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Difference between revisions of "1993 AHSME Problems/Problem 24"

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\text{(E) } 66</math>
 
\text{(E) } 66</math>
  
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== Solution ==
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First let’s try to find the number of possible unique combinations. I’ll denote shiny coins as 1 and dull coins as 0.
  
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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.
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Now, each configuration can be represented by a string of 1s and 0s e.g. 0110100. Notice thata combination can be uniquely determined solely by the placement of their 0s OR 1s e.g. 1 - - 1 1 - - where the dashes can be replaced by 0s. This makes the number of unique combinations 7 choose 3 (if you’re counting w.r.t. shiny coins) OR 7 choose 4 (w.r.t dull coins). Both are equal to 35.
  
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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>.
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Next, observe that, for the event that the third shiny coin is not within your first 4 picks, it has to be within the last three numbers. You can think of this as placing the seven coins in a vertical stack in the box and shuffling that stack randomly. Then, to pick, you extract the first coin on the top and keep repeating. It has the same effect.
  
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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>1/35</math>.
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The sequence can have 1 shiny coin in the last 3 digits (Case 1), 2 shiny coins in the last 3 digits (Case 2) or 1 shiny coin in the last three digits (Case 3).
  
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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
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Case 1:
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Let’s start with the first case of one shiny coin in its last 3 digits.
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0110100
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The first four numbers has 4 spaces and 2 shiny coins therefore the number of combinations is 4 choose 2 = 6. The last 3 digits has 3 combinations for the same reason.
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So, probability for Case 1 to occur is:
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<math>\dfrac{6*3}{35}=\dfrac{18}{35}</math>
  
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4/whatever denominator
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Case 2:
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Using the fact that the combinations are uniquely determined by an order of 0s or 1s and you can just fill the rest, in you can ascertain:
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<math>\underbracket{0100}_{\text{4 combinations}}\underbracket{110}_{\text{3 combinations}}</math>
  
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pennies that are dull.
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So, P(Case 2)=<math>12/35</math>
  
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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>.
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Case 3:
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Trivially, it is 1.
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P(Case 3)=<math>1/35</math>
  
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NOTE: This solution is still flawed because it has answer 68. The writer is trying to figure out the problem, but if you see it, fix it. Thanks!
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Adding all these probabilites together gives you the probability that the third shiny coin will not appear in your first 4 draws:
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<math>\dfrac{18}{35}+\dfrac{12}{35}+\dfrac{1}{35}=\dfrac{31}{35}</math>
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<math>\dfrac{a}{b}=\dfrac{31}{35}</math>
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Since the fraction is irreducible:
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<math>a=31</math>
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,<math>b=35</math>
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<cmath>a+b=66</cmath>
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The answer is E.
  
 
== See also ==
 
== See also ==

Revision as of 11:58, 9 December 2017

Problem

A box contains $3$ shiny pennies and $4$ dull pennies. One by one, pennies are drawn at random from the box and not replaced. If the probability is $a/b$ that it will take more than four draws until the third shiny penny appears and $a/b$ is in lowest terms, then $a+b=$

$\text{(A) } 11\quad \text{(B) } 20\quad \text{(C) } 35\quad \text{(D) } 58\quad \text{(E) } 66$

First let’s try to find the number of possible unique combinations. I’ll denote shiny coins as 1 and dull coins as 0.

Now, each configuration can be represented by a string of 1s and 0s e.g. 0110100. Notice thata combination can be uniquely determined solely by the placement of their 0s OR 1s e.g. 1 - - 1 1 - - where the dashes can be replaced by 0s. This makes the number of unique combinations 7 choose 3 (if you’re counting w.r.t. shiny coins) OR 7 choose 4 (w.r.t dull coins). Both are equal to 35.

Next, observe that, for the event that the third shiny coin is not within your first 4 picks, it has to be within the last three numbers. You can think of this as placing the seven coins in a vertical stack in the box and shuffling that stack randomly. Then, to pick, you extract the first coin on the top and keep repeating. It has the same effect.

The sequence can have 1 shiny coin in the last 3 digits (Case 1), 2 shiny coins in the last 3 digits (Case 2) or 1 shiny coin in the last three digits (Case 3).

Case 1: Let’s start with the first case of one shiny coin in its last 3 digits. 0110100 The first four numbers has 4 spaces and 2 shiny coins therefore the number of combinations is 4 choose 2 = 6. The last 3 digits has 3 combinations for the same reason. So, probability for Case 1 to occur is: $\dfrac{6*3}{35}=\dfrac{18}{35}$

Case 2: Using the fact that the combinations are uniquely determined by an order of 0s or 1s and you can just fill the rest, in you can ascertain: $\underbracket{0100}_{\text{4 combinations}}\underbracket{110}_{\text{3 combinations}}$

So, P(Case 2)=$12/35$

Case 3: Trivially, it is 1. P(Case 3)=$1/35$

Adding all these probabilites together gives you the probability that the third shiny coin will not appear in your first 4 draws: $\dfrac{18}{35}+\dfrac{12}{35}+\dfrac{1}{35}=\dfrac{31}{35}$

$\dfrac{a}{b}=\dfrac{31}{35}$

Since the fraction is irreducible:

$a=31$ ,$b=35$

\[a+b=66\]

The answer is E.

See also

1993 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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All AHSME Problems and Solutions

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