Probability problem? If you can answer it first you get 10 points!?
534974
They're independant events, so it's one in 12.
91512
20 questiosn with five options, the probabilty to choose the correct one is 1/5 per each question.\nThe probability to answer all 20 questions is (1/5)^20.\nThe probability to answer more than 10 is (1/5)^11\nThe probability to answer any eight is (1/5)^8\nThe probability to answer maximun 7 is (4/5)^13
776173
1) this operation is made up of two steps, that means the probablility of the whole operation = the multiplication of the probablility of every step\n a) probaility of a 1 from the first box = 1/6, the other box = 1/7 , so P(Both will be 1's) = 1/7 * 1/6\n\nb) p(Exactly one of the will be a 2) = 1/6 * 6/7 + 1/7 * 5/6 \n\nc) P(Both will be even) = 3/6 * 3/7 \nd) P( Exactly one of them will be even) = 3/6 *4/7 + 3/6 * 3/7\n\ne) P(At least one of them will be even) = 1- both of the odd = 1- (3/6 * 4/7)\n\n>> as for the sample space it would take me forever to write it down! sorry\n\n2) Not sure about this one\n3) a) 5%\n    b) 16% >> dont know wut details to include its just a law
936495
This is the opposite of the probability of getting all no's.  The probability of getting all No's is (9/10)^7.  The probability of getting at least one yes is 1-(9/10)^7.\n approximately 0.522
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
694267
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.
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
455115
ten points if you can figure this out first .. right?? \n\n\nANU!!
484411
49/99\n\nProof:\n\n1) Probability that first passenger (P1) sits in last passengers (P100) seat (S100) is 1/99\n2) probability that P1 sits in S99 , and P99 sits in S100 is (1/99) * (1/2)\n3.1) probability that P1 sits in S98, and P98 sits in S100 is (1/99) * (1/3)\n3.2) probability that P1 -> S98, P98->S99, P99->S100 is (1/99) * (1/3) * (1/2)\n4.1) P1->S97, P97->S100 = (1/99)*(1/4)\n4.2) P1->S97, P97->S99, P99->S100 = (1/99)*(1/4)*(1/2)\n4.3) P1->S97, P97->S98, P98->S100 =  (1/99)*(1/4)*(1/3)\n4.4) P1->S97, P97->S98, P98->S99, P99->S100 = (1/99)*(1/4)*(1/3)*(1/2)\n5.1) P1->S96, P96->S100 = (1/99)*(1/5)\n5.2) P1->S96, P96->S99, P99->S100 = (1/99)*(1/5)*(1/2)\n5.3) P1->S96, P96->S98, P98->S100 =  (1/99)*(1/5)*(1/3)\n5.4) P1->S96, P96->S98, P98->S99, P99->S100 = (1/99)*(1/5)*(1/3)*(1/2)\n5.5) P1->S96, P96->S97, P97->S100 = (1/99)*(1/5)*(1/4)\n5.6) P1->S96, P96->S97, P97->S99, P99->S100 = (1/99)*(1/5)*(1/4)*(1/2)\n5.7) P1->S96, P96->S97, P97->S98, P98->S100 =  (1/99)*(1/5)*(1/4)*(1/3)\n5.8) P1->S96, P96->S97, P97->S98, P98->S99, P99->S100 = (1/99)*(1/5)*(1/4)*(1/3)*(1/2)\n\n1) (1/99)\n2) (1/99)(1/2)\n3) (1/99)(1/3)(1 + ½)\n4) (1/99)(1/4)(1 + 1/3)(1 + ½)\n5) (1/99)(1/5)(1 + ¼)(1 + 1/3)(1 + ½)\n\ni) (1/99)(1/i)product(j=2 to (i-1),(1+1/j))\n\n99) (1/99)(1/99)product(j=2 to 98,(1+1/j))\n\nGrand Sum = (1/99)*(1 + ½ +sum(i=3 to 99,(1/i)*product(j=2 to (i-1),(1+1/j))))\n=(1/99)*(1 + ½ + sum(i=3 to 99,(1/2)))\n=(1/99)*(1 + 98/2)\n= 50/99\n\nProbability that last passenger gets bumped = 50/99\n\nHence Probability that last passenger does NOT get bumped = 1 – 50/99 = 49/99\nA bit less than 50%\n\nWould be 1/2 like others suggest if first passenger was not constrained to sit in the wrong seat as given in the problem statement.
998556
Find the probability of the complement occurring and then subtract that from 1.\nThe complement would be that an odd occurs 0 or 1 times.\nP(odd) = .5 and P(even) = .5\nP(0 odds) = C(10,0)(.5)^0 (.5)^10 = .0009765625\nP(1 odd) = C(10,1) (.5)^1 (.5)^9 = .009765625\n.0009765625 + .009765625 = .0107421875\n1 - .0107421875 = .9892578125
1155743
Mutually exclusive - They both cannot occur at the same time.\nTherefore P(A and B) = 0\nP(A or B) = .5
872621
well in my skool we take it as {<and rule>} and {<or rule>example:there is 5 white marble and 5 black marbles in a bag...u pick 2 randomly without replacing them in the bag.....wat is the probability that u get 2 white marbles==>this is a combination 2 the way it is solved is multiplying the the 2 combined probabilities\nanswer:pr:(white AND white)=(5/10)*(4/9)\n\nnow the set is this way....wat is the probability that u get 2 marbles of different colors in the first 2 picks===>set rule\n\nanswer:there is 2 combinations in which u can get 2 different colors there are {white then black} OR {black then white} u find the 2 combinations then ADD them\npr(WandB OR BandW)={(5/10)*(5/9)}+{(5/10)*(5/9)}
1159931
Here's the formula:\nThe probability that all birthdays are different is\nP(365,r)/365^r, where r is the number of people and P stands for permutation.\nThus, the probability that at least two are the same is\n1 - P(365,r)/365^r\nr = 20\n1 - P(365,20)/365^20\n= 1 - .588561616419\n= .411438383581\nSo there is a .4114 chance that at least 2 people in a group of 20 people will have the same birthday.\nThe birthday problem is pretty interesting. If there are 23 people, the probability that at least two have the same birthday is .507. If there are 50 people, it is .970.
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.
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
696506
There are 10 different digits\nso p(x)=1/10=0.1\nwhere p(x) means the probability of number 'x' being selected.\neg probability of the number 3 being selected,\np(3) = 0.1
1161943
Let's deal with "b" first because it makes the general solution clear.\n\nFor an unbiased coin the probability of "Heads" (H) = probability of "Tails"(T) = 1/2 = 0.5\n\nIf n = 3, then you interested in the probabilty of two people having heads and one tails; or two people having tails and one heads.\n\nThe pobability of a single event where two people have heads and one has tails = 0.5 x 0.5 x 0.5= 0.125. But this can happen 3 ways: THH, HTH, HHT and therefore the probability that 2 people have heads and one has tails = 3 x 0.125 = 0.375. Of course the probability that two will have tails and one heads will also be 0.375. So the probability that a leader would have been chosen after one iteration (vote) = 2 x 0.375 = 0.75. (Note: if 3 coins are tossed there are only 4 possibilities: all heads, all tails, two tails and one head, or two heads and one tail.)\nThe probability that a leader is not chosen the first time = 1-0.75 = 0.25. And the probablty that a leader will not be chosen after two iterations is (0.25)^2 = 0.0625. And therefore the probabilty a leader will be chosen within two iterations = 1-0.0625= 0.9375 or nearly 94%. Now we'll deal with the general case.\n\na) For arbitrary n, and a single iteration, the probability of choosing a leader = 2n(0.5)^n. And the probability of not choosing a leader is 1-2n(0.5)^n. There is a difference between the probability of choosing a leader "in" the kth iteration and "by" the kth iteration. The probability that a leader will not have been chosen after k iterations is \n(1-2n(0.5)^n)^k. And the probabilty a leader will have been chosen is simply 1 minus this value.\n\nSample calculation: assume n = 20 and k=50\n\nProb. leader not chosen = (1-40(0.5)^20)^50 = 0.9981\n\nAnd the probability a leader will have been chosen = 1-0.9981 = 0.0019 which is less than 0.2 percent or 1 in 500. This shows that as n increases, the probability that one person will have a different result than everyone else becomes very small and enormous number of iterations would be required to have a reasonable chance of choosing a leader.