How can I calculate the probability of a number combination?
146975
If you know the area code and nothing else, then it's a 7-digit number and so 10 million combinations.  If you know the area code and the first three numbers then there are 4 digits or 10,000 numbers.
938617
You can read off the outcomes from Pascal's Triangle. For 5 tosses consult the fifth row\n\n........1  1\n.......1  2  1\n......1  3  3  1\n.....1  4  6  4  1\n....1 5 10 10 5 1\n\nThere is \n1 chance in 32 of 5 heads and 0 tails\n5 chances in 32 of 4 heads and 1 tail\n10 chances in 32 of 3 heads and 2 tails\n10 chances in 32 of 2 heads and 3 tails\n5 chances in 32 of 1 head and 4 tails\n1 chance in 32 of 0 heads and 5 tails
998556
Find the probability of the complement occurring and then subtract that from 1.\nThe complement would be that an odd occurs 0 or 1 times.\nP(odd) = .5 and P(even) = .5\nP(0 odds) = C(10,0)(.5)^0 (.5)^10 = .0009765625\nP(1 odd) = C(10,1) (.5)^1 (.5)^9 = .009765625\n.0009765625 + .009765625 = .0107421875\n1 - .0107421875 = .9892578125
590394
In 36 possibles outcomes\nprobability of doubles\n1-1\n2-2\n3-3\n4-4\n5-5\n6-6\n6 in 36 or 1 in 6 or 16,666%\nOf the sum being 4\n1 +3\n2+2\n3+1\n3 in 36 or 1 in 12 or 8,333%\nIf it is combined probability you can sum the two\n16,66+8,33=25%
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).
911469
I'll do some for you to show how they are done. In all cases probability P = #number of good throws/ # of all throws.\nAlso 1-P = #number of bad throws/ # of all throws\nOften you can arrive at a solution by considering the dices one by one.\n1) First dice can be everyting, second dice can be one of the five numbers not yet thrown, third dice can be one of the four numbers not yet thrown, etc. This leads to P = (6/6) * (5/6) * ... *(1/6) =720/46656 = 0.0154\n2) Divide the dice in three pairs, named A,B and C. The probability to throw a pair of two's with pair A is 1/36. Idem for B and C. But we have 6 posibilities to arrange the pairs ABC, ACB, ... And we have 6x5x4x3x2x1 possibilities to arrange the dice into the three pairs. Therefore P = 6 x 720 x (1/36)^3 = 0.0926\n3) First dice 1, second dice 1, third dice 1 gives a probability of (1/6)^3 = 1/216. But there are 20 possibilities to arrange the dice.\nP=20/216 = 0.0926
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: )
727752
Number of all permutations = 8!\n\nNumber of all permutations with 5 boys first (then 3 girls) = 5!3!\nNumber of all permutations with 5 boys last (3 girls first ) = 5!3!\n\nSo the probability in question is 2*5!3!/8! = 1/28
681148
Combination is a very wide word in this situation. 50%-%50 or %30-%70 or something else. \nand most probably there will be nothing.\nI guesss you are waiting for this answer: Spiderman!
460997
If you want a number less than three,then the likely numbers are:1 and 2\nThe probability of throwing a 1 is 1/6\nThe probability of throwing a 2 is 1/6\nThese two events are mutually exclussive,therefore sum up their respective probabilities. \n1/6+1/6=1/3\nSo,the probability of obtaining a number less than 3 is 1/3
540053
In the experiment of tossing a coin ten times.\nTotal number of possible choices = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 \n\n      = 1024 \n => n = 1024\n\nLet A be the event of getting number of heads is double the number of tails.\n\nFor event A\n------------\nNumber of favourable choices  =  Number of chances of getting heads is double the number of tailes.\n\n                  = No chances for occurance of event A\n                  = 0\n           =>  m  = 0\n\nProbablity of occurance of an event A =  Number of favourable choices / total number of possible choices\n     => P(A) =  m/n\n     => P(A) =  0/1024\n     => P(A) =  0\nTherefore the probability is zero.\n\n\n\nFor all about probabilty:\n\nhttp://www.futureaccountant.com
938348
There are many combinations. One of them is 1,2,3,4,5,16. Is there some specific property you wanted these six numbers to have?
292028
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
733920
a) There are 16 combinations and 4 of them include exactly 3 heads (HHHT, HHTH, HTHH, THHH).  So the probability is 4/16 or 1/4.\n\nb) This question results in the same probability (4/16 = 1/4), except it asks for it as 'odds' which is the probability of success divided by the probability of failure.  That is 1 to 3 odds.\n\nc) The odds against heads are just the reverse.  Probability of failure divided by probability of success.  That is 3 to 1 odds against.
939050
Assuming order matters on the first 5 numbers, there are 39 x 39 x 39 x 39 x 39 x 19 combinations of draws.  Only one matches with your selection making the odds\n1 in 1,714,259,781
976734
This will be an exercise in factorials.  Generally, if you choose one from the list of 4 senior partners, there are (4 choose 1) ways to do this, which equals 4!/(1!3!) = (4x3x2x1)/(3x2x1) = 4.\n\nIf you choose 1 senior and 1 junior partner, you multiply the two selection numbers together: specifically 4 x 6 = 24.\n\nHere we go:\n\nOne senior partner trio: SJJ : 4!/(1!3!) x 6!/(2!4!) = 4 x 15 = 60.\nTwo senior partner trio : SSJ: 4!(2!2!) x 6!/(1!5!) = 6 x 6 = 36.\nAll senior partner trio: SSS: 4!/(3!1!) = 4.\n\nTotal = 60 + 36 + 4 = 100 different groups.\n\nHaving seen this you can trust me on your other probability question.
249369
Is the term you are looking for "probability"?\n\nI hope this helps.