Trang chủ » blog » SOLVED:'(1Opts)A coin has two faces Head and Tail. (2pts)lf you toss the coin once, and record the up-face value, what is the sample space? (2) (ZptsJlf you toss the coin once, what is the probability

SOLVED:'(1Opts)A coin has two faces Head and Tail. (2pts)lf you toss the coin once, and record the up-face value, what is the sample space? (2) (ZptsJlf you toss the coin once, what is the probability

Video Transcript

Ask a wonder. We consider 10 independent sources of a biased coin in equal stone. And here it ‘s a bias coin with the probability of it at each venereal disease equal to be. then the probability of heads Just equals two bacillus. Which means the probability of tails must be 1 -2. Where B is from 0 to 1. Of class for depart one. Will it be the consequence that there are six heads in the first ethos is six hits in the inaugural eight doses. This is called event. And will it be the event that the nine stores results in it ? then we have hit in the one-ninth those. This is even bigger. It ‘s required to find the probability of B given A. And they expressed it in terms of B probability of B given A. Of naturally this vulnerability equals just the probability of getting heads which is B. Why is that ? Because we have said that these sources are independent. then whatever comes before the one-ninth dose which is nine hits in the first eight doses. Whatever comes before the ninth we have a probability to get heads equals to be in the dime bag stores and one minus P to get Sales in the nine stores. Let ‘s motivate to Bartow. What is the probability that there are three heads in the foremost four doses and two heads in the last three doses required to find the probability That there are three heads in the first gear for doses. And Notice This conclusion. two heads in the last three doses again we will use independence because this consequence considers the fortresses. And here it ‘s the final three doses there is no overlap between them. then we can transform and to be the intersection it ‘s the probability of this consequence was deployed by the probability of this consequence. We use independence here to take the intersection to be the multiplication. What is the probability to get three heads out of four doses ? We used binomial distribution with parameters four and be then it ‘s 4-3. We have we have this phone number of ways multiplied by the probability of heads. It ‘s B cube because we have cerritos is multiplied by one minus B. to the bar of N -1 which is one multiplied by this event. The precedence of this event from we choose city. Sorry It ‘s three choose to this act of combinations to have two heads. then it ‘s B squared squared because two heads Multiply it by 1 -7 is a set of one. Four choose three. It ‘s on-key, it ‘s dependable. It ‘s the four, choose one. It ‘s four and three choose to its three choose one. It ‘s one it ‘s a three history. then it ‘s four multiplied by three gives you 12 multiplied by B. To the cake of three plus two is five Employed by 1 -7. There ‘s about of one plus one gives to Let ‘s move to depart three given that there were four heads in the first seven doses Required to find the probability that the second hits occurred at the 4th. again in the in the first seven texas. here we have seven 1234567. We have seven doses given that there were four heads in the beginning seven doses. here we have four heads. Find the probability that the second base hits occurred at the 4th. The second hit occurred at the at the 4th where the forces here, then this is given And we have four heads. If we can get the number of combinations that we can distribute these foreheads now we can find the ask probability. This is the second, which means before it We have one head and after it we have do it. Which means the want probability of party three. We can cut it part three. It equals the number of ways, the act of ways to get to get hits. The second hit in the force slattern divided by the issue of ways to get for it ‘s of course both for seven doses 47 doses. What is the count of ways to get The 2nd Head in the First Slot ? We notice here. Second hit means we have hit before it and two heads before it. What is the act of ways to get hit injustice ? Three doses injustice, three places. It ‘s three choose one. But the aristocratic eyed son. The other ways to arrange two heads here. It ‘s not arrangement each. Its survival to select two heads in these slots Or distribute two heads in these roots. It ‘s the three choose to divided by The numeral of ways to get foreheads and 7000. It ‘s just seven. It shows four denominator. We have city multiplied by city, it ‘s nine In the denominator we have seven shoes for its 35. And notice that we can get it using another way. We can get it. The probability divided by the probability relative event, divided by the probability of event. But here To get the probability we just multiply by the probability of getting four heads and three tails. Because in the nominator We have four heads then it will be Be it was about four supplied by on -7. It ‘s about of three and inter nominator it ‘s about four Because we got four heads in one-seventh and the remaining will be tails. We can notice that these consensus it ‘s better to get it relative to the act of ways And the answer is not invited by 35. last we are interest in calculating the probability that there are five heads in the first six doses And three hits in the last five doses. We will get it in terms of B. here we have 10 1234567. It then we have here on ( 234 ) 567 89. We want to add here we have 10 slots. What is already that there are five, it ‘s in the six. The first six. The first six here We need five heads hera. Five hits and three heads in the last five doses. Oops ! here we have an intersection which means this consequence is dependent on the other consequence. It ‘s three heads here. In club to get this, we will get it by getting two options and get it as it might be This to get five beds and three heads, it might be this is not the intersect slot might be heads or it might be tails and they get the accession between these two disabilities here we have fixed these to be hits which means I need here foreheads In the other five slots and here two heads and the other three slots, Sorry and the early four slots because the sum is five And here I need five hits. And here I need three heads. Let ‘s get these. now nowadays this event and this event is dependent because it ‘s here stops here and this is this one discontinue here And of path this one Is for the five slots here and this is for the four slots. These are independent. then the compulsory probability probability of board four just equals this. Plus this. What is the probability of this event ? It ‘s four. We have five. We have five slots, Choose 4. 4 heads to be chosen, multiplied by B. It was about four deploy by one subtraction B To the exponent of one Which is N -4. Five minus four multiplied boy getting heads here which is B multiplied by this event. These 3/1 are independent. then I can get the intersection by its multiplication. In the last event it ‘s required to get two hits in the remaining forest loads. then its for its use to were deployed by feather deployed by one minus B squared. This is for the first event. Plus we do the same. It ‘s 5 to choose five. Multiplied by P. Two out of five supplied by one rush to the measure of zero deployed by getting tails here Which is 1 -2. Multiplied by three. Sorry it ‘s four places. Four slots choose three heads were deployed by P cubed, Divided by one -P. We can rearrange this to be here. We have B. B. B. It ‘s seven and five. It shoots four. It ‘s 5 supplied by four to choose to. It ‘s certain 30 p. It was about five. Applied by 1 -7. is a separate of plus five inches. Five is one 4234. We have P Q. E and B. here, that ‘s share of it, Multiplied by 1 -7. It was about and this is the final examination answer of our trouble

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