If a die is thrown once, what is the probability that a number less than 3 is thrown?
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
765113
Very low. I'm not actually going to do the computation for you but I'll show you how it's set up.\n\nThis would be easiest to figure out on a graphing calculator like the TI-89.\n\nI'm assuming that the 1/3500 is for a single hole, so note that you're playing 50*18 holes in the season, or 900 holes. Getting 3 holes in one and 897 not is going to be the odds of three holes in one((1/3500)^3) mutiplied by the odds of 897 not((3499/3500)^897) multiplied by the ways to picking any three random holes of the 900 for the holes-in-one(this is done by a function called choose, denoted 2c5 for the number of ways to pick any two of five slots.\n\nThe answer is:\n((1/3500)^3) * ((3499/3500)^897) * (3c900)\n\nIt's going to be a very small number indeed. Congratulations on your fantastic season, I hope you do as well again next year.
630810
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?
67102
43 out of 400.\nThere is 120*120 pairs of minuites within which in which they can arrive resturent. And to be within 3 minuite, the diffrence of this of this  two pair have to be 3 or less. If the first one of this is between 4 to 117, there will be 7 possibility of  second one.\nAnd teher will be 6 for 3 and 118, 5 for 2 and 119,and 4 for 1 and 120.\nso total possibolity is 114*8+6*2+5*2+4*2 that is 828\nSo the probability is 828/14400=43/200.
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.
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.
1049750
For any roll, the chances of a particular number on an N sided die coming up is 1 in N.  Therefore, generating a random number from 1-N for each roll would be the answer.
433571
The probability of a prime number with one roll of dice would be 3/6=1/2. In ratio form, it is 1:2. And another way is 1 to 2. To find it out you simply have to find the prime numbers, or numbers that can't be divided by. those numbers are 2,3, and 5. 1 is NOT a prime number nor a composite number because it can be divided by no matter what.
761738
Probability(1st person picks any bill)=8/8\n\nProbability(2nd person picks the same bill)=1/8\n\nProbability(3rd person picks the same bill)=1/8\n\nThe events are independant as the outcome of one event will not affect the outcome of another. Thus allowing multiplication of probabilities.\n\nProb(All 3 draw same bill) = 8/8*1/8*1/8\n\n=1/64
332812
A fair dice is one in which the probability of getting one side is 1/6.\n\nSum of 6 or 11 can be obtained in the following ways: \n(1,5) (2,4) (3,3) (5,1) (4,2) (5,6) (6,5)\n\nThere are a total of seven elements here.\nTotal possible elements = (6 X 6) = 36\n\nHence probability = 7/36
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)
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.
786207
did u mean '93.5% or more'?\n\nif so, there is only 1 chance of that occurring and that is when all 8 coins are tossed and revealing heads. \n\nso that is a probability of 1/2^8 = 1/256 = 0.00390625
1181169
Variation 1\nThe first player is chosen by having each player roll one die. The highest roll is the first player to start. At the beginning of each turn, the player throws all six six-sided dice. If any dice are not lying flat or on the playing surface, they (only the dice not laying flat or on the playing surface) must be thrown over again. After each throw some or all of the points possible must be taken, setting aside each die that is counted. To continue, all of the die not counted must be thrown again. At each throw, points must be made or the turn ends. When no points are scored in a throw, this is called farkle.\n\nWhen a player reaches at least 350 points in a turn and decides to stop, the points for the turn are added to their total score. A player with no points may not end his turn until he has rolled a total of 1000 points in the round.\n\nThe points of each throw are counted separately. When only one die remains, the player has three throws to get points (either a one or a five spot). When all six die have points, all six die are thrown again.\n\n[edit]\nVariation 2\nEveryone starts out "off the table." Each player collects points during his turn, which he may add to his total, or not, depending on how aggressively or cautiously he plays. \nTo begin a turn if the player is "off the table," he rolls all six dice. \nAs he rolls during his turn, the player is looking to score points, in the form of Sets (see figure below). \nIf a roll scores any points, he may set aside each Set he wants to claim points from, and either: \nroll all remaining dice, hoping to score additional Sets. \nor if all six dice are put aside into Sets, roll all six dice again, hoping to score additional Sets. \nor pass play to the next player, if he deems the chance of failure to be too great. \nIf a roll scores no points, the player forfeits the points scored that turn, and play is passed to the next player. \nFor a player who is "on the table," he may start by either: \nrolling all six dice as described above, \nor picking up the rollable dice from the last person. In this case, instead of starting this turn's scoring from zero, as usual, scoring starts from the score of the last person. \nexample: Player 1 stops his turn with 700 points, and opts to not roll his remaining two dice. He adds 700 to his score, and it is now Player 2's turn. Player 2 may pick up those two dice, and if he scores anything with them, he adds those points to 700, as his score. Or Player 2 may opt to start with all six dice, and start his own scoring from zero.\n\nAt the end of a turn, \nIf, after rolling, a player is "off the table," and he has scored at least 500 in a turn, he is "on the table." \nIf by this point, a player is "on the table," he can add whatever score he got that turn, to his total. \n\ncopied and pasted from source home.houston.rr.com/agrant/farkle/farkle
452389
There are 2 rules you need to know. The addition rule and the multiplication rule\n\nIf we roll a die the chance of rolling 6 is 1/6.\n\nthe chance of rolling 5 is also 1/6\n\nthe chance of rolling either a 5 or a 6 is therefore 1/6 + 1/6 = 1/3\n\nAnd the chance of rolling 4, 5 or 6 is 3/6 or 1/2.\n\nWhen the events are alternative outcomes and you want to know the chance of the outcome being one of these outcomes, add the probability of each outcome.\n\nIf you roll a die twice (or two dice at once) the probability of rolling a par of ones is 1/36.\n\nThis involves the multiplication rule. The chance of the first die showing a 1 is 1/6, and the chance of the next die showing a 1 is also 1 in 6. \n\nOnly one-sixth of the time do you get the first die to give you the 1 you want and only one-sixth of those times will the second due give you another 1. So 1/6 x 1/6 = 1/36.\n\nHowever if the question is what is the probability of just one six being thrown in two rolls\n\nthen the probability of 1 on the first throw but not on the second throw is (1/6 x 5/6) = 5/36\n\nAnd the probability of 1 on the second throw but not on the first throw is (5/6 x 1/6) = 5/36 \n\nAnd so throwing a 1 on either roll but not the other roll has a total probability of 5/36 + 5/36 = 10/36.\n\nThe slightly different question "what is the probability of rolling one or more 1s in two rolls of the dice?" is the sum of\n\nprobability of one on the first throw but not the second\n\n+ probability of one on the second throw but not the first\n\n+ the probability of a double one being thrown\n\n5/36 + 5/36 + 1/36 = 11/36\n\nSo the rule is when it is a matter of alternative outcomes to one event, add the probability of each outcome.\n\nWhen it is a matter of the outcome of two events both happening, multiply the probability of each happening to get the probability of BOTH happening.
304947
If you throw dice and get a 6 three times in a row, the odds of it getting another 6 the fourth time are still 1/6, as good or as bad as for any other number. The history has nothing to do with it. Each throw of a dice or lottery draw is a series of random events, the outcome of which is not influenced by previous outcomes.\n\nAs the British magazine The Economist once wrote, lotteries are a voluntary tax on being bad at math ;-)
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.