Statistics- how to compute probability's?
452389
There are 2 rules you need to know. The addition rule and the multiplication rule\n\nIf we roll a die the chance of rolling 6 is 1/6.\n\nthe chance of rolling 5 is also 1/6\n\nthe chance of rolling either a 5 or a 6 is therefore 1/6 + 1/6 = 1/3\n\nAnd the chance of rolling 4, 5 or 6 is 3/6 or 1/2.\n\nWhen the events are alternative outcomes and you want to know the chance of the outcome being one of these outcomes, add the probability of each outcome.\n\nIf you roll a die twice (or two dice at once) the probability of rolling a par of ones is 1/36.\n\nThis involves the multiplication rule. The chance of the first die showing a 1 is 1/6, and the chance of the next die showing a 1 is also 1 in 6. \n\nOnly one-sixth of the time do you get the first die to give you the 1 you want and only one-sixth of those times will the second due give you another 1. So 1/6 x 1/6 = 1/36.\n\nHowever if the question is what is the probability of just one six being thrown in two rolls\n\nthen the probability of 1 on the first throw but not on the second throw is (1/6 x 5/6) = 5/36\n\nAnd the probability of 1 on the second throw but not on the first throw is (5/6 x 1/6) = 5/36 \n\nAnd so throwing a 1 on either roll but not the other roll has a total probability of 5/36 + 5/36 = 10/36.\n\nThe slightly different question "what is the probability of rolling one or more 1s in two rolls of the dice?" is the sum of\n\nprobability of one on the first throw but not the second\n\n+ probability of one on the second throw but not the first\n\n+ the probability of a double one being thrown\n\n5/36 + 5/36 + 1/36 = 11/36\n\nSo the rule is when it is a matter of alternative outcomes to one event, add the probability of each outcome.\n\nWhen it is a matter of the outcome of two events both happening, multiply the probability of each happening to get the probability of BOTH happening.
877981
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
276294
use the formula for confidence interval.. its found on most statistics book.. i would type the formula, but that wouldn't be helping you to be a better student..
1076864
Put the data of each set into bins according to their leading digit. Then according to Benford's law the distribution should be as follows:\nLeading digit Probability\n1 30.1%\n2 17.6%\n3 12.5%\n4 9.7%\n5 7.9%\n6 6.7%\n7 5.8%\n8 5.1%\n9 4.6%\n\nUsing a chi-squared or Kolmogorov test it is possible to determine which of the four data sets differs significantly from Benford's distribution.
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For and event, odds are ususally expressed as the number of possible outcomes that qualify as a success compared to the total number of possible outcomes: 1 in 6, 3 in 10, etc.\n\nProbability is the same thing expressed as a fraction:\n1/6 = .66... 3/10 = .3\n\nWhen you select k objects from a set of n objects, the number of possible ways that you can do this is n!/(k!(n-k)!).\n\n\nIn all of your problems, your event is selecting a set of 3 objects from a set of 16. The number of possible outcomes of this is: \n16!/(3!13!) = 16x15x14/6 = 8x5x14 = 560.\n\nThe number of outcomes that are considered to be a success are:\n\n3 forks (pick 3 forks from the set of 7) = 7!/(3!4!) = 7x6x5/6 = 35\n\n2 forks 1 knife  (pick 2 forks from the set of sevan and 1 knife from the set o 5) = 7!/(2!5!) x 5 = 7x6/2 X 5 = 21x5 = 105\n\n3 spoons (Select 3 spoons from the set of 4) = 4!/(3!1!) = 4\n\n1 fork, 1 knife, 1 spoon  (Select 1 fork from the set of seven, 1 knife from the set of 5 and one spoon from the set of 4) = 7!/(1!6!) x 5!/(1!4!) x 4!/(1!3!) = 7 x 5 x 4 = 140\n\n1. P(3 forks) = 35/560\n2. P(2 forks, 1 Knife) = 105/560\n3. P (3 spoons) = 4/560\n4. P(1 fork, 1 knife, 1 spoon) = 140/560
131088
try this site\n\nhttp://www.futureaccountant.com/probability/
917829
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.
749852
1.Probability of getting a number from 1-4 on throwing the dice\n         = 4/6\n\n2.Probability of a correct throw = 1/4\n\n3.Total number of questions = 160\n\nYour Answer = (4/6 * 1/4) raised to the power 160
627413
OK.  Here are the possibilities.  It is important to keep the order of the dice in mind, so you can think of this as rolling a die 5 times.\n\nSSSSD\nSSSDS\nSSDSS\nSDSSS\nDSSSS\n\nNow each of these will have the same probability, so if I can find the probability of one of them, I can take that and multiply by 5 since there are five combinations.\n\nSSSSD can also occur in different ways.  If S=1, then D could be 2, 3, 4, 5, or 6.  So if S=1, then the probability of this combination is (1/6)^4(5/6).\n\nNow if S=2, then D can be 1, 3, 4, 5, or 6.  So again, (1/6)^4(5/6).\n\nSo hopefully you can see that there are 6 ways that you can do this.  So the probability of SSSSD is 6(1/6)^4(5/6).\n\nSo overall, the probability of getting four of a kind (without getting five of a kind) is\n5*6*(1/6)^4(5/6) = 150/6^5.\n\nSo it looks like your method does work.  Good job; this isn't an easy on.  There is always more than one way to look at probabilities.
933392
Probability in statistics would agree with your claim to a point of certainty
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You would use monte Carlo simulation when you need to find the distribution of a variable, and this depends on other variables which are random, especially when the interact in a complicated way.\n\nRather than trying to work out an answer, you just run the model using lots of different random numbers as inputs.  The outputs should approximate your answer if you have enough inputs.\n\nThese can be checked by using solvable cases.  I.e. consider the average value when you through a dice =(1+2+3+4+5+6)/6=3.5\n\nYou could also attempt to answer this by Monte Carlo methods by throwing a dice lots of times and taking the average of your results.  If you make enough throws the answer should be close enough for all practical purposes.
17841
You might try DAP. You can find it here along with a lot of other free statistical software.\nhttp://members.aol.com/johnp71/javasta2.html
254656
I'm assuming you are talking about a probability problem in Bayes' theory since you haven't specified what any of your symbols mean. This is why there are no responses.\nThe answer is actually quite clear. Draw three circles in a triple intersection for the events R, S and I and label it clearly. The diagram simplifies to a certain double intersection that motivates the result.
1136735
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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P(selecting a person at random and he has the symptom and is detected by the procedure)\n\n= P(patient having it) and P(detecting it using the procedure)\n\n= (0.5/100)*(98/100)\n= 0.0049\n\nBut if the person is not randomly selected and you are sure s/he has the problem and its just a matter of detecting it with the procedure, then Probability = (98/100) = 0.98
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1)  I think you're missing a bit.\n\nGiven that P(W<=4)=0.825 then P(W>4) = 1-0.825 = 0.175\n\nThis is because all probabilities sum to 1.  So since <=4 annd >4 make up all teh possibilities the value could take - then they're total probabilties must be 1.\n\n2)  Expected value\nTo do these you simply have to multiply the probability by the value and total them up.\n\n+36,000×0.7 -6,000×0.3 = $23,400\n\n3)  Binomial\nYou have a bin(6, 1/6) as there are 6 trials and the probability of a success (ie getting a four) is 1/6.\n\nYou require the probability of 2 fours so just using the binomial formula P(X = x) = nCx × p^x × (1-p)^(n-x) you get:\n\nP(X = 2) = 6C2 × (1/6)² × (5/6)^4 = 0.2009\n\n4)  Binomial\n\nthis again is a binomial Bin(10, 0.33) - as we have 10 trials (students) and the probability of each one being a success (ie member of ethnic minority) is 0.33\n\nUse the formula above to get the probability of more than 6:\n\nP(X = 7) = 10C7 × 0.33^7 × 0.67^3 = 0.01538...\nP(X = 8) = \nP(X = 9) = \nP(X = 10) = \n\ntotalling these up gives the probability of more than 6.\nThen the probability of NOT more than 6, P(X <=6) = 1 - P(X>6) using the idea from 1.\n\n5)  The mean of teh binomial is just np (ie the number of trails time the probabilty of success).\n\nSo for your example where we have bin(6, 0.7) it will be 6×0.7 = 4.2 seeds germinating.
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Probability is the measure of how likely an event is. Is is given by the formula: \nP = (The Number Of Ways an Event Can Occur) divided by (The Total Number Of Possible Outcomes) \n\n1) To find the probability for a pen to be selected from the first box, you write the number of pens over the total number of items in the box. So: \n\n2 (number of pens) / 10 ( total items in the box) \n\nwhich reduces to 1/5. \n\n2) To find the probabilty for a marker to be selected from the second box, you do the same thing. \n2 (markers) / 5 ( total items in the box) which cannot be reduced. \n\nThe answers are 1/5 and 2/5. \nHope this helps ;-)