Guest Posted April 14, 2009 Report Share Posted April 14, 2009 I have a bag containing 4 different colors of marbles. The number of marbles is a prime number. The product of the numbers of each color marbles is a 2 digit number The number of marbles in any color is different from the number of marbles in any other color There are twice as many green marbles as there are yellow marbles There are 5 more black marbles than red marbles If I take 3 marbles from the bag, what is the probability that I get at least one green marble? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 14, 2009 Report Share Posted April 14, 2009 61 Marbles Total Red - 10 Yellow - 12 Black - 15 Green - 24 Therefore...... you have a 40% change of getting a Green Marble 1:2.5 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 14, 2009 Report Share Posted April 14, 2009 if "The product of the numbers of each color marbles is a 2 digit number" means that green x red x yellow x red = two digit number, then: black:6 red:1 yellow:2 green:4 multiply:48, sum:13(prime) With total 13 marbles, there are c(13,3)=286 combination to form a 3 set. c(9,3)=84 of them will be formed by 3 non green marbles. Remaining 286-84=202 marbles will include at lest one green marble. Then the odds that at least one green marble is picked=202/286 ??????? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 14, 2009 Report Share Posted April 14, 2009 The product of those numbers is not a two digit number 1 red 2 yellow 4 green 6 black Picking random 3 marbles to take out of the bag will skew any percentage to pull a green once the number of marbles is decreased from 13 to 10. It would be a probability of a probability and you can therefore only estimate the final probability. The current (above) percentages would lead me to estimate 2 black and 1 green to be pulled from the bag leaving: 1 red 2 yellow 3 green 4 black Probability of the last draw being green is 3/10 = ~20% See the catch of estimating an estimation? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 14, 2009 Report Share Posted April 14, 2009 meant = ~30% lol Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 14, 2009 Report Share Posted April 14, 2009 course that's more mathematically correct nobody Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 14, 2009 Report Share Posted April 14, 2009 61 Marbles Total Red - 10 Yellow - 12 Black - 15 Green - 24 Therefore...... you have a 40% change of getting a Green Marble 1:2.5 I have to say that I see about 9 errors in this. The worst being not covering your answers with a spoiler section Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 14, 2009 Report Share Posted April 14, 2009 (edited) if "The product of the numbers of each color marbles is a 2 digit number" means that green x red x yellow x red = two digit number, then: black:6 red:1 yellow:2 green:4 multiply:48, sum:13(prime) With total 13 marbles, there are c(13,3)=286 combination to form a 3 set. c(9,3)=84 of them will be formed by 3 non green marbles. Remaining 286-84=202 marbles will include at lest one green marble. Then the odds that at least one green marble is picked=202/286 ??????? Your answer is absolutely correct, nobody. 1 red, 2 yellow, 4 green, and 6 black is the only combination of marbles that fits all criterias. The chance of failure = the chance of picking 3 non-green in 3 picks = 9/13*8/12*7/11 ~ 0.293706 The chance of success ~ 1 - 0.293706 ~ 0.706294 or ~ 70.63% (which is the same as 202/286) Edited April 14, 2009 by uhre Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 14, 2009 Report Share Posted April 14, 2009 (edited) Probability of the last draw being green is 3/10 = ~20% See the catch of estimating an estimation? You are absolutely correct in the number of marbles. That leaves 2 parts you need to address again: 1) The probability part, and 2) Using spoilers to cover your answers Edited April 14, 2009 by uhre Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 14, 2009 Report Share Posted April 14, 2009 You are absolutely correct in the number of marbles. That leaves 2 parts you need to address again: 1) The probability part, and 2) Using spoilers to cover your answers The number of marbles is correct. The probability can be calculated as it is "1 - P(no green marbes)" ie picking thre non green marbes in succsession. If you assume that you do NOT put the marble back each time then you get.... = 1 - (9/13 x 8/12 x 7/11) = 1 - 0.293706 = 0.706294 If you do put the marble back then the probability is "1 - (9/13)^3" = 0.668184 I THINK!!!!!!! Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 14, 2009 Report Share Posted April 14, 2009 You are absolutely correct in the number of marbles. That leaves 2 parts you need to address again: 1) The probability part, and 2) Using spoilers to cover your answers The probabilty is somewhere around 0.71.. In fractions it is 102:143 MPM Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 14, 2009 Report Share Posted April 14, 2009 there can be many combinations.... here is one of them.. green = 4 yellow = 2 red = 3 black = 8 total = 17 thus probability for green ball is ~24% Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 14, 2009 Report Share Posted April 14, 2009 there can be many combinations.... here is one of them.. green = 4 yellow = 2 red = 3 black = 8 total = 17 thus probability for green ball is ~24% Accept 4x2x3x8=144 which is not a two digit number. And what is with the immense lack of spoiler windows?!? Quote Link to comment Share on other sites More sharing options...
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I have a bag containing 4 different colors of marbles.
The number of marbles is a prime number.
The product of the numbers of each color marbles is a 2 digit number
The number of marbles in any color is different from the number of marbles in any other color
There are twice as many green marbles as there are yellow marbles
There are 5 more black marbles than red marbles
If I take 3 marbles from the bag, what is the probability that I get at least one green marble?
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