Probability with dates?
915215
The winning number is always a three digit number, of which there are 1000, not 100.\n\nThere are 65 days when the winning number cannot possibly match the date. Those days are\n\nOctober 10 through October 31\nNovember 10 through November 30\nDecember 10 through December 31\n\nOn the remaining 300 days, the probability that the winning number will match the date is 1/1000. So I would say that on "any given day" the probability of a match is \n\n(300/365) * (1/1000)\n\nGary H, below, forgot that in October, November, and December, the first nine days of each month are good, so he is short by 27 days. There are 300 good days, not 273.\n\nFor part 2), the probability that we don't get a match on any of the 300 days when a match is possible is 999/1000. So the probability of that happening on all 300 such days is\n\n(999/1000)^300\n\nFor part 3), I agree with Gary H that it is 1 minus the answer to part 2).\n\nIn part 4), Gary H's reasoning cannot be correct. Let's say that there were 1001 states instead of 50. Then his reasoning would lead to a probability of greater than 1, clearly nonsense.\n\nWhat you must do instead is ask, what is the probability of 9-1-1 NOT coming up on September 11 in Maine? It's 999/1000. What is the probability of it NOT coming up in Vermont? 999/1000 again. What is the probability of it NOT coming up in ANY of the 50 states?\n\n(999/1000)^50\n\nWhat is the probability that it comes up in at least one state?\n\n1 - (999/1000)^50\n\nThis reasoning always leads to a probability less than 1, for no matter how many states you have you can never be sure that 9-1-1 will come up on September 11.
518990
perhaps you should improve your question to get improved answers, but based on what you have asked, I would assume it's a simple probability box situation, and the answer would be 7! which equals 5040.\n\nYou have seven books, and seven possible places\n\n{1} {2} {3} {4} {5} {6} {7}\n\nyou can arrange those in 7 different places, ie:\n\nin your first row, you can have either books 1-7, in row two you can either have books 1-7, less the book you already included, this gives the following breakdown:\n\n7 options (any book any where)\n6 options (any remaining book anywhere)\n5 options (any remaining book anywhere)\n4 options (any remaining book anywhere)\n3 options (any remaining book anywhere)\n2 options (any remaining book anywhere)\n1 option (remaining book only has one place)\n\nTherefore:\n\n7*6*5*4*3*2*1 = 7! = 5040.
263161
This is a good probability problem -- I LOVE probability problems.  This is really asking 4 different questions.  First, what are the chances of the first pull being black.  (9 total socks, 5 are black ...).  Then what are the chances of the second one being blue (Keep in mind that now you only have EIGHT socks in the drawer, unless you put the last one back - so 1 blue and eight total socks).  Then chances of a brown sock with only 7 socks left (3 brown, seven total socks).  \n\nWhen you get all those probabilities, in order to see what the chances of this AND this AND this happening, you have to multiply the fractions (if it was "What are the chances of this OR this happening, you'd add them).\n\nGood luck.  This stuff is FUN!  (Especially if you ever want to go to VEGAS!)
19818
try\nhttp://www.peterwebb.co.uk/probability.htm
95999
Don't you think you could have written this more clearly? \n\nIf there is a uniform distribution of the weight w being between 5 and 10, then each of these probabilities is equal:\n\nP5_6=P{5<=w<6}\nP6_7=P{6<= w<7}\nP7_8=P{7<=w<8}\nP8_9=P{8<=w<9}\nP9_10=P{9<=w<10} \n\nThe sum of these is 1, so \n\n1=P5_6+P6_7+P7_8+P8_9+P9_10, and\nP5_6=P6_7=P7_8-P8_9=P9_10\n\nso 1=5xP9_10, so P9_10=0.2\n\nThe probability is 0.2.
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.
506414
I answered this one for you earlier...\n\nYou have X ~ Bin(6, 1/6)\n\nas you have 6 trials, with the proability of success (a four) being 1/6\n\nYou then simply need to stick this in the formula:\n\nP(X = x) = nCx × p^x × (1-p)^(n-x)\n\nIn this case you want the probability of 2 fours:\n\nP(X = 2) = 6C2 × (1/6)² × (5/6)^4 = 0.2009
419180
Day before yesterday = 27\nYesterday = 45\nToday = 135\nTomorrow = 180
792306
If the probability of picking an A is .5 and the probability of picking a B given an A is picked  is .3   \n Find the probability of picking both A and B?\n\nAnswer:     Pr(A) = .5   \n                  Pr(B/A) = .3 \n                  Pr(A∩B) = ? \n \n\n Pr(A∩B) =  Pr(B/A) * Pr(A) \n                = .3*.5\n                =.15    that is the answer.
249369
Is the term you are looking for "probability"?\n\nI hope this helps.
955894
x1*p(x1) + x2*p(x2)\n\nx1 = 1 for  event dots> 3,\nx2 = 4 for event  dots<=3.  \nThese are equally probable because there are 6 dots on the die.  \n\n1*(1/2) + 4*(1/2)
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
1009251
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?
242742
Remember, chance events are not dependent on the previous outcome.  The correct answer, as others have pointed out, is 50%.
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
435037
try this site for help on probabiliity\n\nhttp://www.futureaccountant.com/probability/
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.