Probabilty help! I keep getting the wrong answer?
924497
if     Pr(X) = 1/2*sin X = 0.25 \nthen           sin X = 0.5 \n\nIn the interval 0 < X < Pi :  \nSin X = 0.5 means X = 1/3.Pi \nThat's equal to 1.047197551\n\nClosest answer is  d)
220111
Greetings:\n\nHaving selected one of the four, the probability that person #2 will have a different date of birth is 364/365 (if we seek no repeats, #2 can have any 364 of 365 possible b.days).  Similarly, having selected two, the probability that #3 will have a unique d.o.b. is 363/365 (any 363 of 365 possible dates okay).  Finally, for #4, the probability is 362/365.  Hence by the multiplication property, the probability that all four will have different b.days is given by, (364/365)(363/365)(362/365) = (364)(363)(362)/(365)^3 = 0.984 to three places.\n\nRegards,\n\nNikkor180
783748
Between 1/3 and 2/3 chance\nChance of him on his first step 1/3.\nChance if he first steps back,=  2/27 -  2/3 x 1/3 x 1/3 (ie he has to take 1 step back, then two forward to fall off cliff).\nChance if first two steps back then three \nsteps forward =  4/243\n\nSo its1/3+2/27+4/243 etc etc\n\nThe other options just keep adding up. \n\nThe increments get smaller, but they keep adding on to the probability.
469430
the probility of a particular event is basicly "the amount of favorable outcomes over the total amount of outcomes. for example, \n\nwhats the chances of trowing a 3 with a dice?\n\ntotal favorable outcomes= 1 (because the is only one 3 on a dice)\ntotal outcomes= 6 (because there a six outcomes posible with a dice)\n\ntherefore the probibility of trowing a 3 with a dice is:\n=1/6\n\nthere are other formulars (eg. the  binomial theorm) for more complicated probibilities but this is the basics.
19818
try\nhttp://www.peterwebb.co.uk/probability.htm
554382
Matthew rolled four dice. What is the probability of the total of the spots equaling 14? The answer is 1/6.\n\nJenna decided to repeat Matthew's experiment, except she rolled six dice. What is the probability of the total of the spots equaling 21? The answer is 1/6.\n\nLarry decided to repeat Jenna's experiment, but with ten dice this time. What is the probability of the total of the spots equaling 35? The answer is 1/6.\n\nGive a formula as to show when 2n dice are rolled, which total of dice will have a probability of 1/6 of being rolled.
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!)
1092672
Its 1:220\n\n(4/52)(3/51) = 1/221 which is the probability\nThus the odds are 1/220 or 1:220
232542
I did this on my calculator because you have to do 75 calculations (which would take forever by hand).\nLess than 75 means it landed tails 0, 1, 2,..., or 74 times.  The probability of tails is .5 and the probability of not tails is also .5\nSay you want tails to come up k times where k = 0,1, 2,...,74\nThe probability of x = k is\nP(X=k) = C(n,k)(.5)^k(.5)^(n-k) Add the exponents on .5 to get\n= C(n,k)(.5)^n where n = 200 (because that is the amount of trials)\nk goes from 0 to 74, so we need to take the sum from 0 to 74\nThe sum (k=0 to k=74) of (C(200,k)(.5)^200\n=.000144637632\nI see no other way of doing this except with a calculator or computer.  (I'm sure your teacher does not expect you to do 75 calculations by hand and then add them all up.)
963120
A.  The probability that it will BE solved is 1 - prob that it WON'T be solved.  That is 1.00 - (1.00 - .015 [Tom doesn't]) * (1.00 - 0.50 [Alice doesn't]).  \n\n = 1.00 - (.85) * (.50).  = 1.000 - .425 = .575\n\nB.  If Tom solves it then it is 150 / 575 = 6/23
6514
False.\n\nThe key word in the question is 'always'.  If there is any probability of a different answer, no matter how small, then it can not be 'always'.
1113295
(1, 2, 3,4) or (-1, -2, -3, -4) or (1, 2, -3, -4) or (1, -2, -3, 4)\nYou also have\n(-1, -2, 3, 4), (-1, 2, -3, 4), (-1, 2, 3, -4), and (1, -2, 3, -4)\nSo there are 8 possibilities.\nEach having a probability of 1/8 or occurring.\nFor (1,2,3,4) and (-1,2,-3,4), the prob. of an even natural number is 2/4 = 1/2\nMultiply this by 1/8 to get the conditional prob.\n(1/2)(1/8) = 1/16\nFor each of the following: (1,2,-3,-4), (1,-2,-3,4), (-1, -2, 3, 4), and (-1, 2, 3, -4), the prob. of an even natural number is 1/4\n(1/4)(1/8) = 1/32\nFor (-1,-2,-3,-4) and (1,-2,3,-4), the prob. of an even natural number is 0.\n0(1/8) = 0\nAdd all these together.\n2(1/16) + 4(1/32) + 2(0)\n= 1/8 + 1/8\n= 2/8\n= 1/4\nThis doesn't match your solution, but you can see if this makes any sense to you.
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?
885691
Part (a) of problem 1 is pretty easy.  There are 10 marbles total, and 6 of them are red.  On your first draw, you have a 6/10, or 0.6 (60%) probability of drawing a red sock.  In other words, there are 6 ways of drawing a red sock out of a total of 10 possibilities.\n\nFor part (b), you need to know that the probability of two independent events happening is given by the product of the probabilities of each of the individual events.  In this case, the two independent events are that the first marble drawn is red AND the that the second marble is also red.  We already calculated the probability for drawing a red marble on the first draw in part (a).  After the first draw, the box contains 9 total marbles, 5 of which are red, so the probability that the second marble is red is 5/9.  The probability that both are red is given by the product, 6/10 * 5/9 = 30/90 = 0.3 (30%).\n\nThe reasoning for part (b) is the same as for part (a).  The probability that the first marble drawn is white is 4/10, and the probability that the second marble drawn is also white is 3/9.  The probability that both are white is given by the product, 4/10 * 3/9 = 12/90 = 0.133... (13.333...%).\n\nFor part c, there are two ways one can get a red and a white marble: either the first marble is red and the second white, OR the first marble is white and the second red.  When there is more than one way to achieve the desired result, the probability of getting that result is given by the SUM of the probabilities of each of the different ways to achieve that result.  In this case, the probability of drawing red-white is 6/10 * 4/9 = 24/90, while the probability of drawing white-red is 4/10 * 6/90 = 24/90.  The total probability of getting a red and a white marble is then 24/90 + 24/90 = 48/90 = 0.53333....\n\nUsing the same reasoning, you should now be able to solve parts (a) and (b) of the second problem.  You should find for part (a) that the probability is equal to 0.42424242...., and that the probability for part (b) is 0.09090909....\n\nFor part (c), you would use a similar logic as used for part (c) of the first problem.  The probability of getting a matched pair (either both white or both blue) is given by the sum of the probabilities found in parts (a) and (b).  You should find that the probability is 17/33 = 0.51515151....\n\nFor the third problem, you need to know that the expected "payoff" of a game is given by the sum of the products of the prize value times the probability of winning that prize.  In this case, the probabilities of winning the various prizes are:\n\n$100 : 1/500\n$50 : 3/500\n$25 : 5/500\n\nSo the expected payoff (i.e., the "mean" or "average" winnings for a single ticket) is given by:\n\nP = $100*(1/500) + $50*(3/500) + $25*(5/500)\nP = $100/500 + $150/500 + $125/500\nP = $375/500\nP = $0.75
257322
It's very simple. Every taster has to choose one out of 5 cups, so on every round the probability of a single taster choosing A is 1/5.\nSince the event is repeated 4 times independently, the overal probability will be 1/5 of 1/5 of 1/5 of 1/5.\nThat is, (1/5)^4 = 1/625.
249369
Is the term you are looking for "probability"?\n\nI hope this helps.
97799
i don't believe the person above me got the second part right.  he is right about the beginning - the first probability is 2/5:\n\npossible numbers to draw: 2 4 6 11 13 15\n\nso you can choose any of these six out of the fifteen given (6/15 = 2/5).  the next number you draw must be also a "winning" number.  BUT, you have to have an even number less than 7 AND an odd number greater than 10.  this gives us two scenarios:\n\nIF the first number picked was even and <7, then the possible numbers you can pick from are ONLY: 11 13 15, since picking another even number <7 will not meet your requirements.  this gives us a probablity of 3/14, since there are 3 "winning" numbers out of the 14 numbers left.\n\n\nIF the first number picked was odd and >10, then the possible numbers you can pick from are ONLY: 2 4 6, since picking another odd number >10 will not meet your requirements.  this gives us a probablity of 3/14, since there are 3 "winning" numbers out of the 14 numbers left.\n\nSO, finally, suppose you go with the first scenario.  the probability would be: (6/15)*(3/14) = 3/35  going with the second scenario will give you the same thing, 3/35.  thus adding these two, since you can have either scenario occur and still meet your requirements, gives you: 3/35+3/35 = 6/35.\n\nAnd that would be the odds: 6/35.