it is some theoty in my probability lesson, my teacher once metioned it and i just forgot that,  hope you can answer this problem for me, thank you.
119156
If a random experiment has two possible outcomes, success with proability  p and failure with probability q=1-p. And the experiment is repeated until a success happens. The number of trials before the success is a random variable  with density function\nf(x)=q^(x-1) * p     is called the geometric distribution.\n\n\nhttp://library.thinkquest.org/10030/6dpdgd.htm\n\nhttp://en.wikipedia.org/wiki/Geometric_distribution
222024
Find the probability P(X<19.7), the probability P(X>25.3), add those probabilities together and subtract that sum from 1.  That will be the answer.\nP(X<19.7) = Φ[(19.7-22.0)/2.4] = 0.17\nP(X>25.3) = Φ[(22.0-25.3)/2.4] = 0.08\ntherefore P(19.7<X<25.3) = 1-0.17-0.08 = 0.75
302886
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?
568976
Ahh!  True, 6 consecutive flips of a coin in which each lands heads has a (1/2)^6 probability of happening.  But what are the odds that the seventh flip will be heads, too?
507811
a) Can't sell more than 4, so 1*p(selling 1) + 2*p(selling 2) + 3p(selling 3) + 4p(selling4).\n\nb) Find the number n so that p(x>n) is less than 0.05 for a Poisson "vif" mean 3.
540991
This probability problem can be used using Binomial Distribution.  \n\nGiven a number of 'x' success, the probability in a binomial experiment is:\n\nP(x) = [n choose x]*p^x*q^(n-x)\n\nwhere:\n\nn choose x is a combination problem: n!/[(n-x)!x!]\np = probability of success\nq = probability of failure\nn = total selection\nx = successes\n\nIn this problem, we are looking for at least one case has a success (which is one of the highest 9 numbers).  In essence, we are solving for:\n\nP(x=1) + P(x=2) + P(x=3) + P(x=4) + P(X=5).  \n\nHowever it would be easier to find the probability of the failures and subtract from the total probability because we only have to calculate 1 probability.  This would be:\n\n1 - P(x=0) is what we are looking for.  Now all we need is to calculate the probabilty.\n\nFor P(x=0), we have the following:\n\nn = 5\nx = 0\np = (9/26) = probability of success\nq = (17/26) = probability of failure\n\nPlugging into P(x), we get:\n\nP(x=0) = (5!/5!)(9/26)^0(17/26)^5\n\nSolving, you get P(x=0) = 0.12\n\nTherefore, \n\n1 - P(x=0) is ------>\n\n1- 0.12 = 0.88\n\nBelow is a website devoted to Binomial Distribution exercises\n---------\n\nHope this helps
178452
try this site for help on probability\n\nhttp://www.futureaccountant.com/probability/
310634
Like rand uses between 0 and 1, and all real numbers in between have the same probability to appear the distribution is uniform. If it were not the case then the numbers generated could not be named "random".
400312
I am a teacher of mathematics and in a 31 year career I have had it happen to me exactly 1 time!!! It is a very very low probability
303066
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.
669887
It's not zero, because if there was nothing in the bag, you can't draw two items out and therefore the odds of drawing two thing (and of the same colour) is also zero.\n\nWhen you put 20 balls of some colour into a bag with nothing in it, and draw two balls out, there is a probability of 1 that they will be the same colour.  Since 0≠1, the answer is not zero.\n\nThere are 19 colours in the bag (before the other 20 are added).  Each colour is represented by 10 balls.  Thus there are 19*10=190 balls in the bag.
298222
A coin is flipped three times.\n\nA is the event that there is a head on the first flip.\nB is the event that there is a head on the second flip.\nC is the event that there is exactly one head in three flips.\n\nP(A) = 0.5\nP(B) = 0.5\nP(C) = 3/8 = 0.375 (If you flip a coin three times, there are 8 possible outcomes; of those, three result in exactly one head. You can also use the binomial here.)\nP(A|C) = P(A and C)/P(C)\n= (1/8)/(3/8) (Because only HTT will allow A and C to both happen, which is one outcome out of eight.)\n= 1/3\nP(B|C) = 1/3 (Same reasoning as above.)\n\nSo A and B are independent, but are not independent of C.\n\nP(A and B|C) = P(A and B and C)/P(C)\n\nNow if A and B, happens, then the first two flips were heads, meaning that there are at least two heads. But if C happens, there can only be one head, so\n\nP(A and B and C) = 0\n\nSo,\n\nP(A and B|C) = 0\n\nHowever P(A|C)*P(B|C) = (1/3)^2.\n\nSo I guess that the equality doesn't hold. And that also means that there is no positive constant K that will make it hold since K would have to be zero in this case.
2901
That would be a "random variable". So I guess other words of the puzzle might be wrong.
81652
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/
89291
This cannot be ascertained without knowing how many are being selected in all...\n\n\nhttp://www.futureaccountant.com/probability/\nhttp://www.schoolingkids.com/
525847
P(At least 2 orange) = P(2 orange or 3 orange)\n= P(2 orange) + P(3 orange) (These event are mutually exclusive, meaning that these events cannot occur at the same time.  This uses an addition rule.)\n= P(OON or ONO or NOO) + P(OOO) (OON means orange 1st and orange 2nd and non-orange 3rd.)\n= P(OON) + P(ONO) + P(NOO) + P(OOO) (also mutually exclusive)\n= (3/11)(2/10)(8/9) + (3/11)(8/10)(2/9) + (8/11)(3/10)(2/9)\n+ (3/11)(2/10)(1/9) (Since there are ands here, you would use the multiplication rule)\n\nThis gets you approximately 0.152 as the answer.\n\nThe other people are not answering correctly.  2/11 is the probability of getting an orange candy if you select one.  That is not what was asked.  Three are being selected, not one!  Jacob, just because you rounded correctly doesn't mean your answer is correct.
620429
Most of them got it right.\n\nThis is another way to think about the result.\n\nFirst player = T or H doesn't matter\n\nSecond Player = Must be that same as the First Player\n                        =  only 2 possibility (H or T)\n                        = 1/2\nThird Player = Must be the same as First Player\n                    = again, only 2 possibility ( H or T )\n                    = 1/2\n\nIn order for the Second Player and Third Player to have the same result as First Plaer = 1/2 x 1/2\n                                 = 1/4.