A sample space consists of 161 separate events that are equally likely. What is the probability of each?
365970
1/161\nTake the number of ways the event can occur divided by the total number of events.
735176
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 ;-)
242656
This one is easy-assume a 50% chance each time, so its 1/2 X 1/2 X 1/2 X 1/2, so 1 chance in 16.  There is another way of figuring this, using combinations, like sn777 did, but sn777 made a logical error.  There are these possible combinations: 4 m, 3m 1F, 2m, 2F, 1m 3F, and 4F. So just looking at that, you would think its 25%. BUT, there is only one combination of 4M, while there are 4 combinations that make 3m 1F, and 6 combinations that make 2m and 2F, etc-so the total possible combinations of 4 children is 16, and 4M is only one of those.
600596
There are 6 possible outcomes for the roll. There is only one outcome when the number is 4. Therfore the chances of getting a 4 are 1/6.
834182
sample space=8\nnumber of outcomes with at\nleast 2 heads=4\n\nP(at least 2 heads)=4/8=1/2\n\nodds:even money\n\ni hope that this helps
158553
Ya it is "Sample Space"\n\nhttp://www.futureaccountant.com/probability/study-notes/\nhttp://www.schoolingkids.com/probability/study-notes/
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.
440605
Binomial Distribution, \nPoisson Distribution, \nNormal Distribution\n\nhttp://www.futureaccountant.com/probability/
1143172
Suppose the people are p1, p2, p3, p4, p5, p6, p7.\nThe probability that p1 was born on Monday and nobody else was is:\n1/7 * (6/7)^6 where 1/7 is the probability a person is born on Monday and 6/7 is the probability someone isn't born on Monday.\nThe same is true for the probability that p2 is born on Monday and nobody else is, etc.\nSo the total is\n7 * (1/7)*(6/7)^6\n= (6/7)^6\n= 46656/117649\n= .3965694565
1186248
In this case 'N', which is the sample size, is 120.  The probability of an event occuring is equal to the number of times it occurred in the sample divided by N, so:\n\nprobability of X=0:  20/120 = 1/6 ~ 0.1667\nprobability of X=1:  65/120 = 13/24 ~ 0.5417\nprobability of X=2:  25/120 = 5/24 ~ 0.2083\nprobability of X=3:  10/120 = 1/12 ~ 0.0833\n\nAlso the average value of X would be 1.2082 using this probability distribution through an infinite number of line checks.
458039
In the experiment of testing for the number of errors in a chapter...\n\nLet the event an error appearing in a chapter be a success..\n\nLet "x" represent the number of success.\nThe number of errors in a chapter can be either 0,1,2,...\n==> "x" can carry the values 0,1,2,3,...\n==> "x" is a discrete random variable with range = {0,1,2,..} [Countably infinite]\n\nThe probability distribution of "x" is a poisson distribution. \n[Since there are a large number of trials all of which are identical and independent with only two possible outcomes in each trial, either a success (there being an error) or a failure (there not being an error), with the probability of success in all the trials remaining the same all throught ]\n\nThe probability function of a poisson distribution\n==> P(X = x) = {e^(-m) * m^(x)}/x!\n \nParameter of the distribution\n==> Mean of the distribution\n==> Mean number of errors per chapter =0.8 \n==> m = 0.8\n\nTherefore P(X = x) = {e^(-0.8) * (0.8)^(x)}/x!\n\nProbability that there are less than 2 errors in a chapter \n==> P(x < 2) \n= P(x=0) + P(x=1)\n= {e^(-0.8) * (0.8)^(0)}/0! + {e^(-0.8) * (0.8)^(1)}/1!\n= {e^(-0.8) * 1)/1 + {e^(-0.8) * (0.8)}/1\n= e^(-0.8) + {e^(-0.8) * (0.8)}\n= e^(-0.8) [ 1 + 0.8]\n= e^(-0.8) [1.8]\n\n\nhttp://www.futureaccountant.com/probability/
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)}
624712
This problem is solvable.\nProbability of event A is written P(A).\n\nP(A)= Total # of outcomes that correspond to event A (divided by -or- over) Total # of outcomes in sample\n\nAnswer:\nP(A) = 12/20 = 3/5
242742
Remember, chance events are not dependent on the previous outcome.  The correct answer, as others have pointed out, is 50%.
631161
This is a binomial probability.\nP(success) = 1/5, P(failure) = 4/5.  There are 6 events here, so P(2 successs) = (1/5)(1/5)(4/5)(4/5)(4/5)(4/5).\nP(at least 2 successes) = P(2 successes) + P93 successes) + . . . . P(6 successes) or 1 - P(O successes) + P(1 success) + P(2 successes) which you already computed above.\nLet me know if you don't understand.
855670
Boy, this is tricky.  I think the probability is (8! * 41c2 * 10c2)/49!, where 41c2 = 41!/39!2! and 10c2 = 10!/8!2!  In this number, 8! represents the correct picks, 41c2 the incorrect picks, and 10c2 the number of ways to order the picks.\n\nFor exactly seven correct numbers in 10 picks , only a minor change is needed.  It becomes (7! * 42c3 * 10c3)/49!.  Follow this pattern for 6 of 10, 5 of 10, etc
577657
since it is "at least", use 1-(probability of no aces or no kings), so:\n\n1-((44/52)*(43/51))\n\nfor the second problem, I would use the idea of (probability of no orange + probability of one orange), so:\n\nprob of no orange:\n((5/8)*(4/7)*(3/6)*(2/5))\n\nprob of one orange, assuming you've learned combinations and permutations:\n(3C1*5C3)/8C4\n\nadd the two prob from above.\n\nhope this helps...probabilty can be quite ambiguous.