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Urn I contains 25 white and 15 black balls. Urn II contains 15 white and 25 black balls. An urn is selected at random and five balls are drawn randomly from this urn without replacement. If exactly five of these balls are white, what is the probability that the balls came from Urn I?

So my question derives from the "without replacement" part of the question. I am assuming all five balls are drawn from a urn at the same time rather than one by one? But then it confuses me why they would throw in "without replacement" if you would draw them all at the same time. Anyone have insight on whether it is one-by-one or not?

Given that they are drawn at the same time I get: P(U1|W) = P(U1 AND W)/P(W) which turns into P(W|U1)(P(U1)/ (P(W|U1)P(U1) + P(W|U2)P(U2)) Where W: the five balls drawn are white Ui: Urn i is choosen

is this correct?

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    $\begingroup$ All at once and one at a time give the same probabilities. "With replacement" means we draw a ball, record its colour, and put it back in the urn, then draw again, put back, and so on. So with replacement one might draw the same ball more than once. If the drawing is done without replacement, that cannot happen. $\endgroup$ – André Nicolas Oct 27 '15 at 3:30
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"Without replacement" may mean "all at the same time" or it may mean "draw one by one without replacing." Either way gives the same result.

"With replacement" would be "draw one, add a tally, return to the urn, then draw again"


Your formula is correct.   All that remains is how you measure the probabilities for drawing "without replacement".

So... How will you measure the probabilities for drawing "without replacement"?

$\mathsf P(W\mid U_1) = $ the probability of selecting $5$ of the $25$ white out of all ways to select any $5$ of all $40$ (without replacement).

$\mathsf P(W\mid U_2) = $ the probability of selecting $5$ of the $15$ white out of all ways to select any $5$ of all $40$ (without replacement).

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  • $\begingroup$ Basically, no. @FutureUIUXDeveloper $P(W\mid U_1) = $ the probability of selecting $5$ of the $25$ white out of all ways to select any $5$ of all $40$ . $P(W\mid U_2) = $ the probability of selecting $5$ of the $15$ white out of all ways to select any $5$ of all $40$ . Clearly the more white balls in an urn the more likely the five selected will all be white . $\endgroup$ – Graham Kemp Oct 27 '15 at 3:56
  • $\begingroup$ So for P(W|U1) will be (5/25+4/25+3/25+2/25+1/25)/40 given the draw is done without replacement. And in the case there is replacement then it would be chance of drawing 5 whites from 25 total whites in a pool of 40 balls((5/25)/40) $\endgroup$ – FutureUIUXDeveloper Oct 27 '15 at 4:16
  • $\begingroup$ What. No, @FutureUIUXDeveloper What is the probability of drawing 1 of the 25 white balls from 40, then having done that, drawing 1 of the 24 white balls from the 39 remaining, and so forth...? $\endgroup$ – Graham Kemp Oct 27 '15 at 4:20
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    $\begingroup$ I think it is 25!/(5!*20!)/ (40!/(5!*35!) $\endgroup$ – FutureUIUXDeveloper Oct 27 '15 at 4:40

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