Probability Help, Random Variables?
815085
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
624712
This problem is solvable.\nProbability of event A is written P(A).\n\nP(A)= Total # of outcomes that correspond to event A (divided by -or- over) Total # of outcomes in sample\n\nAnswer:\nP(A) = 12/20 = 3/5
97799
i don't believe the person above me got the second part right.  he is right about the beginning - the first probability is 2/5:\n\npossible numbers to draw: 2 4 6 11 13 15\n\nso you can choose any of these six out of the fifteen given (6/15 = 2/5).  the next number you draw must be also a "winning" number.  BUT, you have to have an even number less than 7 AND an odd number greater than 10.  this gives us two scenarios:\n\nIF the first number picked was even and <7, then the possible numbers you can pick from are ONLY: 11 13 15, since picking another even number <7 will not meet your requirements.  this gives us a probablity of 3/14, since there are 3 "winning" numbers out of the 14 numbers left.\n\n\nIF the first number picked was odd and >10, then the possible numbers you can pick from are ONLY: 2 4 6, since picking another odd number >10 will not meet your requirements.  this gives us a probablity of 3/14, since there are 3 "winning" numbers out of the 14 numbers left.\n\nSO, finally, suppose you go with the first scenario.  the probability would be: (6/15)*(3/14) = 3/35  going with the second scenario will give you the same thing, 3/35.  thus adding these two, since you can have either scenario occur and still meet your requirements, gives you: 3/35+3/35 = 6/35.\n\nAnd that would be the odds: 6/35.
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.
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.
911469
I'll do some for you to show how they are done. In all cases probability P = #number of good throws/ # of all throws.\nAlso 1-P = #number of bad throws/ # of all throws\nOften you can arrive at a solution by considering the dices one by one.\n1) First dice can be everyting, second dice can be one of the five numbers not yet thrown, third dice can be one of the four numbers not yet thrown, etc. This leads to P = (6/6) * (5/6) * ... *(1/6) =720/46656 = 0.0154\n2) Divide the dice in three pairs, named A,B and C. The probability to throw a pair of two's with pair A is 1/36. Idem for B and C. But we have 6 posibilities to arrange the pairs ABC, ACB, ... And we have 6x5x4x3x2x1 possibilities to arrange the dice into the three pairs. Therefore P = 6 x 720 x (1/36)^3 = 0.0926\n3) First dice 1, second dice 1, third dice 1 gives a probability of (1/6)^3 = 1/216. But there are 20 possibilities to arrange the dice.\nP=20/216 = 0.0926
590072
Make a tree that shows all possibilities you have.\nOn the first level there is the choice of the box with 0.5 probability each. On the next level, box A, you have two choices again, with the prob. is 3/7 to find a red ball, 4/7 to find a white ball, etc.\n\nWhen you have finished to tree, use it to make a list and\ncalculate the prob. for picking red-red, red-white, etc.
499164
P(A and B and C) = P(A)P(B|A)P(C|A and B)\n\nNow P(Acceptance) = 0.08,\nP(Receive housing|Acceptance) = 0.6, and\nP(At least one roomate|Acceptance and Receive housing) = 0.8, meaning that\nP(No roommate|Acceptance and Receive housing) = 1-0.8 = 0.2.\n\nSo P(Acceptance and Receive housing and No roomate) = 0.08 * 0.6 * 0.2
1169384
The previous responders are incorrect. \n\nLet the expected length of the game (in tosses) be L. We can classify each game into one of three starting sequences (in the following table P=probability,E=expectation value for total length of game):\n\nH --> P = 1/2 , E = L+1\nTH --> P = 1/4 , E = L+2\nTT --> P = 1/4 , E = 2\n\nThe first two cases are equivalent to starting the game over after 1 and 2 tosses respectively.\n\nSumming P*E for these three events gives L:\n\nL =(1/2)*(L+1) + (1/4)*(L+2) + (1/4)*(2)\n= (3/4)L + 3/2\n\n--> L = 6
731902
There are 20 students in the class. There are 4 possibilities when you choose the 3 random students. I've included the probabilities of these 4 possibilities. 1) and 3) are the answers to your questions.\n\n1)   3 boys = (4/20)(3/19)(2/18) = 0.0035\n\n2)    2 b, 1 g = (4/20)(3/19)(16/18)x3(=3C2) = 0.0842\n\n3)    2 g, 1 b = (16/20)(15/19)(4/18)x3(=3C2) = 0.4211\n\n4)     3 g = (16/20)(15/19)(14/18) =  0.4912\n\nFor the probabilities to be correct they must sum to 1:\n\n0.0035+0.0842+0.4211+0.4912 = 1
95999
Don't you think you could have written this more clearly? \n\nIf there is a uniform distribution of the weight w being between 5 and 10, then each of these probabilities is equal:\n\nP5_6=P{5<=w<6}\nP6_7=P{6<= w<7}\nP7_8=P{7<=w<8}\nP8_9=P{8<=w<9}\nP9_10=P{9<=w<10} \n\nThe sum of these is 1, so \n\n1=P5_6+P6_7+P7_8+P8_9+P9_10, and\nP5_6=P6_7=P7_8-P8_9=P9_10\n\nso 1=5xP9_10, so P9_10=0.2\n\nThe probability is 0.2.
405425
Probability of getting one head = 1/2\nProbability of getting two heads = (1/2) * 2 = 1/4\nYou have a 25% chance of getting a head two times...\nAnd 12.5% chance of getting a head 3 times...\nAnd 6.25% chance of getting a head 4 times...\n.\n.\n.\nAnd 0.09765625% chance of getting a head 10 times!!! :)
543159
Zero probability. Anything more than that is technically *possible*.
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
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?
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
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%