<p>u have 30 marbles in a jar</p>
<p>10 red, 10 blue, 10 green</p>
<p>u draw 21 marbles</p>
<p>what is the probablitity that of the remaining marbles in the jar (9), there is atleast 1 marble of each color?</p>
<p>wait a second, in the actual oct SAT, didnt it ask what the probability that you choose atleast one of each color???</p>
<p>xindianx must be remembering the problem correctly. that would make it a fairly easy question (the answer would be 1). they would never ask the question the way sonic pwr described it. that's much too hard a question for the SAT.</p>
<p>wait a second, it says at least one of each color, so the answer has to be 21, cos you could draw 20 marbles and get only two colors, but with 21 you're sure to have all colors.</p>
<p>yea that's what i thought 2400, that's what i put as my answer on the test, and bluesapphire, it's asking for the probablity, and the probability is between 0 and 1.</p>
<p>the question is solvable.. but its jsut kinda tedious to write it out.. since the only possibility of not having a marble of each color left in the 9 marbles is if u drew 20 marbles of any two colors.. so its sth along the lines of 20/30x 19/29 x 18/28.. until 1/11 (it doesnt matter wat u draw for the 21st marble) and then u will needa repeat this 3 times.. for the combination of red blue n green.. and then just 1- this probability, which equals to the probability of getting at least 1 marble of each color...i gues its sth like this.. could be wrong tho</p>
<p>You can reword the problem, and get the same results: "What's the probability of drawing 9 marbles, and having at least one of each color in the 9?"</p>
<p>Prob( 3 colors in 9 marbles) = 1 - P( 2 colors in 9 marbles) - P( 1 color in 9 marbles)
= 1 - {P( No red in 9) + P( No blue in 9) + P(No green in 9)}
- {P( all reds in 9) + P( all blues in 9) + P( all greens in 9)}
= 1 - 3[ (20/30)(19/29)...( 12/22)]
- 3 [ (10/30)(9/29)...(2/22)]</p>
<p>The logic is pretty straightforward, but the numbers are messy.</p>
<p>Im impressed that you actually bothered to work out the answer rather than trying to make it into an easier problem or bumping it up until someone does come and solve it. ;)</p>
<p>Kjoodles:
If you were referring to me - I can't help it, it's a reflex action... :) .</p>
<p>You are a genius :)</p>
<p>so the answer is 21/30 , great tats what i guessed, i thought it had to be greater than 20/30 but less than 1.
y yes the question i wrote was partially wrong i guess, i think xindianx is correct</p>
<p>was THIS THE EXPERIMENTAL SECTION i hope not</p>
<p>Y wouldn't it be...</p>
<p>1- 3(10C9<em>10C0</em>10C0/(30C21) - 3(2(10C1<em>10C8</em>10C0) + 2(10C2<em>10C7</em>10C0) + 2(10C3<em>10C6</em>10C0) + 2 (10C4<em>10C5</em>10C0))/30C21 = .9647833426 (calc won't do the fraction)</p>
<p>it's all combinations (order doesn't matter I hope)...</p>
<p>unless the question was "what are the odds you get 0 of one color in the 21 chosen", which is of course 0.</p>