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I have the following question:

John has ten single dollar bills of which 3 are counterfeit. If he selects 4 of them at random, what is the probability of getting 2 good bills and 2 counterfeit bills?

At first I did 7/10 * 6/9 * 3/8 * 2/7 which didn't work and I'm not sure why. I equated this question to the marble questions - essentially 7 red marbles and 3 blue marbles. This process works for the marble questions, but not here and I'm not sure why.

mike
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1 Answers1

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Your result $7/10\cdot6/9\cdot3/8\cdot2/7=1/20$ is the probability to get the four bills in a specific order. But there are $4!/(2!2!)=6$ ways to pick the four bills, so the probability is $6/20=0.3$.

EDIT

If you want to get your result right in terms of probability, in my opinion the best way to avoid errors is that of using a probability tree:

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To address your other questions about the right way to count the number of combinations, the rule p=(favorable cases)/(total cases) works only if all the cases one considers are actually a priori equally likely to happen. For the problem at hand John selects at random one out of the possible $10\cdot9\cdot8\cdot7=5040$ ways to extract four bills (just think of the bills as numbered), so these are our elementary events as they are all alike. Of these, only $(7\cdot6)\cdot(3\cdot2)\cdot6=1512$ will contain exactly two counterfeit bills, so the probability we are after is $1512/5040=0.3$.

Notice that I counted all the possible outcomes considering them different even if they contain the same bills in a different order. As favorable cases do not differ for the order of bills, we may as well consider two outcomes to be the same if they contain the same bills in a different order. In that case, John selects at random one out of the possible $\binom{10}{4}=210$ ways to extract four bills and of these only $\binom{7}{2}\binom{3}{2}=63$ will contain exactly two counterfeit bills. Needless to say, the probability is still $63/210=0.3$.

Intelligenti pauca
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  • I really appreciate your response. The text book found the total combinations of picking 4 and then multiplied each individual combination and divided that product by the total combination. I do see your point about order, but the approach I Just mentioned now also works in the marble scenarios. I used that approach to determine the probability of picking 2 red marbles where there are 3 red marbles and 5 blue marbles. The order approach also worked so 3/8 * 2/7, both equal 3/28. So 3C2/8C2 also equaled 3/28. Can you tell me why? – mike Nov 28 '15 at 23:32
  • Also, if 7 of the bills are legitimate and 3 are counterfeit, why does it make sense to do 7C2 and 3C2 to determine the numerator when the bills are the same respectively. So for example 7C2, why are there 21 combinations when each of the picks is a legitimate $1 bill there is no difference in any of the combinations. Wouldn't that make it 1? So all 21 combinations look like this - LL, doesn't this mean just 1 combination? – mike Nov 28 '15 at 23:47
  • I'm not sure I understand your questions, see anyway my edited answer. – Intelligenti pauca Nov 29 '15 at 10:40
  • Again, thank you. I understand the substance of the explanations, both of the ones you were kind enough to explain. Here's my hurdle: when using combinations we are looking for unique groupings; if i have 7 of the same things (legitimate bills) and 3 of the same things (counterfeit) how could i have 21 unique combinations of groups of 2 legitimate bills. Similar confusion for the counterfeit bills. Unless each bill is unique i'm not seeing how i can have unique combinations of indistinguishable bills. – mike Nov 29 '15 at 19:26
  • If i had 7 boys and had to pick two, i would use the same combination formula you used for the legitimate bills. But unlike the bills, the boys are unique so the use of the formula make sense here. That's where i'm stuck. And if your response is that they are unique then is that the standard assumption unless told otherwise? Because the question didn't say they are unique. – mike Nov 29 '15 at 19:29
  • If i'm not mistaken, when you corrected my initial approach you multiplied my result by 6 and you obtained that 6 by determining the permutations of 2 groups of indistinguishable bills, legitimate and counterfeit. If the bills were distinguishable your correction would have been 4! I hope if made my confusion clear. – mike Nov 29 '15 at 19:39
  • I'd suggest you to look carefully at the probability tree: from there you may understand where that factor of 6 comes from and why you must consider bills as distinguishable. By the way each bill has a unique serial number, but even if it hadn't you could write a number on all of them. Every set of macroscopic objects is formed by distinguishable items. – Intelligenti pauca Nov 29 '15 at 19:45