Difference between revisions of "2011 AMC 10A Problems/Problem 21"
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Note that in this solution, all four slots on the scale are deemed to be distinguishable. In essence, this "overcounts" all numbers by a factor of <math>8</math>, since you can switch the coins on the left side, you can switch the coins on the right side, or you can switch sides of the scale. But since all numbers are increased 8-fold, it will cancel out when calculating the probability. | Note that in this solution, all four slots on the scale are deemed to be distinguishable. In essence, this "overcounts" all numbers by a factor of <math>8</math>, since you can switch the coins on the left side, you can switch the coins on the right side, or you can switch sides of the scale. But since all numbers are increased 8-fold, it will cancel out when calculating the probability. | ||
− | == Solution | + | == Solution 2== |
Place the two coins from the first pair in a line followed by the two coins from the second pair followed by the six left-over coins. We can do that in <math>\binom{10}{2} = 45</math> different ways. | Place the two coins from the first pair in a line followed by the two coins from the second pair followed by the six left-over coins. We can do that in <math>\binom{10}{2} = 45</math> different ways. | ||
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Of these, <math>\binom{6}{2}=15</math> cases has both counterfeit coins in the left-over. The probability of having chosen four genuine coins therefore is <math>\boxed{\frac{15}{19} \ \mathbf{(D)}}</math>. | Of these, <math>\binom{6}{2}=15</math> cases has both counterfeit coins in the left-over. The probability of having chosen four genuine coins therefore is <math>\boxed{\frac{15}{19} \ \mathbf{(D)}}</math>. | ||
− | == Solution | + | == Solution 3== |
WLOG, allow for all the coins to be distinguishable. | WLOG, allow for all the coins to be distinguishable. | ||
We split this up into cases. Case <math>1</math> being the weight of <math>2</math> genuine coins together and Case <math>2</math> being the weight of <math>1</math> genuine coin and <math>1</math> counterfeit coin. | We split this up into cases. Case <math>1</math> being the weight of <math>2</math> genuine coins together and Case <math>2</math> being the weight of <math>1</math> genuine coin and <math>1</math> counterfeit coin. |
Latest revision as of 15:04, 9 June 2024
Contents
Problem
Two counterfeit coins of equal weight are mixed with identical genuine coins. The weight of each of the counterfeit coins is different from the weight of each of the genuine coins. A pair of coins is selected at random without replacement from the
coins. A second pair is selected at random without replacement from the remaining
coins. The combined weight of the first pair is equal to the combined weight of the second pair. What is the probability that all
selected coins are genuine?
Solution 1
If we pick indistinguishable real coins from the set of
real coins, there are
ways to pick the coins. If we then place the coins in four distinguishable slots on the scale, there are
ways to arrange them, giving
ways to choose and place
real coins. This gives
desirable combinations.
If we pick real coins and
fake coins, there are
ways to choose the coins. There are
choices for the first slot on the left side of the scale. Whichever type of coin is placed in that first slot, there are
choices for the second slot on the left side of the scale, since it must be of the opposite type of coin. There are
choices for the first slot on the right side of the scale and only
choice for the last slot on the right side.
Thus, there are ways to arrange the coins, and
sets of possible coins, for a total of
combinations that are legal, yet undesirable.
The overall probability is thus .
Note that in this solution, all four slots on the scale are deemed to be distinguishable. In essence, this "overcounts" all numbers by a factor of , since you can switch the coins on the left side, you can switch the coins on the right side, or you can switch sides of the scale. But since all numbers are increased 8-fold, it will cancel out when calculating the probability.
Solution 2
Place the two coins from the first pair in a line followed by the two coins from the second pair followed by the six left-over coins. We can do that in different ways.
We need to exclude those cases where the weight shows a difference. There are two cases where a pair has both counterfeit coins and there are cases where one counterfeit coin is chosen and one is in the left-over. That leaves
cases.
Of these, cases has both counterfeit coins in the left-over. The probability of having chosen four genuine coins therefore is
.
Solution 3
WLOG, allow for all the coins to be distinguishable.
We split this up into cases. Case being the weight of
genuine coins together and Case
being the weight of
genuine coin and
counterfeit coin.
Case : All Genuine coins chosen.
This happens in
ways
Case : Genuine coin and Counterfeit coin both chosen.
This happens in
ways.
Hence, the answer is .
Video Solution by the Beauty of Math
https://www.youtube.com/watch?v=W5NlV2B_83U&feature=push-u-sub&attr_tag=XAAsHxmGZAHEdZgT%3A6
~Icematrix
See Also
2011 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.