If choosing an arbitrary triangle from all possible triangles have the same probability, what is the probability of choosing a triangle with one angle=45 degrees? The complete answers (with solution) will give the best answer bonus.
217908
The probability would be zero.  Since the angles of a triangle in this case would be considered a continuous random variable (meaning that the angle could be anything in the range of (0, 180)), the probability that you would find a triangle with an angle that is exactly 45 degrees is impossible.\n\nThis is analogous to finding a person who is exactly 6 feet tall.  You may find a person who is close to being 6 feet tall, but their height will always be off by a fraction of an inch.  Same is true with the triangles; you may be able to find a triangle that has an angle very close to 45 degrees, but that angle will always be off by a fraction of a degree.
357155
The probability is the number of desired outcomes divided by the total number of outcomes.\n\nThere are 4 vowels out of a total of 11 letters.  Each letter has equal probability so the chance of a vowel is 4/11 (or 36.3636%)\n\nSimilarly there are 7 consonants out of a total of 11 letters.  The probability of a consonant is 7/11 or 64.6464%\n\nFortunately these add up to 11/11 or 100%, so that confirms our answers.\n\nIn summary:\nThe chance of drawing a vowel is 4/11 or 36.3636%
297736
Probability\n= P(boy and anyday) and P(boy and that same day)\n= (1/2)(1) * (1/2)(1/365)\n= (1/2)*(1/730)\n= 1/1460
555946
Actually the only way it would be possible is if you use Elliptic or Riemannian geometry, and would require a manifold.  A representation of such a construct would be similar to drawing a 4 dimensional shape in 3 or even 2 dimensions.\n\nEuclidian geometry doesn't allow triangles with a sum of all angles to be greater than 180 degrees.
427734
Not every problem should be solved with an elephant gun.\nJust ask yourself a different question. What is the probability that x2 is closer to x1 than to x3. since the answer to that question is the same as the question you ask and the only 2 possibilities are certain\nand mutually exclusive, the answer should be the same, 1/2.\n\nWhat about if x2 is exactly in the middle you ask?\n\nIt's like asking what is the probability that a random number in an interval is the midpoint. 1/infinity=0
477328
Here's the answer you want:  With a compass set to the radius of the circle, start marking off radii around the circle, dividing it into 6 equal parts.  Connect any 3 alternate points and you'll have your triangle.\n\nNO MATH REQUIRED!
513971
Use the fact that the sum of the angles of a triangle add up to 180 degrees. Let A, B, C be the angles of the triangle. The m<A + m<B + m<C =180. If you know m<A and m<B, you can find m<C.\n\nWhen I write m<A, I mean the measure of angle A, not m is less than A.
522518
From symmetry, you have all the angles\n69 for two, and 180-69= 111 for the other two.\nUse the law of cosines: for triangle abc with opposing angles A,B,C we have the length of the 3rd side c for given a,b from:\n\nc = sqrt(a^2 +b^2 -2abcos(C))\n\nIn your case: a=4, b=6 and C = 69 so\n\nBD = sqrt(36+16-48(cos(69)) = 5.9\n\nFor C = 111 we have:\n\nAC = sqrt(36+16-48(cos(111))= 8.32
74795
For your first pick, you have 52 possible successes out of 52 possible draws, or 52/52, on the second draw, you have 39 possible successes out of 51, or 39/51. On the third draw, you have 26 possible successes out of 50, or 26/50.  On your last draw, you have 13 possible successes out of 49, or 13/49.  By probability, you multiply all of these together to find the total probability - (52/52) * (39/51) * (26/50) *( 13/49) = 685464/6497400 = .105498 or 10.5498%.
525847
P(At least 2 orange) = P(2 orange or 3 orange)\n= P(2 orange) + P(3 orange) (These event are mutually exclusive, meaning that these events cannot occur at the same time.  This uses an addition rule.)\n= P(OON or ONO or NOO) + P(OOO) (OON means orange 1st and orange 2nd and non-orange 3rd.)\n= P(OON) + P(ONO) + P(NOO) + P(OOO) (also mutually exclusive)\n= (3/11)(2/10)(8/9) + (3/11)(8/10)(2/9) + (8/11)(3/10)(2/9)\n+ (3/11)(2/10)(1/9) (Since there are ands here, you would use the multiplication rule)\n\nThis gets you approximately 0.152 as the answer.\n\nThe other people are not answering correctly.  2/11 is the probability of getting an orange candy if you select one.  That is not what was asked.  Three are being selected, not one!  Jacob, just because you rounded correctly doesn't mean your answer is correct.
490268
The x's will all be the same, and the y's will be the difference between 4 & the number:\n\nA=(2,5)\nB=(0,3)\nC=(5,2)\n\nAnother way to do this is to actually draw the triangle and the line y=4. This will give you a visual of what should happen.
189358
Initial case n = 3\n180(3-2) = 180 which is known to be true for a triangle\n\nAssumption: Angle sum of n sided polygon = 180(n-2)\nProve: Angle sum of n+1 sided polygon = 180((n+1)-2) = 180(n-1)\n\nNow, take any n+1 sided polygon, and split it into an n sided polygon and a triangle by drawing a line between 2 vertices separated by a single vertex in between. The total angle sum of the n+1 polygon will be equal to the angle sum of the n sided polygon plus the triangle.\n\nUsing the assumption, the angle sum of the n-sided polygon is 180(n-2). The total angle sum therefore is\n\n180(n - 2) + 180 = 180(n - 2 + 1) = 180(n - 1)    QED\n\nTherefore since it is true for n = 3, and if it is true for n it is also true for n+1, by induction it is true for all n >= 3.
317566
Okay, intially, you pick one marble, out of 12. and there are 3 yellows, then your initial chances are 1/4 of picking a yellow. Now, you've picked one, so there are 11 marbles left in the bag. There are 4 blue marbles in the bag, so you have a 4/11 chance of picking a blue ball at tthis point. \nTo Calculate Probability based  on an event that 1) Has already happened ( the yellow ball was picked first) and that is an event depending on the results of the , you simply multiply the chances. ( 1/4*4/11) or 4/44, or 1/11 chances that that will happen.
278634
Memorize that for any triangle having sides of length a, b, and c and angle C opposite side c,\n\nc^2 = a^2 + b^2 - 2abcos C .\n\nNote that for C = 90 degrees where cosine of 90 degrees is zero, the rule reduces to the pythagorean theorem.
386030
In total there are C(5,2) = 5!/2!3! = 10 ways to choose two officers among 5 people.\n\n1) There are C(3,2) = 3 ways to choose two women (among the three) so the probability is 3/10 = 0.3.\n\n2) There is only C(2,2) = one way to choose two men, so the probability of having two men as officers is 1/10.  Threrefore 0.9 for having at least one woman.\n\n3) This derives from the previous two answers.  There are 9 ways of having at least one woman officer, and 3 ways of having two.  So, 3/9 = 0.333.\n\n4) Knowing that Beth is an officer, this leaves one person to choose among two men and two women for the other position, so it's 2/4 or 0.5.
350677
C - 3 possible answers\n\nIn any given right triangles with the two other angles being a and b,\n\nsin a = cos b AND sin b = cos a,\n\n\nas such, the first and fourth combination will give one answer, the other two will either give square trigonometric values of each angle i.e. sin^2 a and cos^2 a  OR sin^2 b and cos^2 b.\n\nThe only condition that will not satisfy this is when the angle is 45 degrees, which means a = b and contradicting the given statement of a being less than b.\n\nThus the answer is C.
513848
I am making the assumpition that the points you are given form the probability density distribution curve p(x) for the random variable.   This means that P(a<x<b) is ∫[a to b)p(x)dx.  In this type of distribution  ∫[-∞ to +∞]p(x)dx = 1\n\nStart by plotting the distribution from the points given.  The probability of a point being within a range is the area under the curve within that range .  Use geometry to compute these areas: break up the figure into rectangles and triangles.  First get the full area, call that A (this should come out 1).  The width of the area you want is the limit of the values given; in case i, that would be the distance from x = 1 to the right-hand limit of the curve at x = 3, or ∆x = 2.\n\nTo show you how, here is the way to get the full curve area A:\n\nRectangle from (0,.3 to (3,.3): Base 3 height .3, area .9\nTriangle from (1,.3) to (1.5,.5) to (2,.3): Base 1, altitude (.5-.3) = .2, area = .5*1*.2 = .1\n\nTotal area = 1 (as it should).\n\nArea for P(x>1) rectangle base 2, height .3, area .6; triangle base 1 altitude .2, area .1, then P(x>1) = .7, the sum of the areas.\n\nFrom a probabiliy distribution, you cannot get the probability of a single value, only a range.\n\nYou will have to interpolate altitudes for points that do not coincide with the ones given.\n\nEDIT: I have made a drawing of the distribution if that will help.  Find it at http://img218.imageshack.us/img218/8930/distributionhg6.png