a spinner is spun twice. what is the probability that the spinner stops on 4 and then on 1?
154434
The problem cannot be attended to without knowing the total number of possibilities.\n\nhow many sections are there on the spinnner..\n\n\nhttp://www.futureaccountant.com/probability/\nhttp://www.schoolingkids.com/
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
1012943
Since the first coil is replaced, the two events are independent.  So both times you have a 88/100 = 0.88 chance of getting a good coil.\n\nP(2 good coils) = P(1st is good and 2nd is good)\n= P(1st is good)P(2nd is good) (from independence)\n= (0.88)(0.88)\n= 0.88^2\n= 0.7744
1101632
4/52*4/51*4/50 = 482.65 times in one million trys or 0.048% chance.\n\nhope that helps.\n\nm
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?
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.
917829
When you are computing the probability of a given event such as 2 heads, 2 tails, or one of each, you are essentially counting the number of ways a given event can occur and dividing this number by the number of all possible outcomes.\n\nIf you toss 2 coins once ( or one coin twice) there are 4 possible outcomes:\n\nHH = both heads\nHT = 1st heads, 2nd tails\nTH = 1st tails, 2nd heads\nTT = both tails\n\nAssuming the coins are "fair" each outcome is equally likely.\n\nThere is one way in four that two heads (or 2 tails) can happen so the probability for each of these outcomes is 1/4\nBut there are two ways you can have one head and one tail and so the probability of this happening is 2/4 =1/2. The sum of all the probabilities of the distinct outcomes will always equal 1. \n1/4 +1/4 + 1/2 =1\n\nHow about if you have 3 tosses? The 8 possible outcomes are as follows:\n\nHHH \nHHT\nHTH\nTHH\nHTT\nTHT\nTTH\nTTT\nWe see there is only one way to have 3 heads (or 3 tails) and so the probability of 3 heads (or 3 tails) = 1/8\nHowever there are 3 ways you can have 2 heads and one tail, so the probability of this happening is 3/8. And there are also 3 ways you can have one head and 2 tails, so 3/8 is the probability for this too.
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.
354101
This is easy, just say there are 100 shaft holders.  So of A, 50 of them are processed there, and 3% don't work, so 1.5 out of 50.  For B, it is 30 x 4%=1.2, and C is 20 x 5%=1.  \n\nSo out of a total 100, 3.7 don't work, the the probability that this one chosen at random wont work is 3.7%.  \n\nHopefully the way i calculated the answer helps you out in the future.
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.
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
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
733920
a) There are 16 combinations and 4 of them include exactly 3 heads (HHHT, HHTH, HTHH, THHH).  So the probability is 4/16 or 1/4.\n\nb) This question results in the same probability (4/16 = 1/4), except it asks for it as 'odds' which is the probability of success divided by the probability of failure.  That is 1 to 3 odds.\n\nc) The odds against heads are just the reverse.  Probability of failure divided by probability of success.  That is 3 to 1 odds against.
261322
In the experiment of rolling two dice\n\nTotal number of possible choices = 36 ==> n=36\n\nLet "A" be the event of rolling the dice such that the sum on the dice is 7\n\nFor Event "A"\n\nNumber of favorable choices = 6  ==> m = 6\n[(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)]\n\nProbability of getting a sum of 7 on rolling the dice \n==> Probabilty of occurance of event "A" = m/n\n==> P(A) = 6/36\n==> P(A) = 1/6\n\nhttp://www.futureaccountant.com/\nhttp://www.schoolingkids.com/probability/
657146
Time for Pascal's Triangle.\n.......1\n......1 1\n.....1 2 1\n....1 3 3 1\n...1 4 6 4 1\n..1 5 10 10 5 1\n1 6 15 20 15 6 1\n\n1 chance in 64 of 6 Heads and 0 Tails\n6 chances in 64 of 5 Heads and 1 Tail\n15 chances in 64 of 4 Heads and 2 Tails\n20 chances in 64 of 3 Heads and 3 Tails\n15 chances in 64 of 2 Heads and 4 Tails\n6 chances in 64 of 1 Heads and 5 Tails\n1 chance in 64 of 0 Heads and 6 Tails\n\nChance of exactly 3 tails: 20 in 64\nChance of 3 or more tails: 42 in 64\nChance of more than 3 tails (4 or more) : 22 in 64\n\nAnyone who think the chance of 3 tails in 6 tosses is 50% is generalising wildly.\n\nThe chance of 1 tail in 1 toss is 50%\nThe chance of 1 head & 1 tail in 2 tosses is 2 chances in 4\nThe chance of 2 heads & 2 tails in 4 tosses is 6 chances in 16. (37.5%)\nThe chance of 3 heads & 3 tails in 6 tosses is 20 chances in 64. (31.25%)\nThe chance of 4 heads & 4 tails in 8 tosses is 70 chances in 256. (27.34%)\nThe chance of 5 heads & 5 tails in 10 tosses is 252 chances in 1024. (24.61%)\n\n(read the middle number in each row of Pascal's Triangle, and extend it downwards)\nThe bigger the number of tosses, the probability of tossing exactly half of the tosses as tails keeps on falling. This should not surprise us. There are so many more possible outcomes.
296539
This is a good question, and is the basis behind the theory of statistical thermodynamics. \n\nWhen you roll one die (i'm assuming a standard, fair,  six sided die) the probability of getting any particular number is 1/6. That is pretty obvious, each number is equally likely.\n\nInteresting things start to happen when you roll multiple dice however. Lets take all the possible outcomes for rolling two standard dice.\n\nDie 1   -   Die2   -   Total\n  1          1           2\n  1          2           3\n  1          3           4\n  1          4           5\n  1          5           6\n  1          6           7\n  2          1           3\n  2          2           4\n  2          3           5\n  2          4           6\n  2          5           7\n  2          6           8\n  3          1           4\n  3          2           5\n  3          3           6\n  3          4           7\n  3          5           8\n  3          6           9\n  4          1           5\n  4          2           6\n  4          3           7\n  4          4           8\n  4          5           9\n  4          6           10\n  5          1           6\n  5          2           7\n  5          3           8\n  5          4           9\n  5          5           10\n  5          6           11\n  6          1           7\n  6          2           8\n  6          3           9\n  6          4           10\n  6          5           11\n  6          6           12\n\nNow each die is independent of the other one. This means that getting 1-6 is a different result to getting 6-1. Thus it's easy to see that there are 36 possibile outcomes as shown above (6*6). Take a look at the totals of the outcomes though.\n\nThe first total i've written down is '2'. This comes from getting a 1 and another 1. You'll notice that this total does not appear anywhere else at all. The only possible way to get a total of 2 is if you get a 1 on both dice.\n\nAlso look at the last total, '12'. This is from getting a 6 and another 6, and is also the only place you'll see this total. There is only 1 possible way to get 12 as a total.\n\nHowever if you look at the total '7', you'll notice this appears 6 times. There are 6 different ways to get a 7 (1-6, 2-5, 3-4, 4-3, 5-2, 6-1). Therefore you are 6 times more likely to get a total of '7' than you are to get a total of either '1' or '12'! If you were to plot the average score received after rolling the two dice and plot them on a probability distribution curve (just graph the average scores along the bottom with the probability of getting that score up the side) you will see that it starts of with a low probability for getting an average of 1, rises to a peak at getting an average of 3.5, and then drops back to a low probability got getting an average of 6.\n\nNow with two dice there is still a reasonably good chance of scoring an average of 1, and a reasonably good chance of scoring an average of 6 (you'll see it happen every now and then if you roll the dice for a few minutes), but what happens if we go to 3 dice? Well the number of possibile outcomes is multipled by 6 (don't worry, I won't list them all!) But you'll notice there is STILL only 1 possible way of getting an average score of '1' (1-1-1). And there is STILL only one possible way of getting an average score of '6' (6-6-6). Therfore, the probability of you getting these numbers as an average of rolling 3 dice is much less than with two. Also the probability of getting an average of 3.5 is much higher (i'll leave that to the reader to confirm). If you plot the results for 3 dice onto the probability curve you'll notice that it is a bit steeper. It still has it's peak at 3.5, but it is much sharper.\n\nThe really interesting things happen when you go to large numbers of dice. Imagine that you have one die for every atom in a sample of some substance. Ther
791422
Each individual toss is 1 / 2, a 50% probability.\n\nThe possibilities of all outcomes is 1 / (2 x 2 x 2), a 12.5% probability.\n\nOnly 2 possible outcomes do not meet the desired outcome (all heads, all tails).\n\n2 x 12.5% is 25% (but that's the opposite of what you want)\n\n100% - 25% = 75%