<p>Suppose I have a bag with 1,000 marbles, out of which 5 are Red and rest are White. I randomly take out one marble at a time out of the bag. How many marbles do I have to take out to have a 95% probability that no Red marble remains in the bag?</p>
<ol>
<li>You would get a 95% probability for every one hundred marbles</li>
</ol>
<p>@Davinci07: Thanks. Could you please briefly elaborate how you got to the answer?</p>
<p>^Don’t think its that easy…</p>
<p>I’d like to see a full solution to this problem as well. Here’s what I have so far: If you take x marbles, there are 1000Cx ways to do that. If you must have all 5 red ones, then there are x-5 other marbles remaining for you to take and 995 to choose from. So that makes 995C(x-5) ways to do that.</p>
<p>So to have a .95 chance of having all the red marbles…</p>
<p>.95 =( 995 C (x-5) ) / (1000 C x )</p>
<p>I can simplify that but I land on a 5th degree polynomial. Mathematica says that the answer rounds to 990 – I am surprised it is so high and not nearly 100% sure that I did this right.</p>
<p>Was this level 5 on an SAT? Kidding…</p>
<p>I was about to type a solution, but realized that I got exactly what pckeller got. So his solution looks correct to me. My intuition does tell me that the answer should be quite large (although if I were to guess I would have guessed lower than 990). I didn’t double check the mathematica results however - I generally lose interest in problems once technology gets involved.</p>
<p>I wonder if this is less a probability problem than a badly written simple algebra problem. </p>
<p>The probability of no red marbles is the inverse of the probability of always picking a white marble.</p>
<p>Right now, the probability of picking a white marble is 995/1000 or 99.5%. You need to reduce the number of marbles to get to a 95% probability of picking a white marble.</p>
<p>So (995-x)/(1000-x)= .95. or 900 marbles.</p>
<p>If this is really an SAT problem, you should be wary of things that seem to difficult to solve.</p>
<p>You are not assuming that the problem is poorly worded. You are assuming that there is a word missing, namely the word “white” in the last sentence. This gives a completely different question than the one the original poster has given.</p>
<p>This is definitely not a SAT problem.</p>
<p>I’m not totally convinced that pckeller’s solution is right. The process of taking out the marbles one by one is subtly different from taking “5”. It’s an issue of truly independent events. But given that M (the number of marbels is large) it’s a likely close approximation.</p>
<p>I would approach the counting/probability by recursion. Let k be the number of marbles that I need to take out (in order) to have a 95% probability of picking m red marbles. I’ll eventually set m to 5.</p>
<p>Define this probability as P(m,1000). This is equal to P(m-1,1000)*[(k-(m-1))/1000]. In other words if you know the probability for picking m-1 marbles (in k steps) then you can figure out the probability for picking m marbles in k steps. That m’th marble must be in one of the first k non-red slots if you’re to pick it in those k steps – and we can easily compute the probability of it being is one of those slots – i.e [(k-(m-1))/1000]. Continue the recursion for m-2, m-3, etc. The probability for m-5 is one (all the marbles are white).</p>
<p>So the final equation is: [k/1000]<em>[(k-1)/1000]</em>[(k-2)/1000]<em>[(k-3)/1000]</em>[(k-4)/1000] = .95. That’s very close to k<strong>5 = (1000</strong>5)<em>(.95). Now (.95)</em>*(.2) is .989 so I then get k=989.</p>
<p>I’m going to need to be more awake to work through your solution!</p>
<p>But I will say that I disagree with your complaint about my solution: the reason for using the combinatoric approach is that the events are NOT independent. Choosing one after another WOULD change the probabilities for each marble. That’s exactly why I looked at the number of overall combinations instead of taking a one-at-a-time approach.</p>
<p>Still, I would not be shocked or offended to find that my answer is wrong…</p>
<p>As i said, I am going to let this go for now…but when I look at it again, I am going to rescale it to a more manageable size:</p>
<p>Say you have 20 marbles, 5 red and 15 white. You pull out a handful. How many must be in the handful to be 90% confident that you have removed all the red ones?</p>
<p>I know it’s the same problem, but with smaller numbers you can check on the intermediate results using a calculator.</p>
<p>Of course the events are not independent. I don’t assume that they are. The dependence is imbedded in the recursion. My intuition is that the approach of looking at all combinations of “5” misses the true probability perhaps (and I need to think about that) by mis-counting successful events. Because the number of actual marbles is high you still end up with a close approximation doing it this way.</p>
<p>That said it’s possible that your approach also works. Your results are very similar to mine.</p>
<p>Anyway this is very far off from being a SAT problem, or for that matter an introductory probability course problem.</p>
<p>Thank you everybody for your input. I greatly appreciate it.</p>
<p>Sorry, it is not from a SAT book or test. I was discussing SAT probability questions with my daughter when this came up during discussion. We spent quite a bit of time on it, trying different algebra methods and reversing the problem. We got answers ranging from 770 to 950. The question seems simple enough, but now I know that answer isn’t :). Thx.</p>