Statistics probability question?
815956
An area of 0.99 has a z-szore of 2.33\n\nAll you have to do is to calculate the mean given the value of x, s and z.\n\nz = (x - mean)/s\nmean  = x - sz\nmean = 6 - (0.2)(2.33)\nmean = 5.534 oz.
799156
Use Binomial distribution:\n\n(n nCR x) p^x (1-p)^(n-x)\n\nn nCR x = n!/(x!(n-x)!)\np=probabiliy (.10)\nn=patients (10)\nx= desired\n\na) P(X<=2) = P(X=0) + P(X=1) + P (X=2)\nb) 1- P (X<=2)... 1-prevoius answer
515436
plz complete the question. i like to solve the questions on probability, but yur question is incomplete.
765221
CASE 1:\nThe probability that 2 green balls are selected from box A:\n1/2 (chance of picking box A)\nChance of picking first ball green (3/6)\nChance of picking second ball green (2/5)\n\nSo the combined odds are: 1/2 x 3/6 x 2/5 = 1/4 x 2/5 = 1/10\n\nCASE 2:\nThe probability that 2 red balls are selected from box B:\n1/2 (chance of picking box B)\nChance of picking first ball red (4/9)\nChance of picking second ball red (3/8)\n\nSo the combined odds are: 1/2 x 4/9 x 3/8 = 12/72 = 1/6
1169384
The previous responders are incorrect. \n\nLet the expected length of the game (in tosses) be L. We can classify each game into one of three starting sequences (in the following table P=probability,E=expectation value for total length of game):\n\nH --> P = 1/2 , E = L+1\nTH --> P = 1/4 , E = L+2\nTT --> P = 1/4 , E = 2\n\nThe first two cases are equivalent to starting the game over after 1 and 2 tosses respectively.\n\nSumming P*E for these three events gives L:\n\nL =(1/2)*(L+1) + (1/4)*(L+2) + (1/4)*(2)\n= (3/4)L + 3/2\n\n--> L = 6
906595
C.\nThis is a binomial probability problem.\nThe binomial formula says \nP(r out of n) = (nCr)(p^r)(q^(n-r)), where n is the number of trials (here 6 rolls), r is the number of successes (here 0, 1, 2, and 3 twos), p is the probability of success (here 1/6), and q is the probability of failure (here 5/6)\nYou need to plug in 0, 1, 2, and 3 for "r" into this formula.  If you do, you get:\nP(0 twos) = .3348979767\nP(1 two) = .401877572\nP(2 twos) = .200938786\nP(4 twos) = .05358367627\nYou add up these answers to get your final answer, about .9913
522796
a) Sensitivity = P(Test + | Has disease), so out of the people who have the disease, 93.4% would test positive for it.\n\nSpecificity = P(Test - | Doesn't have disease), so out of the people who don't have the disease, 98.6% would test negative for it.\n\nb) This is again a Bayes' Rule problem\nP(T+|D) = 0.934\nP(T-|ND) = 0.968\nP(D) = 0.002\n\nP(D|T+) = P(D)P(T+|D)/(P(D)P(T+|D) + P(ND)P(T-|ND))\n= (0.002)(0.934)/(0.002(0.934) + 0.998(0.968))
354101
This is easy, just say there are 100 shaft holders.  So of A, 50 of them are processed there, and 3% don't work, so 1.5 out of 50.  For B, it is 30 x 4%=1.2, and C is 20 x 5%=1.  \n\nSo out of a total 100, 3.7 don't work, the the probability that this one chosen at random wont work is 3.7%.  \n\nHopefully the way i calculated the answer helps you out in the future.
630810
a) probability of rolling a 2 is 1 out 6. for 5 balanced dice\n(1/6)(1/6)(1/6)(1/6)(1/6) = (1/7776)\n\nb) isnt this the same as part a?
672123
The probability of A is 26%.  Probability of B is 42%.  Both A and B is 7%.  Probability of neither = 100% - [(26%+42%)-7%] = 39%.  Or in your format, 0.39.
883654
If the claim is correct, then the mean of samples of size 25 has a normal distribution with a mean of 10 (the population mean) ounces and a standard deviation of 0.5/sqrt(25) = 0.1.\n\nz = (10.15-10)/0.1 = 1.5\n\nAccording to the standard normal table, the probability is 0.0669.\n\nThis is due to the fact that the population is normal.\n\nedit: You guys, it is not three sd's above.  You have to think in terms of standard error.  It is 1.5 standad errors above.  Do the math.
284048
In the experiment of rolling two dice\n\nTotal number of possible choices = 36 ==> n=36\n\nLet "A" be the event of rolling the dice such that the sum on the dice is 5\n\nFor Event "A"\n\nNumber of favorable choices = 4  ==> m = 4\n[(1,4), (2,3), (3,2), (4,1)]\n\nProbability of getting a sum of 5 on rolling the dice \n==> Probabilty of occurance of event "A" = m/n\n==> P(A) = 4/36\n==> P(A) = 1/9\n\nhttp://www.futureaccountant.com/\nhttp://www.schoolingkids.com/probability/
1097980
0.8037 or 80.37%\n\nThis is why:\n\nP(E)=1-P(not E); so the probability of at least 2 people having the same number is 1- the probability that each person has a unique number.\n\nThe probability that each person has a unique number:\n\nFirst person has 100 choices out of 100 numbers, so 100/100.\nSecond person can't choose the first person's number, so only 99 choices out of 100, so 99/100;\nThird only 98/100;\n. . .\n18th person, has (100-17)/100=84/100\n\nTherefore the probability that none of these people have the same number is (100/100)•(99/100)•(98/100)•(97/100)•. . .•(84/100) ≈19.63%\n\nSo the probability that at least two people have the same number is 100%-19.63%=80.37%.
1119536
1) If no one is injured, then no subs ever get to play.  The probability is (.9)^11 = .3138106 or about 31.38%.\n\n2) (.9)^9 * (.1)^2 = .003874205 or about 0.39%\n\n3) "At least" is best calculated as the complement of "no substitutes were needed."  In the first calculation we determined the probability of not needing any substitutes.  So the probability of needing at least one sub would be .6861894 or about 68.62%\n\n4) The trick here is that the two injured could both be players, or one could be a player and the other could be the sub.  (They could not both be subs, however.)  The calculation in #2 (above) is part of this--the probability that two of the players were injured.  For the other part we need to know the probability that one player is injured (while 10 are not) and that the sub is also injured.  .1 * (.9)^10 * .15 = .0052302.  So the overall result for either of these outcomes (add this result with the result from #2) is 0.00910438 or about 0.91%.
770080
Let us call x the minutes hand on the clock.\nIf 0<x<=5, then it is train B. (5 minutes)\nIf 5<x<=12, then it is train A. (7 minutes)\nIf 12<x<= 24, then it is train A. (12 minutes)\nIf 24<x<= 25, then it is train B. (1 minute)\nIf 25<x<= 45, then it is train B. (20 minutes)\nIf 45<x<=48, then it is train A. (3 minutes)\nIf 48<x<=60, then it is train A. (12 minutes)\n\nWe add it up to get that for 34 of the minutes, he gets on A, and that for of the 26 minutes he gets on B. There are 60 minutes in an hour, so the probability is 34/60, which reduces to 17/30.
882956
Probability is 3 out of 5
777500
1) \n    a) The probability that the first one is not right is 7/8, the next is\n6/7 since there are 6 chances of getting the wrong one out of the remaining 7. The next is 5/6, then 4/5, 3/4,2/3,1/2. The probability of all these events happening is\n 7!/8! = 1/8\nMy colleagues who answered prior to my answer fail to\nrealize the difference between dependent and independent\ndistribution. In this case we have dependent distribution since each time a letter and envelope is matched we have 1 less\nof each to choose from.\n\n\nb) The probability that at least one envelope right is the additive\ninverse of the probabilty of no envelope getting it right based on\nall probabilities which 1. So 1 - 1/8 = 7/8.\n\n2)  The probability that no 2 have the same birthday is  computed like this. Based on a 365 day year just to make it reasonable. \n  First person - any day. \n Second - 364/365 'cuz 1 day is eliminated by the first one\nThird -363/365   'cuz 2 days are eliminated by the first two\n etc.\n The probability of no two the same is therefore\n (364)(363)(362)...(346)/365^19 = .588