Difference between revisions of "1993 AHSME Problems/Problem 24"
Expilncalc (talk | contribs) (→Solution: There is some flaw in my solution that causes the answer to be 68 instead of 66. I am trying to figure it out.) |
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Let's look at the problem and its numbers. We want to find the probability that it takes strictly more than 4 tries to get all 3 shiny pennies without replacement. Should we calculate the probably or 1 minus it? Of course the latter, because either we get all 3 shiny pennies in 3 draws (easy to calculate) or 4 (just one case). The probability and its not happening are mutually exclusive, so we can use the 1-P and P approach. | Let's look at the problem and its numbers. We want to find the probability that it takes strictly more than 4 tries to get all 3 shiny pennies without replacement. Should we calculate the probably or 1 minus it? Of course the latter, because either we get all 3 shiny pennies in 3 draws (easy to calculate) or 4 (just one case). The probability and its not happening are mutually exclusive, so we can use the 1-P and P approach. | ||
| − | We know that if we draw in 3, the denominator is <math>7*6*5 = 210 | + | We know that if we draw in 3, the denominator is <math>7*6*5 = 210</math>. If 4, then <math>210 * 4 = 840</math>. |
| − | Let's start with drawing all 3. We only have < | + | Let's start with drawing all 3. We only have <math>3*2*1</math>, drawing one shiny penny each time, and the probability is <math>\frac{1/35}</math>. |
| − | Now, on to four pennies. We know that no matter what, we will still incorporate the < | + | Now, on to four pennies. We know that no matter what, we will still incorporate the <math>3*2*1</math>, but this time, we multiply by a 4 because we can afford to take one of the |
4/whatever denominator | 4/whatever denominator | ||
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pennies that are dull. | pennies that are dull. | ||
| − | Since this probability is also 1/35, we have our probability that it takes 4 or less: 2/35. One minus this result is 33/35, or answer choice < | + | Since this probability is also 1/35, we have our probability that it takes 4 or less: 2/35. One minus this result is 33/35, or answer choice <math>\fbox{E}</math>. |
== See also == | == See also == | ||
Revision as of 20:50, 26 June 2017
Problem
A box contains 
shiny pennies and 
dull pennies. One by one, pennies are drawn at random from the box and not replaced. If the probability is 
that it will take more than four draws until the third shiny penny appears and is in lowest terms, then 
Solution
Let's look at the problem and its numbers. We want to find the probability that it takes strictly more than 4 tries to get all 3 shiny pennies without replacement. Should we calculate the probably or 1 minus it? Of course the latter, because either we get all 3 shiny pennies in 3 draws (easy to calculate) or 4 (just one case). The probability and its not happening are mutually exclusive, so we can use the 1-P and P approach.
We know that if we draw in 3, the denominator is 
. If 4, then 
.
Let's start with drawing all 3. We only have 
, drawing one shiny penny each time, and the probability is $\frac{1/35}$ (Error compiling LaTeX. Unknown error_msg).
Now, on to four pennies. We know that no matter what, we will still incorporate the , but this time, we multiply by a 4 because we can afford to take one of the
4/whatever denominator
pennies that are dull.
Since this probability is also 1/35, we have our probability that it takes 4 or less: 2/35. One minus this result is 33/35, or answer choice 
.
See also
| 1993 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 23 |
Followed by Problem 25 | |
| 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 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
