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Problem said:

Urns I, II and III contain three pennies and four dimes, two pennies and five dimes and three pennies and one dime, respectively. One coin is selected at random from each urn.

(a) What is the probability that all three selected coins have the same denomination?

P(a:same denomination)=(3/7)(2/7)(3/7)+(4/7)(5/7)(1/4)= 19/98

(b) If two of the three coins are dimes, what is the probability that the coin selected from urn I was a dime?

My concern: I have no clue where I am suppose to start to solve this exercise, any help will be apprecited. Thanks!

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    $\begingroup$ Small typo, first term should be $(3/7)(2/7)(3/4)$. For the second problem, you can do it with a "tree" calculation, or by a formal conditional probability calculation. Let $T$ be the event two out of three were dimes, and let $D$ be the event the coin from Urn I was a dime. We want $\Pr(D\mid T)$, which is $\Pr(D\cap T)/\Pr(T)$. Which way one answers the problem for you depends on what approach is the more familiar. $\endgroup$ – André Nicolas Dec 13 '15 at 22:20
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Of the possible $7\times7\times4=196$ events, you get exactly 2 dimes in $4\times5\times3+3\times5\times1+4\times2\times1=83$ cases. Of these, a dime is selected from urn I in $4\times5\times3+4\times2\times1=68$ cases. The probability that the coin selected from urn I was a dime is then $68/83$.

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