X of An is back to nothing. then she gets ahead. So the streak is one. then a irregular heading. The stripe is, besides 1/3 promontory. The streak is three, and then she hit the tales. So the streak goes back to zero, so we can call this except one that ‘s the size of the head mottle after one flip, except to except three and so on. And the begin charge was X of zero, and we ‘re told that that is equal to zero. therefore after one flip except N X of one is one. then the street gets reset back to nothing. then the stripe becomes one. Increases to two increases to three gets reset back to zero, and it ‘s up to one again. So these numbers here are our adam of N values for N is equal to nothing. Up to seven for Part B were asked if this is a mark of chain and the answer is yes. And the reason is is because it follows the Markov place, which basically means that in order to know where we ‘re going to go in the following passage, we entirely know we only have to know where we are in the stream state. We do n’t have to consider the past. entirely the current state of matter is needed in order to know what the possibilities are for, what will happen future and what they ‘re probabilities are. For exemplar, say we ‘re at except for in this situation, and our head streak is excessively well, there ‘s only two things that can happen hera. We can flip another heads, in which event we increase to three. Or we can flip a tales in which, in which character, the street goes down to zero. And if we know that probabilities for heads and tails, then we know what all the possibilities are for the future transition and what those probabilities of those possibilities are. And it does not matter what has happened in the past. All we need to know is our current ten of em. If the current state is zero, we know that merely two things gon na happen sabotage increase to one if we flip a heads or we can stay at zero if we flip the tails. thus next we have see where were asked to identify the state of matter distance of the chain. And so the mottle size could be anywhere from zero up to four. We ‘re told that once Michelle hits a streak of four, she stops flipping so it barely stays at four perpetually thereafter, and then for party were told that the probability of flipping a head is P and were asked to determine the one step transition probabilities for the chain and to draw the department of state diagram for that chain. I think it ‘s actually easier to do this in the inverse ordering this time, so let ‘s let ‘s start with the diagram so we can have a stripe of zero stripe of one, 23 and four. thus if we ‘re at zero, the probability of the following footstep being have a streak of one is the probability that we flip ahead, which is given to us as p. And the early matter that could happen is that we could flip it tails, in which case our mottle transitions to zero. It remains at zero, and the probability of tales is consequently one minus p. If our streak is one, the probability of increasing it to two on the future flip is P. And we can say that the wholly way up the chain. Any time that we flip a tails, we go binding to zero. So we know that there ‘s transition from 120 that is flipping a chase when you ‘re at one and from 2 to 0 and from 3 to 0. And these air all one minus phosphorus because they ‘re all pendent on flipping a narrative. once Michelle gets to a streak of four heads, she stops flipping. So we stay at four everlastingly then, so that probability is one. So that ‘s TheStreet diagram where this Jane, So to write out the passage probabilities we have a probability from 0 to 0 is adequate to one subtraction p. That ‘s this justly here. The probability of going from 0 to 1 is P. It ‘s this probability veracious hera. And those were the only two possibilities from zero probability of going from 1 to 0 is equal to one subtraction phosphorus, and you can besides go from 1 to 2 phosphorus and you can not stay at one. You can either only go back to zero or go to two. So that Cyril So these breeze the three possibilities for transitioning out of one. A one head streak going from 2 to 1 is adequate to one minus p. indeed nowadays for a streak of four, the probability of going from four instinct at four is one. Because by definition the game is over and Michelle stops flipping coins. So the most recent stripe remains at four. And therefore all other transitions are zero. And in truth, we should be writing Oh, transitions, all probability, one step probabilities. So we can besides go from 0 to 2. That would be zero 0 to 30 then I would just write the check, just positively recording and then write the rest and then show it to you. And so this is all the one tone transition probabilities. And remember, if you have a system with five states, which is what we have here 0123 and four, the number of one footfall transition probabilities should be five times 55 squared. So we have 25 one step transition probabilities.
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