161–169 of 169 posts
Never liked the way these problems are worded. `You take a random ball out of the urn—it’s red—and discard it.` How normal people read it: Given this specific instance where you just discarded a red ball from this urn, what's the probability of the next ball? How it expects you to read it: Given infinitely many random samples from the urn. For cases where you get red, remove it, then take a second sample. What's the…
Earlier quoted context omitted.
How would normal distribution change the result? I am imaging a random number generated from truncated normal distribution with mean 50, clipped 0 and 100: the answer will still lean right. How would binomial work? You have 101 urns from which to choose.
Normal: it is very close to 50/50. It does still lean, but only just . It's close enough that in practice any guess is fine. Binomial: it is 50/50. The reason these change the situation dramatically is that under normal or binomial, you are waaaay more likely to be in a situation that is close to 50/50 to begin with because they are way more likely to occur. The quirk of this question is that a 0 green 100 red urn is…
How do you narrow 101 choices into two?
Seems to be fully depending on your interpretation of "more likely". If you take that as "more likely than it was before", then the answer is obviously "more likely to be green". If you take it as "more likely to be green or red this pull", then that answer would depend fully on the unknown distribution and the closest least wrong answer would be "don't know". Backtranslating from the interpretation/answer pairs, if…
But there is no unknown distribution. There is a latent parameter (the n that's uniformly sampled) whose distribution is known; so the distribution of colors for the next draw is well defined. "Depend fully on the unknown distribution" is a statement about the distribution conditioned on that latent parameter; even in that case, the answer isn't "don't know" but "whatever the ratio of remaining balls given that param…
The second interpretation says that given any selected n, the probability of drawing green of the next ball from that urn can only increase or stay the same (in the all red balls urn case) compared to drawing from that urn before.
It is the absolute vs relative interpretation of what the 'more' refers to.
I'll grant that the inclusion of 'uniform' for n in the question suggests the absolute interpretation as the relative interpretation would be independent of the uniformity of n.
The question could have eliminated ambiguity by just asking for the probability of the next ball color, but given the context of a Google recruitment test they might be more interested in how candidates reason under uncertain conditions rather than their dry probability skills.
Earlier quoted context omitted.
Is that true? You'd still have evidence that the distribution of balls tilts one way, wouldn't you?
Flip three coins. First one lands heads. Does that mean the rest are more likely to be heads? No.
Earlier quoted context omitted.
Flip three coins. First one lands heads. Does that mean the rest are more likely to be heads? No.
If I flip three coins and use the results to decide which color balls to put in a bag, I'll have one of these distributions: RRR,RRG,RGG,GGG. If I draw a red ball, I am more likely to have drawn from RRR or RRG than RGG (or GGG), so I have learned something about the distribution more than just how it was generated.
Let’s ignore RRG because it becomes a 50-50 anyway.
RRR has a 100% chance of yielding Red. But is 3X less likely than RGG.
RGG has a 33% chance of yielding Red but was 3X more likely.
Edit: so still 50/50 Given a draw of red, you have exactly the same probability of having been in RRR and RGG. So you were equally likely to have 100% chance or a 0% chance.
There are 100 balls. First the host chooses N uniformly from [0, 100] (unknown to us) and makes N of them red and 100-N of them green. But it's annoying to paint 100 balls, right? So he does it this way instead: the host takes 100 white balls and puts them all in a row. Then he takes a black ball and puts it between balls N and N+1, and he declares to himself that all the white balls to the left of the black one will be "red," and all to the right will be "green." Then he has us randomly select two white balls without replacement, and depending on which side of the black ball they were, he'll paint just those two balls red or green accordingly before showing us! Since we never see the other 98 unpainted white balls, we'll never know the difference between this setup and the original.
(I think it's intuitively obvious that this setup is mathematically equivalent to the original, i.e. generates the same distribution, so I won't try to justify that rigorously.)
Actually though, the host does it slightly differently, again without changing the underlying distribution: you see, he actually starts with 101 white balls, all in a row, and he begins by randomly selecting one of them and paints it black, and then proceeds as above.
But this gives the whole problem away: there are 101 white balls in a row, and three are drawn uniformly at random without replacement. The first one is declared "black," and the other two are red or green depending on their positions relative to the black one. All three-ball sequences are equally likely, so by symmetry, the probability that the black ball is between the second and third is exactly 1/3. The other two balls, the ones we draw, are different colors if and only if the black ball is between them, so the probability that we get two balls of the same color is exactly 2/3.
That's the answer: whatever is the color of the first ball we draw, the second one has that color with probability 2/3. Thus, the second ball is red with probability 2/3.
I guess I'm late to the party, but here's my take. There are 100 balls. First the host chooses N uniformly from [0, 100] (unknown to us) and makes N of them red and 100-N of them green. But it's annoying to paint 100 balls, right? So he does it this way instead: the host takes 100 white balls and puts them all in a row. Then he takes a black ball and puts it between balls N and N+1, and he declares to himself that al…
Off the top of my head and using my fingers the probability is: 4901:4900 in favor of red. (100*98/2 = 100*49. Plus one in favor of red) If the number of distribution were odd, the odds would be even. Am I right?
Where did you get 98 from? Let's try it as a set of 3 balls. You have 3 balls in an urn, random number of red v green, the first one you pull is red, what's the likelihood that the second one you pull is also red?
So I added one to one side instead of 50 and 49 to the other.
So 4950:4949.
... is that right? Im pretty sure red is favored, and Im pretty sure if the number of dists were odd the odds would be even. But are the odds right?