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A gold urn contains 3 red and 4 white balls and a silver urn contains 5 red and white balls. A die is rolled and, if a six shows, one ball is selected at random from the gold urn. Otherwise a ball is selected at random from the silver urn. Find the probability of selecting a red ball. The ball selected is not replaced and a second ball is selected at random from the same urn. Find the probability that both balls are white.

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  • $\begingroup$ How many white balls in the silver urn? $\endgroup$ – scott Nov 3 '16 at 15:39
  • $\begingroup$ @scott Oh, it's 2. $\endgroup$ – Azrael Nov 3 '16 at 15:40
  • $\begingroup$ @Bram28 I'm quite bamboozled, though I've tried letting A=event where six is the outcome of the roll, B=event where a red ball is gotten in the gold urn, C=event where a red ball is gotten in the silver urn but now I'm stuck with putting everything together. $\endgroup$ – Azrael Nov 3 '16 at 15:43
  • $\begingroup$ OK, that's a good start! OK, the chance of getting a red ball equals the chance of getting a red ball from the silver urn + the chance of getting a red ball from the gold urn, i.e. P(B)+P(C). (this is all talking about the first ball you pick, by the way). Do you see that? $\endgroup$ – Bram28 Nov 3 '16 at 15:49
  • $\begingroup$ if you roll a 6 - what is the probability that you then get a red ball? Then what is the prob of getting a 6 AND a red ball - that is half of it - you can also separately get not6 then a red ball $\endgroup$ – Cato Nov 3 '16 at 15:51

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