document.write( "Question 1196684: 1) There are 4 golden coins and 8 iron coins in a bag. You select one coin from the bag, if it is a golden coin, you keep it; but if it is an iron coin, you put it back in the bag. Find the probability of earning exactly 2 golden coins after:
\n" ); document.write( "a) Two consecutive selections
\n" ); document.write( "b) Three consecutive selections
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Algebra.Com's Answer #829637 by greenestamps(13200)\"\" \"About 
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\n" ); document.write( "**** NOTE ****

\n" ); document.write( "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't say that.

\n" ); document.write( "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.

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\n" ); document.write( "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.

\n" ); document.write( "(a) probability of 2 gold coins after 2 selections

\n" ); document.write( "Both selections must be gold.

\n" ); document.write( "P(gold,gold) = (4/12)(3/11) = 1/11

\n" ); document.write( "ANSWER: 1/11

\n" ); document.write( "(b) probability of 2 gold coins after exactly 3 selections

\n" ); document.write( "The third selection has to be gold; the first two can be either gold then iron or iron then gold.

\n" ); document.write( "P(gold,iron,gold) = (4/12)(8/11)(3/11) = 8/121
\n" ); document.write( "P(iron,gold,gold) = (8/12)(4/12)(3/11) = 2/33

\n" ); document.write( "The probability of 2 gold on exactly 3 selections is

\n" ); document.write( "8/121 + 2/33 = 24/363+22/363 = 46/363

\n" ); document.write( "ANSWER: 46/363

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