Math...? Probability...? Please show your work?
812753
a.  820 is 10 min lates sooo... 25% prob\nb. 800 is 10 min early sooo..... 15%\n\nc.  and.. then hmmm i gues you need to find out what "mutauilly exclusive means and compare the two answers\nd. same for d
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.
770080
Let us call x the minutes hand on the clock.\nIf 0<x<=5, then it is train B. (5 minutes)\nIf 5<x<=12, then it is train A. (7 minutes)\nIf 12<x<= 24, then it is train A. (12 minutes)\nIf 24<x<= 25, then it is train B. (1 minute)\nIf 25<x<= 45, then it is train B. (20 minutes)\nIf 45<x<=48, then it is train A. (3 minutes)\nIf 48<x<=60, then it is train A. (12 minutes)\n\nWe add it up to get that for 34 of the minutes, he gets on A, and that for of the 26 minutes he gets on B. There are 60 minutes in an hour, so the probability is 34/60, which reduces to 17/30.
554599
4.2\n\nfor a binomial distribution the mean is the number * the probability
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
515436
plz complete the question. i like to solve the questions on probability, but yur question is incomplete.
440605
Binomial Distribution, \nPoisson Distribution, \nNormal Distribution\n\nhttp://www.futureaccountant.com/probability/
792306
If the probability of picking an A is .5 and the probability of picking a B given an A is picked  is .3   \n Find the probability of picking both A and B?\n\nAnswer:     Pr(A) = .5   \n                  Pr(B/A) = .3 \n                  Pr(A∩B) = ? \n \n\n Pr(A∩B) =  Pr(B/A) * Pr(A) \n                = .3*.5\n                =.15    that is the answer.
1136735
P(Last is W) = P{(2nd is W and 1st is W) or (2nd is B and 1st is W)}\n= P(2nd is W and 1st is W) + P(2nd is B and 1st is W) (since these are mutually exclusive events)\n= P(2nd is W)P(1st is W|2nd is W) + P(2nd is B)P(1st is W|2nd is B) (both use the general multiplication rule)\n= (6/10)(7/11) + (4/10)(6/11)\n= (42+24)/110\n= 66/110\n= 0.6\n\nWhich is the probability of getting a white from either box, which I think makes sense, since both have the same proportion of whites.
728828
1) AA*aa = 100% Aa\n\n2 & 3) Aa*Aa = 50% Aa, 25% aa (also 25% AA)\n\nFor 4-6, figure the probability of the A gene result, then multiply that by the probability of the B gene result.\n\n4) Aa*Aa = 50% Aa (.5); Bb*BB = 50% BB (.5)\n.5*.5 = .25 = 25% AaBB\n\n5) 50% Aa; 50% Bb = 25% AaBb\n\n6) 25% AA; 50% BB = .25*.5 = .125 = 12.5%
587245
If you have n events where the number of outcomes for the first event is p1 and the number of outcomes for the second event is p2 and so on up to the nth event (pn), then the number of total ways the event can occur is\n\n(p1)(p2)(p3)...(pn).
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.
967346
This probability problem can be used using Binomial Distribution.  \n\nGiven a number of 'x' success, the probability in a binomial experiment is:\n\nP(x) = [n choose x]*p^x*q^(n-x)\n\nwhere:\n\nn choose x is a combination problem: n!/[(n-x)!x!]\np = probability of success\nq = probability of failure\nn = total selection\nx = successes\n\nIn this problem, we are looking for at least one case has a success (which is one of the highest 9 numbers).  In essence, we are solving for:\n\nP(x=1) + P(x=2) + P(x=3) + P(x=4) + P(X=5).  \n\nHowever it would be easier to find the probability of the failures and subtract from the total probability because we only have to calculate 1 probability.  This would be:\n\n1 - P(x=0) is what we are looking for.  Now all we need is to calculate the probabilty.\n\nFor P(x=0), we have the following:\n\nn = 5\nx = 0\np = (9/26) = probability of success\nq = (17/26) = probability of failure\n\nPlugging into P(x), we get:\n\nP(x=0) = (5!/5!)(9/26)^0(17/26)^5\n\nSolving, you get P(x=0) = 0.12\n\nTherefore, \n\n1 - P(x=0) is ------>\n\n1- 0.12 = 0.88\n\nBelow is a website devoted to Binomial Distribution exercises\n---------\n\nHope this helps
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
430637
First choose which number it is.  There are 6 numbers, so there are C(6,1) = 6 choices.\nThe probability of that number is 1/6\nSince there are three\n(1/6)(1/6)(1/6) = 1/216\nNow multiply this times 6\n6/216 = 1/36
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.
1135408
OH NOES THIS IS LIKE A MATH PROBLEM!!!! i can't handle any more variables :( sry.\n\n\nhahahaha.