a die is rolled five times and the number of fours that come up is tallied. What is the probability of getting exactly zero fours
211830
The chance of not rolling a four is 5 out of 6. So five times:\n\n5/6 ^ 5 = 0.4019\n\nor about 2 in 5.
338826
Probability choose red the first time: 5/8\nAFter choosing red, have 4R and 5W. Therefore, probability of choosing Red again is 4/9. So, probability of choosing red and then red, is 5/8 * 4/9 = 5/18\n\nIf choosing white the first time: 3/8 chance of choosing white first. Then there are 7R and 2W, so probability of choosing red is 7/9. Probability of choosing white then red is 3/8 * 7/9= 7/24.\n\nAdd these two to find answer for first part. 5/18 + 7/24 = .56944\n\nFor second part, keep the proabailty of finding red and then red, and find the probabilty of finding white then white. Probabilty of finding white first time is 3/8. Then have 7R and 2W, so probabilty of finding white now is 2/9. Multiply the two 3/8 * 2/9 = 1/12.\n\nAdd 5/18 + 1/12 = .36111 as the answer for the second part.
672449
a) Try counting the possibilities.\n     There is only one way to roll a 12...(6, 6)\n     and there are 36 different possible rolls.\n     so the chances are 1/36\n\nb) There are 2 possibilities for this one (3, 6) and (6, 3)\n     so it's 2/36 or 1/18\n\nThe difference between the book the you is notation.\n1:35 means that for every one success there are 35 failures.\n1/36 means that there is one success for every 36 tries.
508814
If the die were perfectly balanced, six-sided cube, and had the each of the numbers 1 through 6 on it, then you don't need the tally.\n\nThe odds of any number coming up are 1 in 6.\n\nThe odds of any number coming up five times in a row are:\n1/6 * 1/6 * 1/6 * 1/6 * 1/6 * 6= 6/6^5  = 6/7776, or,  1 in 1296\n\nHope this helps!
239953
Start with the first box. The probability of pulling a white card is 5/9 and a black card is 4/9.\n\nNow, if the first box resulted in a white (a 5/9 chance), then the probability of pulling a white card in the second box is 8/18. (Overall probability of these events is 5/9 * 8/18 (0.247))\n\nIf the first box resulted in a black (a 4/9 chance), then the probability of pulling a white card in the second box is 7/18. (Overall probaibily of these two events is 4/9 * 7/18 (0.173).\n\nEither of the two above events will result in a final white card (and are the only ways a final white card could happen). So simply add the two probabilities together...\n\n0.247 + 0.173 = 0.42 or 42%
650826
I'm assuming each person rolls a 20-side die.\n\nFor two to win, two must roll 15, 16, 17, 18 or 19 points more than one. (20+8) - (10+4) = 14\n\n20-1=19\n20-2,19-1=18\n1+2+3+4+5=15 --> # of ways for two to win\n\nTotal combinations: 20*20 = 400\n15/400 * 100 = 3.75%
603235
To get your answers, imagine what has to happen.  \n\na) First, you have to pick 1 of the 4 white balls out of the 12 balls there.\n\n2nd, you have to pick 1 of the 3 white balls left out of the 11 balls there.\n\nSo, in order to get a red and then a red, you need to get 4 out of 12 followed by 3 out of 11 or 4/12 times 3/11 or 12/132 = 1/22.\n\nb) First, how many ways can 5 people line up?  5!.  How many ways are correct (assuming they all have different names, only 1).  So, the prob is 1/5! or 1/120.
121373
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
321726
For that to happen, 1 must come on each dice.\n\nThus P(Sum is 2) = P(1 appears)*P(1 appears)\n\n=(1/6)^2 = 1/36
137686
its easy. figure out in this way there are 3 digits and since its an odd number the one's place or the last digit can be 1,3,7,9(5 is excluded).the no. of chances are 4. for the second digit the digits can be 0 to 9 leaving 5 so no. of chances are 9. for the first digit 0 and 5 are not possible and the no. of chances are 8.\n\nmultiply the no. of chances for every digit it ll give u the answer i.e 8*9*4 = 288. so u have 288 different numbers.
126071
if the order you draw them is important its a permutation.\n\nthe answer is basically; a 1 in 55 chance for the first card, a 1 in 54 chance for the second card (you've taken the first card out) and a 1 in 53 chance of getting the final card. all three events are independant of eachother. so the answer is 55 x 54 x 53 = 157410\nso 1 in 157410 chance.\n\nif the order you draw them is unimportant them its a combination.\nfor the first card there is a 3 in 55 chance\nthe second card has a 2 in 54 chance\nand the final card has a 1 in 53 chance\nthe answer is: (3/55)  x (2/54) x (1/53) = 1 in 8745
283536
The probability that sum will be greater than 5 is 206/216. 6*6*6 = 216, so there are 216 possible outcomes, out of which there are only 10 combinations that makes the sum less than or equal to 5, so the rest 206 are greater than 5. \n\nThe combinations with less than or equal to 5 are \n\n1 1 1\n1 1 2\n1 2 1\n2 1 1\n1 1 3\n1 3 1\n3 1 1\n1 2 2\n2 1 2\n2 2 1\n\n:)
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?
420760
No one throws 3 dice, try this in Vegas just one time.\n\nI think 16 ways.
10512
Sometimes, it's easier to find the probability of the opposite case you are looking for.  This is one of those times.  Try to find the probability that you don't get the right one.\n\nAfter taking you two chances, one of the three objects is left.  What are the chances that this one that is left is the right one? The one that is left is one of three, so there is a 33.3% chance that it is left behind- which means there is a 33.3% chance that you DID NOT get the right one.  Therefore, there is a 66.7% chance that you WILL get the right one. \n\nIf you are still not convinced, it is easy to show that your reasoning that got you to 83% is wrong.  What if you used the same reasoning with 4 objects and 3 chances to get the "right" one.  Your logic would say there is a 25% chance the first time, a 33% chance the second time, and a 50% chance the third time.  If you just added them (as you did to get 83%), you would find that 25% + 33% + 50% = 108%.  Clearly you can't have a percentage above 100%, which illustrates the problem with that reasoning.\n\nHere's why that reasoning dosn't work; 33% of the times you play this game, your second try has a 0% probability of getting the right one because you already got it on the first try.  The other 66% of the times you play the game, your second try has a 50% probability.  So the AVERAGE probaility of the second try is (33.3% x 0%) + (66.7% x 50%), or 33.6%.  Add this 33.3% to the 33.3% probability of the first try, and you get the correct answer of 66.7%.\n\nIf you still aren't convinced, come over some time for a "friendly" game of dice.
272863
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.
317566
Okay, intially, you pick one marble, out of 12. and there are 3 yellows, then your initial chances are 1/4 of picking a yellow. Now, you've picked one, so there are 11 marbles left in the bag. There are 4 blue marbles in the bag, so you have a 4/11 chance of picking a blue ball at tthis point. \nTo Calculate Probability based  on an event that 1) Has already happened ( the yellow ball was picked first) and that is an event depending on the results of the , you simply multiply the chances. ( 1/4*4/11) or 4/44, or 1/11 chances that that will happen.