1993 AHSME Problems/Problem 24


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 that a 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 3 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.

Example: 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: $\dfrac{18}{35}+\dfrac{12}{35}+\dfrac{1}{35}=\dfrac{31}{35}$


Since the fraction is irreducible:

$a=31$ ,$b=35$


The answer is E.

Solution 2

Using complementary probability, we can reduce the problem into two cases- the third shiny penny is drawn on the third draw, or on the fourth draw. For the first case, there is only one way to have the shiny pennies as the first three coins drawn, out of a possible 35 drawings (7 choose 3). For the second case, the third shiny penny has to be the fourth penny drawn, which leaves three possible orderings for the first three coins drawn (NS S S, S NS S, S S NS), out of 35 (7 choose 4). Adding these two probabilities together gives $\dfrac{4}{35}$, and subtracting this from one yields $\dfrac{31}{35}$, which makes a 31 and b 35, which sum to $\fbox{E}$.

See also

1993 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
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