You Flip a coin and toss a number 1-6  dice \nwhat are ur chances of getting heads and a 2 \non the dice?
554035
1/6 multiplied by 1/2 = 1/12\n\nthe law of independant probabilties.
242656
This one is easy-assume a 50% chance each time, so its 1/2 X 1/2 X 1/2 X 1/2, so 1 chance in 16.  There is another way of figuring this, using combinations, like sn777 did, but sn777 made a logical error.  There are these possible combinations: 4 m, 3m 1F, 2m, 2F, 1m 3F, and 4F. So just looking at that, you would think its 25%. BUT, there is only one combination of 4M, while there are 4 combinations that make 3m 1F, and 6 combinations that make 2m and 2F, etc-so the total possible combinations of 4 children is 16, and 4M is only one of those.
422184
There are 5 ways of getting a 6 (1-5, 2-4, 3-3, 4-2, 5-1) and two ways of getting 11 (6-5, 5-6), so the probability is 7/36.
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?
817262
Experimental probability is obtained by carrying out some trials to see what fraction of the time the event you are looking for occurs.  For example, if you want to know the probability of scoring a total of 8 when you throw a pair of dice, one way of doing it would be to throw a pair of dice five times and see how many times you get 8.  If it happened twice, you would say the experimental probability is \n2/5, or 0.4\nNot a very good result, though, too few trials.  If you did it 100 times, and 8 showed up 15 of those times, the experimental probability would be \n15/100, or 0.15\n\nTheoretical probability is found by considering how many equally likely outcomes there are, and in how many of those outcomes the event you are looking for occurs.  In the example I just gave, there are 36 different ways for the dice to land, and the total is 8 in 5 of those ways, so the theoretical probability is\n5/36, or 0.13888...\n\nWe expect a very small number of trials will not give a very good picture of the real probability.  Choose your answer from that.
514377
There are 20! possible orderings for picking the numbers from the hat. Only one has them in order, so 1/20!.\n\nAnother way to look at it is that when there are 20 numbers in the hat, your chances are 1/20 of picking the 1. Then they are 1/19 of picking the 2, then 1/18 of picking the 3 etc.\n\nSo the chances are (1/20)(1/19)(1/18)...(1/2)(1/1)\n\nIf you threw the numbers back in each time, the probablity of pulling them out in order would be 1/(20^20).
694491
each die has 6 ways it can land\nonly one of them is a 1\nso chance of one die getting 1 is 1/6\nsame with the other one\nSo the odds of rolling "snake eyes" is 1 in 36\n\nchance of being a girl is 17 out of 30 or 17/30\nchance of being an A student is 5 out of 30 (5/30) or 1/6
632203
a. first draw doesnt matter since it's replaced; you're back at 3 in 10 chance\n\nb. two ways this could happen: first, 3/10 chance first ball is red, leaving you 2/9 chance then of second ball being red on second draw.  second, 7/10 chance you draw a white ball, leaving you with 3/9 chance of second ball being red.  (3/10*2/9)+(7/10*3/9) which leaves you with 3/10 chance. again.\n\n\nEDIT:\n\nsurprises me how many people can't do math.\n\nfor A), the second draw is completely independent from the first.  Doesn't matter if I took a dump in the bowl in the first draw; as long i fished it out, you're left again with the same start on the second draw-3 reds and 7 whites.  your probability doesn't change because of what happened in the first draw as the conditions are RESET.\n\nfor B), let's run through allll possible scenarios\nNumber the Balls\n\n1 - red\n2 - red\n3 - red\n4 - white\n5 - white\n6 - white\n7 - white\n8 - white\n9 - white\n10- white\n\nPossible draws (X denotes red second ball outcome)\n1,2 - X\n1,3 - X\n1,4\n1,5\n1,6\n1,7\n1,8\n1,9\n1,10\n(2 fav outcomes if ball 1 is drawn first)\n\n2,1 - X\n2,3 - X\n2,4 \n...\n2,10 \n(2 fav outcomes if ball 1 is drawn second)\n\n3,1 - X\n3,2 - X\n3,4 \n...\n3,10 \n(2 fav outcomes if ball 1 is drawn second)\n\n4,1 - X\n4,2 - X\n4,3 - X\n4,5 \n...\n4,10\n(3 fav outcomes if ball 4 is drawn first)\n\n...\n\n10,1 - X\n10,2 - X\n10,3 - X\n...\n10,8\n10,9\n(3 fav outcomes if balls 5-10 are drawn first)\n\n\nthere are 90 outcomes\nof which there are 2+2+2+3+3+3+3+3+3+3 favorable outcomes = 27.\n\n27/90 = 3/10\n\n\nSecond edit: guy below, 2*3 + 7*3 = 6 + 21 = 27, not 28
147164
Ah, a variation on the Monty Hall problem:\n\nhttp://en.wikipedia.org/wiki/Monty_hall_problem\n\nIt is a probabilistic curiosity, this one. The highly unintuitive answer is that you actually **can** improve your chances by changing your mind.\n\nYour problem appears to only be different in that it starts with 5 boxes. The above link should give you sufficient information to solve the problem.
248227
50% or .5 or 1/2\nThere are two possibilities when you flip the coin, and there's only one way it will be tails.\nTherefore, the probability is 1/2.
599149
The probability of getting heads on one toss is 1/2 (or 50%). To figure out the probability of getting heads on 5 consecutive tosses you just multiply the probability of a single toss 5 times:\n\n1/2 x 1/2 x 1/2 x 1/2 x 1/2 = 1/32\n\nThe probability of getting all tails is the same, for the same reasons: 1/32.\n\nTherefore, the probability of getting either all heads or all tails is simply 1/32 + 1/32 = 2/32 = 1/16
889348
As the previous poster said, you don't have enough information.  If you have n-sided polyhedra, are the sides numbered 1 to n?  Is the expected value in reference to a throw of the dice?  If so, are we talking about two dice?\n\nFor two dice with the sides numbered 1 to n, the answer is n + 1.
74975
Which something?\n\nJust using roulette, a chance of winning on a single spin varies from 47.4% to 2.63% depending on which kind of bet you make: color, odd, even, a specific number, or a pair or quad of numbers, etc.\n\nFor some other event, to calculate the odds would require more information.
412043
I can't believe that this question is being asked in the sixth grade.  Maybe half of my college students could answer this.\n\nThe best way to answer this question is to make a table.  Make the rows the results of the first die and the columns the results of the second die.  Then put the product of the two roll inside of the table.\n            1   2   3   4   5   6\n          ------------------\n        1| 1   2   3   4   5   6\n        2| 2   4   6   8  10  12\n        3| 3   6   9  12  15  18\n        4| 4   8  12  16  20  24\n        5| 5  10  15  20  25  30\n        6| 6  12  18  24  30  36\n\n(Sorry I can't get the columns to even out!)\n\nNotice that there are 36 possible rolls of the two dice.  To get the two probabilities, just count the number of times you get the event in the table and divide by 36.\n\nProbability of an even product = 27/36\nProbability that the product is less than or equal to six = 14/36\n\nI see a lot of incorrect answers up there.  I teach statistics at the college level.  Trust me, this is right.
978546
The expected value you get for rolling a normal die is 3.5, but there is no 3.5 on a die.
466671
Flipping a coin 4 times means that there can be 16 different posibilites of Heads and Tails occuring (2^4 = 16).\n\nThus by dividing 1 (max. probability) by 16 we get .0625. So each combonation has a probability of 6.25%. Now there is only one possibilty (TTTT) where there are no heads thus the possibility of no heads is simply .0625 or 6.25%\nWhere as one head has the following possibilites:\n\nTTTH\nTTHT\nTHTT\nHTTT\n\nThus multiplying .0625 by 4 we get .25 which is possibilty of only 1 head occuring in four flips.\n\nSo in conlusion:\n\nFor four flips of a coin:\n\nProbability of only one head occuring = .25 or 25%\nProbability of no head occuring       = .0625 or 6.25%
698606
Hmm, interesting question, actually, and a practical application of math.  Hey, all you folks who ask what this stuff is good for, here is a real-life situation where you can use it!\n\nGetting to the problem.  Obviously, the contestant to go last spins again if his score is lower than the best score.  If tied, there is a spin-off where his winning chance is 50%; so, if there is a tie, he should respin if he has 45 or less (giving him at least 11 out of 20 good numbers), not spin if he has 55 or more (giving him at most 9 out of 20 good numbers - better to go to the spin-off, where he has a 50% chance), while 50 is a "push".   \n\nNow, for the contestant to go 2nd to last: First, let's assume the first player busts (of course, the second player respins if his score is lower than the first player), which essentially makes this a two-player game.  Based on what number the second player stops on, this is the winnning percentage I calculate for the last player:\n\n100 - 4.875%\n95 - 14.375%\n90 - 23.375%\n85 - 31.875%\n80 - 39.875%\n75 - 47.375%\n70 - 54.375%\n65 - 60.875%  \n60 - 66.875%\n55 - 72.375%\n50 - 77.375%\n\nSo, you may think the second player should respin whenever his odds are under 50%, but that is actually incorrect.  For example, if the second player has 70, giving him a 45.625% chance of winning, respinning would give him a 70% chance of going bust - he is better off stopping with the 45.625% chance.  To determine whether or not respinning improves the second player's chances, a second calculation is needed that considers each of the 20 scores after the second spin - some of them bust the player but the others improve his chance of winning.  Based on that, here is a new table, giving the last player's chance of winning if the second player respins with the score given:   \n\n95 - 95.24375%\n90 - 90.9625%\n85 - 87.13125%\n80 - 83.725%\n75 - 80.71875%\n70 - 78.0875%\n65 - 75.80625%  \n60 - 73.85%\n55 - 72.19375%\n50 - 70.3125%\n45 - 69.68125%\n\nSo, comparing the two tables, the critical points are 55 and 60.  If the second player has 60, he should stop - the last player's chance of winning improves if the second player spins.  At 55, however, the second player should spin, improving his odds by about 0.2%.\n\n\nMoving on to the three player game... now it gets really messy.  The player to go first needs a score that can beat the other two players.  Does he stop on 55 again, hoping the other two players both bust?  He'll need a better score than that to have a 50% chance of winning, but, as was noted in the two-player analysis, stopping with a 30% chance can be better than making a second spin.  As for how to do a complete analysis, uh, beats me.  This may require a computer to do all the math.\n\n* * * * * *\n\nThinking about the three-player game, maybe it isn't as messy as I initially thought - the only additional consideration for the first player, as opposed to the second player, is that either player will stop if the first player's score is beaten.  This can be calculated from the first table.  For this table, listed is the chance of the first player losing (i.e. either player beats him) if he stands on the score given:\n\n100 - 9.51% \n95 - 26.68%\n90 - 41.29%\n85 - 53.59%\n80 - 63.85%\n75 - 72.31%\n70 - 79.18%\n65 - 84.69%\n\nSimilar to the two-player game, I can now calculate the odds if the first player decides to respin.  Since I’m getting a little tired crunching these numbers, for this last table I’ll only list the critical ones:\n\n70 - 83.4%\n65 - 82.3%\n\nIn other words, the first contestant should respin on 65, stop on 70.\n\nIn conclusion, I think the optimum strategies are.....\n\nFirst player: spin on 65, stop on 70\n\nSecond player: spin on 55, stop on 60 (assuming his score at least matches the first player)