How do you calculate probabilities when you know the variables?
407360
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.
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/
171633
probability is a number from 0 to 1 (or 0% to 100%) that describes how likely an event is to occur\nexample~ jenny had 10 marbles. 3 were red. 6 were blue. and 1 was white. she put them all in a bag. she randomely drew one out of the bag. what is the probability that she will pick blue?\n            answer= 6/10 which deducts to= 3/5\n\nactual events is what really did happen. \n\nprobability is guessing based on passed expirements or facts. actual event is a fact.
202632
The level of the student you are asking the question is to be taken into consideration for this....\n\n\ntry this site for your need...\n\n\nhttp://www.futureaccountant.com/probability/\n\n\n\nhttp://www.schoolingkids.com/
872621
well in my skool we take it as {<and rule>} and {<or rule>example:there is 5 white marble and 5 black marbles in a bag...u pick 2 randomly without replacing them in the bag.....wat is the probability that u get 2 white marbles==>this is a combination 2 the way it is solved is multiplying the the 2 combined probabilities\nanswer:pr:(white AND white)=(5/10)*(4/9)\n\nnow the set is this way....wat is the probability that u get 2 marbles of different colors in the first 2 picks===>set rule\n\nanswer:there is 2 combinations in which u can get 2 different colors there are {white then black} OR {black then white} u find the 2 combinations then ADD them\npr(WandB OR BandW)={(5/10)*(5/9)}+{(5/10)*(5/9)}
943382
If l,w and y are separate variables then\n(lw - lwy)/lw=lw/lw - lwy/lwy = 1 - y\n\nIf lw and lwy are each a vaiable then\n(lw - lwy)/lw= lw/lw - lwy/lw = 1-lwy/lw\n\nIf lw and y are variables then\n(lw - lwy)/lw = lw/lw - lwy/lw = 1 -y
152796
try this site...\n\nhttp://www.futureaccountant.com/probability/
114803
If you have two events "A" and "B"\n\nThe probability that\n\na) Both would occur = P(A intersection B)\nb) Only one would occur = P (A intersection B') + P (B intersection A')\nc) At least one of them would occur = P (A union B)\nd) None of them would occur = P (A' intersection B')\n\nTry this site for a good explanation on probability\n\n\nhttp://www.futureaccountant.com/probability/\nhttp://www.schoolingkids.com/
515436
plz complete the question. i like to solve the questions on probability, but yur question is incomplete.
953152
The binomial distribution is given by:\n\n    N!\n----------- x (p)^n x (1-p)^(N-n)\nn!(N-n!)\n\nWhere N is the number of trials, n is the number of successes and p is the probability of success\n\nIn this question, n, the "number of successes" is the number of failures caused by operator error. p, the "probability of success" is the probability that the failure is caused by operator error (i.e. 30%).\n\nSo, in the questions you need to substitute the following for each variable:\n\ni) N=20, p=0.3\n\nAs this question uses the phrase "at least", you will need to use the equation more than once and add all the results. Repeat the calucation for each n for n=10 to n=20 and add the results.\n\nI got the following values for each n (rounded to 8 d.p)\n\nn=10, 0.03081708\nn=11, 0.01200665\nn=12, 0.00385928\nn=13, 0.00101783\nn=14, 0.00021811\nn=15, 0.00003739\nn=16, 0.00000501\nn=17, 0.0000005\nn=18, 0.00000004\nn=19, 0.00000000\nn=20, 0.00000000\n\nTotal: 0.0479619 or 4.80% chance that at least 10 are caused by operator error\n\nii) N=20, p=0.3\n\nThis question is similar to q1, however you need to repeat the calculation for each n for n=0 to n=4 and add the results.\n\nI got the following values for each n (rounded to 8 d.p)\n\nn=0, 0.00079792\nn=1, 0.00683934\nn=2, 0.02784587\nn=3, 0.07160367\nn=4, 0.13042097\n\nTotal: 0.23750778 or 23.75% chance that no more than 4 are caused by operator error\n\niii) In the random sample, only 5 are errors. Work out the probability of this being the case using N=20, n=5, p=0.3. If it seems reasonable, then the 30% figure is probably right, and the sample just contains a low number of failures due to operation error. If the probability is impossibly small, then the 30% is not correct.\n\nThe probability I calculated for n=5 was 0.178863051. This indicates a 17.89% chance that there would be 5 errors in the sample. This is not so outrageous to suggest that 30% is an incorrect percentage of operator errors. It's just that in this sample, the number of actual errors was less than the number of expected errors (which is 30% x 20, or 6 errors).
460739
Here are the formulas:\nIf P(E) = a/b, then the odds for E are a to (b-a)\nTo find the odds against E just reverse the order.\nOdds against E are (b-a) to a\n\nIf the odds of E are a to b, then P(E) = a/(a+b)\n\nFor example,\nif P(R) = 1/10, then the odds for R are\n1 to (10-1) or\n1 to 9.  \nThus, the odds against R are 9 to 1.\n\nAnother example,  say the odds of R are 5 to 3.  \nThe probability of R is 5/(5+3) = 5/8.\nThe probability of not R would then be \n1 - 5/8\n= 3/8
430637
First choose which number it is.  There are 6 numbers, so there are C(6,1) = 6 choices.\nThe probability of that number is 1/6\nSince there are three\n(1/6)(1/6)(1/6) = 1/216\nNow multiply this times 6\n6/216 = 1/36
696155
Whats the probability that the problem is *not* solved? Tom doesn't solve it (0.75) and Alice doesn't (0.5) so its 0.75*0.5 = 0.375. So the probability it is solved is 1-0.375 = 0.625.\n\nNow, we are given that it is solved. What is the probability that Tom solves it? 0.25. \nSo:\nP(Tom | solved) = P(Tom and solved) / P(solved) = P(Tom) / P(solved) = 0.25/0.625 = 0.4.\n\n(edit - note that the reason the below answer is not quite correct is because it is possible for *both* of them to solve it - so the probability Tom solves it when solved + probability Alice solves it when solved should *not* equal 100% - it should be more.)
911284
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.
887044
to calculate just time the total of first choice and the total of second choice or until the total of n choice.\nor you can try this: if you toss n coins total probability is 2^n\nif you toss n dice total probability is 6^n\n\nor if you wanna know the probability just arrange it and try this\nex: the probability if you take two alfabets from A - I are\nAA,AB,AC,ETC\nBA,BB,BC,ETC\nETC\n\n: )
861788
There is no accurate way to do this because too many variables are involved. You can do like I do though and run a series of tests and then average them.\n\nhttp://www.toast.net/
867606
In order to measure probabilities, mathematicians have devised the following formula for finding the probability of an event.\n\nProbability Of An Event \nP(A) =    The Number Of Ways Event A Can Occur    \nThe Total Number Of Possible Outcomes  \n \nThe probability of event A is the number of ways event A can occur divided by the total number of possible outcomes.