A fair die is rolled.  Find the probability that the number obtained is not greater than 4.?
360679
1, 2, 3, 4 are not greater than 4.  There are 6 numbers on a die.  4/6 or 66.6% chance that the number is not greater than 4.
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
1108302
In Theory, it is 1/6.\n\nIn Practice, the dice will have some bias.\n\nIf you observe the dice that has its numbers carved out, you will noticed that the number 1 is a bigger circle and the number 6 has smaller circles hopefully to remove the same amount of material from the dice so that the weight is evenly distributed and the center of gravity is maintained at the true center of the dice.\n\nFurthermore, the table on which the dice roll may be biased as well, uneven surface, minute sharp points, etc. Also, your hand has sweat and different amount of stickiness when you release the dice. The dice may also experience uneven wear and tear with usage. You can go on and probe all the "real" world imperfections and these may add up to skew the probablities of the dice and bias the results.\n\nSo how likely is likely?\n\nIn the real world, it is unlikely to be as likely as you would likely like it to be.
985314
its 1/36 chance of any two particular numbers coming up on each roll of the dice.  (1/6 is for one die)
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.
553417
if there are only 10 numbers to chose from, say 0-9 then the odds are 10*10*10, if only nine number say 1-9 then it is 9*9*9. that help ya?
981521
A dice has 6 faces.\nProb that the two dice show equal = 1/6.\nSo Prob that the two dice show unequal = 5/6.\nSo the expected value of X is\n\nE(X) = 10*5/6 + 100*1/6 = 150/6 = 25.
749852
1.Probability of getting a number from 1-4 on throwing the dice\n         = 4/6\n\n2.Probability of a correct throw = 1/4\n\n3.Total number of questions = 160\n\nYour Answer = (4/6 * 1/4) raised to the power 160
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.
469430
the probility of a particular event is basicly "the amount of favorable outcomes over the total amount of outcomes. for example, \n\nwhats the chances of trowing a 3 with a dice?\n\ntotal favorable outcomes= 1 (because the is only one 3 on a dice)\ntotal outcomes= 6 (because there a six outcomes posible with a dice)\n\ntherefore the probibility of trowing a 3 with a dice is:\n=1/6\n\nthere are other formulars (eg. the  binomial theorm) for more complicated probibilities but this is the basics.
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
191504
In the experiment of testing for the number appearing on the spinner \n\nTotal number of possible choices\n= Number of ways in which a number can appear on the spinner \n= C(15,1)\n=15\n[Any one of the 15 numbers may appear]\n\nLet "A" be the event of the spinner landing on a number less than 7\n\nFor Event "A"\n\nNo. of favorable choices \n= No. of ways in which the spinner would stop at a number less than 7\n==> m(a) = C(6,1) = 6  [1,2,3,4,5,6]\n\nProbability of the spinner landing on a number less than 7\n==> Probability of occurance of event "A" = m/n\n==> P(A) = 6/15 \n= 2/5\n=0.4\n\nhttp://www.futureaccountant.com/probability/study-notes/\n\nhttp://www.schoolingkids.com/
731770
You have 5 dice, with 6 numbers each, for a total of 6^5 possible rolls on your first roll.\n\nThere are 6 ways you can get a Yahtzee:\n1-1-1-1-1\n2-2-2-2-2\n3-3-3-3-3\n4-4-4-4-4\n5-5-5-5-5\n6-6-6-6-6\n\nTherefore your chances of getting a Yahtzee on your first roll is 6/6^5 = 6/7776 = 1/1296 ≈ 0.00077\n\n~ ~ ~\n\nIf you already have a 5-5-5, you only roll two dice -- with two dice you have 6^2 = 36 possible outcomes.\n\nP(yahtzee on next roll) = P(5-5 on next roll) = 1/36\n\nP(full house on next roll) = P(1-1 or 2-2 or 3-3 or 4-4 or 6-6 on next roll) = 5/36\n\n{ notice I left out 5-5, but technically 5-5-5-5-5 is a full house too, in addition to a yahtzee, so depending on how your teacher interprets "full house" the answer to this might be 6/36 rather than 5/36 }\n\nP(4 of kind) = P(5-1 or 5-2 or 5-3 or 5-4 or 5-6) = 5/36\n\n~ ~ ~\n\nYou have 2,3,4,5 and are rolling one die -- you have 6 possible outcomes for rolling that one die\n\nP(large straight) = P(1 or 6) = 2/6 = 1/3\nP(small straight) = 100% since you already have a small straight with your 2-3-4-5\n\n~ ~ ~\n\nYou have 3-4-5 and are rolling 2 dice -- you have 36 possible rolls with those 2 dice\n\nP(full house) = 0, since a full house is 3 of one number and 2 of another, and you already have three different numbers showing!\n\nP(3 of a kind) = P(3-3 or 4-4 or 5-5) = 3/36 = 1/12\n\nP(large straight) = P(1-2 or 2-6) = 2/36 = 1/18\n\n~ ~ ~ \n\nI hope you understood how I approached and solved these, so that you can do them on your own when it's time to take your test on this material.  Good luck!\n\n~ ♥ ~
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
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.
627413
OK.  Here are the possibilities.  It is important to keep the order of the dice in mind, so you can think of this as rolling a die 5 times.\n\nSSSSD\nSSSDS\nSSDSS\nSDSSS\nDSSSS\n\nNow each of these will have the same probability, so if I can find the probability of one of them, I can take that and multiply by 5 since there are five combinations.\n\nSSSSD can also occur in different ways.  If S=1, then D could be 2, 3, 4, 5, or 6.  So if S=1, then the probability of this combination is (1/6)^4(5/6).\n\nNow if S=2, then D can be 1, 3, 4, 5, or 6.  So again, (1/6)^4(5/6).\n\nSo hopefully you can see that there are 6 ways that you can do this.  So the probability of SSSSD is 6(1/6)^4(5/6).\n\nSo overall, the probability of getting four of a kind (without getting five of a kind) is\n5*6*(1/6)^4(5/6) = 150/6^5.\n\nSo it looks like your method does work.  Good job; this isn't an easy on.  There is always more than one way to look at probabilities.
263161
This is a good probability problem -- I LOVE probability problems.  This is really asking 4 different questions.  First, what are the chances of the first pull being black.  (9 total socks, 5 are black ...).  Then what are the chances of the second one being blue (Keep in mind that now you only have EIGHT socks in the drawer, unless you put the last one back - so 1 blue and eight total socks).  Then chances of a brown sock with only 7 socks left (3 brown, seven total socks).  \n\nWhen you get all those probabilities, in order to see what the chances of this AND this AND this happening, you have to multiply the fractions (if it was "What are the chances of this OR this happening, you'd add them).\n\nGood luck.  This stuff is FUN!  (Especially if you ever want to go to VEGAS!)