Mathematics: You have two bags,every with 100 marbles and containing only atmosphere max sale red alternatively white marbles. The 1st bag has 40 red / 60 pearly The 2nd sack distribution namely nameless You lack a red cruel and have only an event to elect nike atmosphere max 2013 Which bag do you choose from and why?
Got this brain teaser in an interview. I deem my answer seems also simple and deficiency to know whether I missing something.
I figured, assuming the 2nd sack has a uniform distribution (the chances of getting 10 Red / 90 White are the nike atmosphere max same as getting 66 Red / 34 White),that the expected # of Red marbles within the 2nd sack was 50. This namely higher than 40,air max uk, so I have a better accident at Red whether work with the 2nd bag.
Is namely logic sound?
There is no algebraic answer to this question. If you mulberry bags assume Y = 0.five,then it is greater than X. If you assume Y = 0.4 it mulberry tote namely equal to X. If you assume Y = 0.three it namely less than X.
But there is no basis as any of these assumptions over any other.as what can occur whether you make also many assumptions of normality amid the financial industry.
But yes,is assumption namely equivalent to saying Y=0.5 as whether the redcount namely distributed uniformly between 0 and 100 inclusive, and P(red|redcount=x) = redcount/100,after P(ruddy)=0.five.
However, I don think this is necessarily the only way to solve the problem. What if necessary an alternatively two of the marbles in sack #1 were red -- would you still forecast FIFTY of the ones within sack #2 to be red What whether the bags had a billion marbles each and sack #1 had only an or two namely were red out of the plenary sack -- would you still expect sack #2 had FIVE HUNDRED MILLION ruddy marbles?
An assumption just as legitimate namely is bags one and 2 were every sampled (uniformly along random) from some unseen sack zero"is represents the overall distribution of red and pearly marbles.
What namely the proportion of ruddy marbles within sack zero? We don know. We COULD assume is this parameter was uniformly distributed a priori -- an assumption just as unsupported as the others. We mulberry bags ebay have observed an example of 100 marbles from bag zero (those within sack #1),nike free, of which 40 were red and 60 pearly By applying Bayes theorem, we can calculate a conditional probability distribution (a posterior) aboard the proportion of red marbles within sack zero, given our observation of bag #1.
The answer is a Beta[41,61] distribution,timberland australia,
What,nike free running,after namely the probability namely a randomly-selected bloody from sack #2 want be red The answer is the expected merit of a draw from that distribution,nike free uk,or about 40.2%.
So your chances of getting a red marble are SLIGHTLY higher in sack #2, since we anticipate you have a 40.2% probability of getting a ruddy bloody there (with considerable mulberry factory uncertainty), versus a known 40% probability of getting a ruddy marble amid bag #1.
I am never arguing namely my assumption of prior uniformity aboard the overall proportion of red marbles namely any more defensible than your assumption of uniformity aboard the distribution of red marbles amid sack #2 -- prefer mulberry tote is it namely feasible to obtain alter answers to the problem based on which reasonable-sounding assumptions (even assumptions of uniformity!) you toss in there. These assumptions are necessary to be capable to produce an answer by always.
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