<p>A bag contains exactly 4 blue marbles, 7 green marbles, and 8 yellow marbles. Fred draws marbles at random from the bag without replacement, one by one. If he does not look at the marbles he draws out, what is the smallest number of marbles he will have to draw out before he knows for sure that on his next draw he will have marbles in every color?</p>
<p>This is a Grid-in btw</p>
<p>Three…I think</p>
<p>I think it is 15.
The only way to be sure to have all colors(assuming you’ve drawn two) is to be sure that you cannot pick those first two colors. There is a (very small) chance of picking 7 green then 8 yellows. Because this situation exists, you must pick 15 to be sure you will have all three colors on the next draw.</p>
<p>I think 15.</p>
<p>You can pull all yellow and green at 15 and on the next draw will get a blue.</p>
<p>was this question on the may 7th sat?</p>
<p>It says how many must you draw so that the next one is your third color.</p>
<p>I don’t remember it from the May SAT</p>
<p>15 is correct. The 16th you are guaranteed to have all three. It is impossible to draw 16 of only two colors, bit it is possible to draw 15 of only two colors.</p>
<ol>
<li>If he draws all the blue and all the green the next draw must be a yellow. This situation is the SMALLEST number possible and since he is looking at these marbles after drawing them (I assume) he can tell when he has 4 blues and 7 greens.</li>
</ol>
<p>I’m confused on why the answer is not 11. Can’t you draw out 4 blues and 7 greens (11 draws) and the next draw will always be yellow? Maybe I misinterpreted the question.</p>
<p>@flacowade</p>
<p>The question states that he does not look at the marbles and needs to be certain without knowing what he has already drawn.</p>
<p>@bioGen</p>
<p>I still believe the answer is 15. There is a possibility of drawing 4 blues and 7 greens, but there is also a possibility of drawing 7 greens and 8 yellows. Because that possibility exists, you cannot be 100% sure that your next draw will be the third color until it is impossible for anything else to happen.
(or at least that is how I interpreted the question)</p>
<p>Where did you get the question? If it was from an SAT prep book, there should be answers.</p>
<p>The question is from Gruber’s Complete SAT Guide 2011 and it says the answer is 15. The answer explanation isn’t clear though, so that’s why I posted the question. </p>
<p>I think I see why 11 draws is wrong now. Would 11 draws be right if the question stated that Fred could see the marbles?</p>
<p>OK.</p>
<p>Think of it this way. In your 11 argument, you stated that you could draw all the blues and greens, leaving only yellows. However, if your eyes are closed, you don’t know what you drew. This leaves the possibility of having drawn other combinations, like 8 yellows, and 3 greens, leaving a chance of drawing a green next instead of the blue you need. Therefore, it is not 11</p>
<p>Because you Must Be Sure, there must be no way possible to have only two colors after your next draw. 15 is the only number that under ALL circumstances will lead to you drawing or already having a third color (in other words, if you have 2 colors after 15 draws, your next draw MUST be the third color, all other combinations of 15 contain all colors)</p>
<p>All numbers after 15 fulfill these guidelines, but 15 is the lowest that makes the chance of only having 2 colors 0%(on your next draw)</p>
<p>Thanks, I understand the problem now.</p>
<p>Oh, I am sorry, misunderstood
It’s fifteen, I’m sure, because you would draw seven green plus eight yellow, so there’d be noway 4 the next marble not to be red</p>