Billy had fifty coins, totaling of $1.00.He dropped one coin. What is the probability that it was a penny?
152796
try this site...\n\nhttp://www.futureaccountant.com/probability/
598521
What were the odds?\n\n4 to 1?\n\n8 to 1?\n\n20 to 1?\n\n**********************************\nEDIT\n\nIf the odds were 6 to 1, then you won $120.
1161943
Let's deal with "b" first because it makes the general solution clear.\n\nFor an unbiased coin the probability of "Heads" (H) = probability of "Tails"(T) = 1/2 = 0.5\n\nIf n = 3, then you interested in the probabilty of two people having heads and one tails; or two people having tails and one heads.\n\nThe pobability of a single event where two people have heads and one has tails = 0.5 x 0.5 x 0.5= 0.125. But this can happen 3 ways: THH, HTH, HHT and therefore the probability that 2 people have heads and one has tails = 3 x 0.125 = 0.375. Of course the probability that two will have tails and one heads will also be 0.375. So the probability that a leader would have been chosen after one iteration (vote) = 2 x 0.375 = 0.75. (Note: if 3 coins are tossed there are only 4 possibilities: all heads, all tails, two tails and one head, or two heads and one tail.)\nThe probability that a leader is not chosen the first time = 1-0.75 = 0.25. And the probablty that a leader will not be chosen after two iterations is (0.25)^2 = 0.0625. And therefore the probabilty a leader will be chosen within two iterations = 1-0.0625= 0.9375 or nearly 94%. Now we'll deal with the general case.\n\na) For arbitrary n, and a single iteration, the probability of choosing a leader = 2n(0.5)^n. And the probability of not choosing a leader is 1-2n(0.5)^n. There is a difference between the probability of choosing a leader "in" the kth iteration and "by" the kth iteration. The probability that a leader will not have been chosen after k iterations is \n(1-2n(0.5)^n)^k. And the probabilty a leader will have been chosen is simply 1 minus this value.\n\nSample calculation: assume n = 20 and k=50\n\nProb. leader not chosen = (1-40(0.5)^20)^50 = 0.9981\n\nAnd the probability a leader will have been chosen = 1-0.9981 = 0.0019 which is less than 0.2 percent or 1 in 500. This shows that as n increases, the probability that one person will have a different result than everyone else becomes very small and enormous number of iterations would be required to have a reasonable chance of choosing a leader.
898818
There are sixty-six possible outcomes in the first two scenarios, and fifty-five possible outcomes in the third (Mary does not win):\n\nWhat is the probability a married couple wins the prizes? \n9.09% (6/66 x 100)\n\nWhat is the probability one prize goes to a man and one to a woman? \n54.5% (36/66 x 100)\n\nIf Mary does not win a prize, then what is the probability that John will? \n18.2% (10/55 x 100)\n\n(or 0% if John never made it to the party!)
876730
A quarter and a nickel.\n\nONE of them is not a nickel...its a quarter!
1022123
A bit is a one or a zero, so decide on a convention such as "Heads is 1 and tails is 0" then flip a coin.\n\nIf you need to produce a random bit in some other fashion, you will need to provide more details about your problem.
343331
What math test do you take?  I'm in Algebra 1 and I only remember 1 question.  It's impossible.\n\nSomeone has a total of 100 coins in dimes and quarters.  The total value adds up to $14.05. How many quarters does she have?\n\nA) 27 (I chose this-it's a dollar off)\nB) 40\nC) 56\nD) 79 (I forget-it's way too high)\n\nWell, that's all I have.  Bye bye!\n\nThanks to the guy who solved that.  I won't get a zero now!
1042197
50/50.  The odds don't change no matter how many times the coin is flipped.
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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
1092672
Its 1:220\n\n(4/52)(3/51) = 1/221 which is the probability\nThus the odds are 1/220 or 1:220
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.
501856
a penny weighs 2.5 g.  there are 454 g in a lb.\n\n454/2.5 = 181.6, so there are 182 pennies, or thereabouts.\n\nROFL @ the other answer...
786207
did u mean '93.5% or more'?\n\nif so, there is only 1 chance of that occurring and that is when all 8 coins are tossed and revealing heads. \n\nso that is a probability of 1/2^8 = 1/256 = 0.00390625
955894
x1*p(x1) + x2*p(x2)\n\nx1 = 1 for  event dots> 3,\nx2 = 4 for event  dots<=3.  \nThese are equally probable because there are 6 dots on the die.  \n\n1*(1/2) + 4*(1/2)
242742
Remember, chance events are not dependent on the previous outcome.  The correct answer, as others have pointed out, is 50%.
135086
back a while in history, the equivalent of the dollar coin was breakable into eight pieces, each piece was a bit.\n\ni'm not sure if my memory is correct, but i think the english, the spanish, and early americans all did this.\n\nSo two bits is 2/8, or 1/4 of a dollar, so... $0.25
1120768
He should set the ratio of silver to gold as follows:\nUrn 1: 1 silver ball, 0 gold balls.\nUrn 2: 49 silver balls, 50 gold balls.\n\nFirst he is going to pick an urn, then he is going to pick a ball.  Half the time he will pick the first urn and have a 100% chance of picking a silver ball.  The other half of the time he will have a 49/99 chance (49.4949%, or almost 50%) of picking a silver ball.\n\nFiguring the weighted averages:\n0.50 x 1.00 + 0.50 x 0.4949\n= 0.50 + 0.247474\n= 0.747474\n\nSo with this arrangement he has a nearly 75% chance of surviving which is better than 50/50!