Probability and Density?
353110
This sounds like an excellent application of random graph theory (each person is a random vertex, and each edge has a weight giving each path a weight as well). I read a few paragraphs on this a while back, but its been too long to fashion an application for you. I think you will be interested in it.\nCheck out Random Graphs, by Bollobas or The Strange Logic of Random Graphs, by Spencer.
712323
p is a probability. So you're almost right - the question is, can you have a p value of 0 or 1?\nAnd the answer is yes. Although nearly every single statistical test you will use will never give 0 or 1, it is theoretically possible.
631161
This is a binomial probability.\nP(success) = 1/5, P(failure) = 4/5.  There are 6 events here, so P(2 successs) = (1/5)(1/5)(4/5)(4/5)(4/5)(4/5).\nP(at least 2 successes) = P(2 successes) + P93 successes) + . . . . P(6 successes) or 1 - P(O successes) + P(1 success) + P(2 successes) which you already computed above.\nLet me know if you don't understand.
794682
Suppose X is a continuous random variable whose density function f is defined by f(x) = x/2 whenever 0<x<2 and f(x) = 0 otherwise. Then,\nE[X] = integral (x * f(x), x=0..2 ) dx\n     = integral (x * x/2, x=0..2 ) dx\n     = (1/2) integral (x^2, x=0..2 ) dx\n     = [(1/6)*x^3, x=0..2] \n     = (1/6)*2^3 - (1/6)*0^3 \n     = (8/6)\n     = (4/3).\n\nNext, \nE[ | X - E[x] | ] = E[ | X - 4/3 | ] \n     = integral ( | x - 4/3 | * f(x), x=0..4/3 ) dx + integral ( | x - 4/3 | * f(x), x=4/3..2 ) dx\n     = integral ( - (x - 4/3) * f(x), x=0..4/3 ) dx + integral ( (x - 4/3) * f(x), x=4/3..2 ) dx\n     = integral ( (4/3 - x) * x/2, x=0..4/3 ) dx + integral ( (x - 4/3) * x/2, x=4/3..2 ) dx\n     = integral ( (2/3)x - (1/2)x^2), x=0..4/3 ) dx + integral ( (1/2)x^2 - (2/3)x, x=4/3..2 ) dx\n     = [(1/3)x^2 - (1/6)x^3, x=0..4/3 ] dx + [(1/6)x^3 - (1/3)x^2, x=4/3..2]\n     = ( (1/3)(16/9) - (1/6)(64/27) - 0 + 0) + ( 8/6 - 4/3 - (1/6)(64/27) + (1/3)(16/9))\n     = ( 16/27 - 32/81 ) + ( 16/27 - 32/81 )\n     = 2(16/27 – 32/81)\n     = 32/81.\n\n--- \n\nSome notation used:\nintegral (f(x), x = a..b) dx means we evaluate the definite integral of f(x) bounded by x = b and x = a.\n\n[ F(x), x =  a.. b ] means we evaluate the difference, F(b) - F(a).\n.
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!)
979650
For any random variable X, if a is a positive number, it is clear that the probability that X will be larger than a goes to 0 as a grows without limit. This is because a probability distribution has integral 1, and therefore the area under the probability density curve (continuous distribution) to the right of a, or the sum of the probabilities for values of X larger than a (discrete distribution) have to go to 0 as a goes to infinity.\nHope this helps -
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.
1099243
No. This is what we call probability. Take a statistics course.
162268
The density of the fluid and the density of the object.
255172
Density = mass/volume\nDensity = 615/105 gm/cm3 = 5.86 gm/cm3
704806
For discrete events, the basic procedure is to list all possible outcomes, and the probability of each (which must add up to 1). The probability of being in one of an identifiable smaller group of outcomes (called a subset) is simply the sum of the individual probabilities of the outcomes within that group.\n\nFor example, if you have two distinguishable dice, there are 36 different ways they can be rolled. If the dice are unweighted, the probability of each is 1/36. Of those, 6 add up to 7. That means the odds of getting 7 are 6/36=1/6.
152796
try this site...\n\nhttp://www.futureaccountant.com/probability/
551202
1) Find the probability of getting a red ball when a bag contains 2 red balls and one blue ball.\n\n2) There is a red ball, a blue ball and a green ball in a bag. Find the probability of getting either a red ball or a blue ball.\n\n3) There are 10 yellow chips and 20 purple chips in a sack. When one chip is selected at random, find the probability of NOT getting yellow chips.
767048
Random chances are the same as probabilities. If you are so interested, you should look into Quantum Theory. It describes it in depth.
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
435037
try this site for help on probabiliity\n\nhttp://www.futureaccountant.com/probability/
1143752
http://cis.jhu.edu/~bruno/s06-311/\n\nhttp://www.eng.wayne.edu/page.php?id=1434\n\nhttp://web.njit.edu/~dhar/math333/fin333_sp02A.pdf#search='sample%20Probability%20and%20Statistics%20exam'