1. A suspect is arrested. There is a crime stain of blood group B. The suspect is blood group B. The probability of a random person being blood group B is 0.1.\n\nLet E be the evidence that the suspect's blood group is the same as the crime stain, and G be the hypothesis that he is guilty.\n\na)How would you estimate P(E | G) ? Is this foolproof? \n\nb)Find the odds that the suspect is guilty, given the evidence that his blood group is the same as the crime stain, in terms of the prior odds that he is guilty.\n\nc)If there are 10^6 people in the relevant population, assuming equal probability of any person being guilty, what is the posterior probability of the suspect being guilty?\n\nd)If it is assumed that the prior probability of guilt equals the prior probability of innocence, what would be the posterior probability of guilt?\n\ne)What is the assumed prior probability of being guilty if  it is assumed that \nP (not G / E)= P(E / not G )?\n\nmany thanks...
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Here is a Bayes Theorem Calculator\nhttp://psych.fullerton.edu/mbirnbaum/bayes/BayesCalc.htm\nhere is another one\nhttp://members.aol.com/johnp71/bayes.html\n\nHere is a Bayes theorem tutorial\nhttp://people.hofstra.edu/faculty/Stefan_Waner/RealWorld/tutorialsf3/unit6_6.html
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D)6\n\nNote the pattern:\nIf f(1) = a, f(2) = b, then\nf(3) = b-a\nf(4) = -a\nf(5) = -b\nf(6) = -b+a\nf(7) = a\nf(8) = b\n\nAt f(7), the pattern starts repeating.  Therefore, the six values are f(1) through f(6) above.\nThis really isn't a permutation/combination problem, but you can often find good answers here on Yahoo Answers!  If you have specific problems with which you have difficulty, why dont you post a few related ones in one question here?  I'm sure I or others should be able to help.\n\nYOUR FOLLOW-UP:\nThis is two small problems in one.\nFirst, find the # of ways to choose the order of the backgrounds.  We have 3 choices for the first one, then 2 choices for the second, then 1 left for the third.  Therefore, there are 3*2*1=6 ways to choose the order of the backgrounds.\n{ (b, e, r), (b, r, e), (e, b, r), (e, r, b), (r, b, e), ( r, e, b) }\nSecond, there would be 3*2*1 = 6 ways to interview the business people.  There would be 2*1 = 2 ways to interview the education people, and just 1 way to interview the recreation person.  So that would make 6*2*1 = 12 ways to interview the people after the background order is set.\n\nThe total number of ways is then 6*12=72 ways to interview all 6.
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5 out of the 8 items in Box A are non-defective.\n3 out of the 5 items in Box B are non-defective.\n\nThe probability that an item from Box A and an item from Box B are drawn and are both non-defective is:\n\n5/8 * 3/5 = 3/8\n\nThe probability that an item from Box A and an item drawn from Box B and both defective is:\n\n3/8 * 2/5 = 3/20.\n\nThis means that the probability that one of the two items drawn is defective is:\n\n1 - p(both are not defective) - p(both are defective)\n=1 - 3/8 - 3/20\n= 19/40\n\nWe can see this by showing all possible outcomes. The items in Box A are shown on the top edge, and the items in Box B are shown on the right edge\n\nN  N  N  N  N  D  D D\nA  A  A   A  A  B  B  B  N\nA  A  A   A  A  B  B  B  N\nA  A  A   A  A  B  B  B  N\nB  B  B  B  B  C  C  C D\nB  B  B  B  B  C  C  C D\n\nThe events marked "A" represent 2 non-defective items. The events marked "B" represent 1 defective and 1 non-defective item. The events marked "C" represent 2 defective items.\n\nOut of 19 events where there is 1 defective and 1 non-defective item, the defective item came from box A 9 times.\n\nSo if 1 item is defective and 1 is not, the probability is 9/19 that it came from box A
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P(Last is W) = P{(2nd is W and 1st is W) or (2nd is B and 1st is W)}\n= P(2nd is W and 1st is W) + P(2nd is B and 1st is W) (since these are mutually exclusive events)\n= P(2nd is W)P(1st is W|2nd is W) + P(2nd is B)P(1st is W|2nd is B) (both use the general multiplication rule)\n= (6/10)(7/11) + (4/10)(6/11)\n= (42+24)/110\n= 66/110\n= 0.6\n\nWhich is the probability of getting a white from either box, which I think makes sense, since both have the same proportion of whites.
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For the first part, per hour the rate is 30.  so we have lambda = 30, so we know the mean and variance (look at the answer before mine).  To find the Probability the 240th person enters in the 8th hour, find the probability 240 or more people arrive in an 8 hour day, then times that by the probability less than 240 arrive in 7 hours.\n\nso (1-P(<240 in 8))*P(<240 in 7)\n\nRemember, The poisson process is memoryless.\n\nSteve
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From the experimental tests we know the following.\n\nP(test positive| is really positive) = 0.90\nP(test negative| is really positive) = 0.10\nP(test positive| is really negative) = 0.15\nP(test negative| is really negative) = 0.85\n\nWe are told the 0.40 of the individuals is cancer free.\n\nWe want to know what the probability that a randomally chosen individual who tests positive is re
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a) Can't sell more than 4, so 1*p(selling 1) + 2*p(selling 2) + 3p(selling 3) + 4p(selling4).\n\nb) Find the number n so that p(x>n) is less than 0.05 for a Poisson "vif" mean 3.
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.................... / ....... &#92;\n................ 0.85 ..... 0.15 (Pass test 1)\n.............. / ....... &#92;\n.......... 0.85 ..... 0.15 (Pass test 2)\n........ / ....... &#92;\n... 0.85 ..... 0.15 (Pass test 3)\n(Fail all 3)\n\n1) How many outcomes does your tree show?\n\nThere are four outcomes:\nP(Pass Test 1) = 0.15 = 15%\nP(Pass Test 2) = 0.85 * 0.15 = 12.75%\nP(Pass Test 3) = 0.85 * 0.85 * 0.15 = 10.8375%\nP(Fail all 3) = 0.85 * 0.85 * 0.85 = 61.4125%\n\n2) What is the probability she will eventually pass an exam.  \n\nThis is the sum of all outcomes where she passes:\nP(Pass Test 1) + P(Pass Test 2) + P(Pass Test 3)\n= 38.5875%\n\nAlternatively you could take the 1 - P(Fails all 3) and get the same answer.\n\n3) What is the probability she will take the exam three times?\n\nIt didn't say whether she passes or fails, so the chance she has to take the test a third time is 0.85 * 0.85 = 72.25%\n\n4) What is the expected number of times she will take the exam?\n15% of the time, she'll take it once\n12.75% of the time, she'll take it twice\n72.25% of the time, she'll take it three times.\n\nSo the expected number of times is:\n0.15 * 1 + 0.1275 * 2 + 0.7225 * 3 = 2.5725\n\nSo she's expected to take it 2.5725 times.\n\n5) What is the expected number of times she will fail the exam?\n15% of the time she will pass test 1 (failures = 0)\n12.75% of the time she will pass test 2 (failures = 1)\n10.8375% of the time she will pass test 3 (failures = 2)\n61.4125% of the time she will fail all 3 (failures = 3)\n\n0.15 x 0 + 0.1275 x 1 + 0.108375 x 2 + 0.614125 x 3 = 2.186625\n\nSo she's expected to fail it 2.186625 times.
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They have to do this because only certain parts of DNA are used to ID people. And so much of DNA is similar they prove identity with statistics. So the likelihood of the DNA being the specific person is 1 in 20000 and that linked with other evidence is often more than enough to prove guilt or whatever. There is a law of probability and when things are put together it is very reliable.\n\nOk I am adding on in response to your question. I have consulted a textbook and think that the test described above is pretty conclusive. The textbook describes running a gel with the DNA fragments on it. They space out and are very obviously different to each other and are compared to the original. (see example of gel here http://gallery.hd.org/_c/natural-science/_more1999/_more05/DNA-genetic-fingerprinting-dark-bands-in-gel-are-almost-unique-for-a-given-individual-tweaked-1-AJHD.jpg.html) . Another thing mentioned in my textbook is taking samples from commnunities to compare blood samples. So one could suggest that they took samples of blood from 20000 people in the neighbourhood and only the one persons blood dna fingerprint matched up.
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Count the total number of groups:\n9 choose 5 = 9*8*7*6/(4*3*2*1) = 9*2*7 = 126.\n\nb) Count the number of groups where 3 women where chosen and 2 men.\n(4 choose 3) * (5 choose 2) = 4 * 10 = 40.\n\nThe prob that a random committee has exactly 3 women = 40/126\n\nc) If there were no women then the number of ways this can happen is:\n(4 choose 0) * (5 choose 5).  As you could probably already see, there is only 1 way to have a committe of 5 men and 0 women.  The prob = 1/126.\n\nf) >=3 women is the same as saying 3 or 4 women as 4 women is the most you can have.\n\n(4 choose 3)*(5 choose 2) + (4 choose 4) * (5 choose 1)=\n4*10 + 1*5 = 45.  Prob = 45/126.
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yea, multiplying the single event probabilities of two independent events to get the prob of the independent events conjunction is correct. In other words (to avoid you going to grab an Advil already): p(A&B)=p(A)*P(B) assuming A and B two independent events. That's BASIC stats.\n\nIn excel, use the function POWER. First argument the probability, second argument the number of occurence in a row.
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a. first draw doesnt matter since it's replaced; you're back at 3 in 10 chance\n\nb. two ways this could happen: first, 3/10 chance first ball is red, leaving you 2/9 chance then of second ball being red on second draw.  second, 7/10 chance you draw a white ball, leaving you with 3/9 chance of second ball being red.  (3/10*2/9)+(7/10*3/9) which leaves you with 3/10 chance. again.\n\n\nEDIT:\n\nsurprises me how many people can't do math.\n\nfor A), the second draw is completely independent from the first.  Doesn't matter if I took a dump in the bowl in the first draw; as long i fished it out, you're left again with the same start on the second draw-3 reds and 7 whites.  your probability doesn't change because of what happened in the first draw as the conditions are RESET.\n\nfor B), let's run through allll possible scenarios\nNumber the Balls\n\n1 - red\n2 - red\n3 - red\n4 - white\n5 - white\n6 - white\n7 - white\n8 - white\n9 - white\n10- white\n\nPossible draws (X denotes red second ball outcome)\n1,2 - X\n1,3 - X\n1,4\n1,5\n1,6\n1,7\n1,8\n1,9\n1,10\n(2 fav outcomes if ball 1 is drawn first)\n\n2,1 - X\n2,3 - X\n2,4 \n...\n2,10 \n(2 fav outcomes if ball 1 is drawn second)\n\n3,1 - X\n3,2 - X\n3,4 \n...\n3,10 \n(2 fav outcomes if ball 1 is drawn second)\n\n4,1 - X\n4,2 - X\n4,3 - X\n4,5 \n...\n4,10\n(3 fav outcomes if ball 4 is drawn first)\n\n...\n\n10,1 - X\n10,2 - X\n10,3 - X\n...\n10,8\n10,9\n(3 fav outcomes if balls 5-10 are drawn first)\n\n\nthere are 90 outcomes\nof which there are 2+2+2+3+3+3+3+3+3+3 favorable outcomes = 27.\n\n27/90 = 3/10\n\n\nSecond edit: guy below, 2*3 + 7*3 = 6 + 21 = 27, not 28
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That's a huge fish story.....true or not. My first question comes from the first arrest as why you didn't go down to the local police station and have your vehicle checked by a drug dog to avoid the second incident. I then wonder why, after the first arrest, you didn't head to your local doctor and get a blood test proving you were not using. This doesn't appear to have been done for the second arrest either. So, what you are left with is a mountain of evidence against you. Considering these are both felonies, any deferment, agreement, plea deal you make will instantly make you a convicted felon, a label that lasts a lifetime. Tread carefully grasshopper, you are in very dangerous waters.
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Okay, keep in mind that this problem is asking the probablility that AT LEAST 8 were married - that means the probability that 8 were married plus the probability that 9 were married plus the probablility that 10 were married.\n\nSo let's start with the probability that 8 were married:\n\nIf the probability that someone was never married is .15, that means the probability that they were married is 1 minus .15, or .85. And if 8 people in the sample were married, that means that 2 were not. So what you're going to do is multiply .85^8 (for married people) x .15^2 (for unmarried). \n\nThe principle is the same for 9 married people and ten married people: \n\nFor 9 married, you multiply .85^9 x .15^1\n\nFor 10 married, you multiply .85^10 x .15^0\n\nFinally, add all three answers together, which gives you the probability that either 8, 9, or 10 people in the sample were married, which is what the question was asking! Hope that helped!
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When you are computing the probability of a given event such as 2 heads, 2 tails, or one of each, you are essentially counting the number of ways a given event can occur and dividing this number by the number of all possible outcomes.\n\nIf you toss 2 coins once ( or one coin twice) there are 4 possible outcomes:\n\nHH = both heads\nHT = 1st heads, 2nd tails\nTH = 1st tails, 2nd heads\nTT = both tails\n\nAssuming the coins are "fair" each outcome is equally likely.\n\nThere is one way in four that two heads (or 2 tails) can happen so the probability for each of these outcomes is 1/4\nBut there are two ways you can have one head and one tail and so the probability of this happening is 2/4 =1/2. The sum of all the probabilities of the distinct outcomes will always equal 1. \n1/4 +1/4 + 1/2 =1\n\nHow about if you have 3 tosses? The 8 possible outcomes are as follows:\n\nHHH \nHHT\nHTH\nTHH\nHTT\nTHT\nTTH\nTTT\nWe see there is only one way to have 3 heads (or 3 tails) and so the probability of 3 heads (or 3 tails) = 1/8\nHowever there are 3 ways you can have 2 heads and one tail, so the probability of this happening is 3/8. And there are also 3 ways you can have one head and 2 tails, so 3/8 is the probability for this too.
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I got ans for (b).\n\n\nNone of the urns is empty. This is possible if every urn have one ball each. (n balls and n urns).\nSo nth urn should have nth ball (all others can't fall in nth urn). (n-1)th urn should have (n-1) ball (all less balls  can't fall in (n-1) urn and n is in nth urn). So ith ball will be in ith urn.\n\np(none empty) = p(all each one ball each)\n              = P(1 urn have 1st ball) * p (2 urn have 2nd ball) * .... * P(nth urn have nth ball)\n              = 1 * 1/2 * 1/3 * ... * 1/n\n              = 1/n!\n\nGood questions!\n\nWill think about (a) and let u know if I get the answer.
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Probability(1st person picks any bill)=8/8\n\nProbability(2nd person picks the same bill)=1/8\n\nProbability(3rd person picks the same bill)=1/8\n\nThe events are independant as the outcome of one event will not affect the outcome of another. Thus allowing multiplication of probabilities.\n\nProb(All 3 draw same bill) = 8/8*1/8*1/8\n\n=1/64