<PRE><B>Question 1196684</B>
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My solution to part (b) is different than that by tutor @ikleyn.  Probably her answer is correct.  I read part (b) to mean that exactly 2 gold coins are obtained in EXACTLY 3 selections -- but the statement of the problem doesn&#39;t say that.&lt;br&gt;
So look at the solutions using both interpretations to see how they are different.  They both show good mathematical methods; but they solve different problems.&lt;br&gt;
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Note that each time a gold coin is selected it is not returned to the bag, so the number of coins left in the bag decreases by 1; but when an iron coin is selected it is returned to the bag, so the number of coins remaining in the bag does not decrease.&lt;br&gt;
(a) probability of 2 gold coins after 2 selections&lt;br&gt;
Both selections must be gold.&lt;br&gt;
P(gold,gold) = (4/12)(3/11) = 1/11&lt;br&gt;
ANSWER: 1/11&lt;br&gt;
(b) probability of 2 gold coins after exactly 3 selections&lt;br&gt;
The third selection has to be gold; the first two can be either gold then iron or iron then gold.&lt;br&gt;
P(gold,iron,gold) = (4/12)(8/11)(3/11) = 8/121
P(iron,gold,gold) = (8/12)(4/12)(3/11) = 2/33&lt;br&gt;
The probability of 2 gold on exactly 3 selections is&lt;br&gt;
8/121 + 2/33 = 24/363+22/363 = 46/363&lt;br&gt;
ANSWER: 46/363&lt;br&gt;
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