Difference between revisions of "2000 AIME II Problems/Problem 3"
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== Solution 2 == | == Solution 2 == | ||
− | Instead of counting the cases and doing <math>\frac{\text{cases wanted}{\text{total amount of cases}}</math> we can use probability directly. | + | Instead of counting the cases and doing <math>\frac{\text{cases wanted}}{\text{total amount of cases}}</math> we can use probability directly. |
+ | |||
For sake of simplicity, WLOG, assume that a pair of ones were removed from the deck of forty cards. We can split it into two cases: | For sake of simplicity, WLOG, assume that a pair of ones were removed from the deck of forty cards. We can split it into two cases: | ||
+ | |||
Case 1: The pair is ones. | Case 1: The pair is ones. | ||
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The probability that any other number is chosen is <math>\frac{36}{38}.</math> The probability that a number that is equal to this number of chosen (for example, if two was chosen originally then another two being chosen) is <math>\frac{3}{37}.</math> Therefore, the probability that the pair is <math>2-10</math> is <math>\frac{36}{38} \cdot \frac{3}{37}.</math> | The probability that any other number is chosen is <math>\frac{36}{38}.</math> The probability that a number that is equal to this number of chosen (for example, if two was chosen originally then another two being chosen) is <math>\frac{3}{37}.</math> Therefore, the probability that the pair is <math>2-10</math> is <math>\frac{36}{38} \cdot \frac{3}{37}.</math> | ||
− | + | Adding these two probabilities gives <math>\frac{2}{38} \cdot \frac{1}{37} + \frac{36}{38} \cdot \frac{3}{37} = \frac{110}{38 \cdot 37} = \frac{55}{703}</math>, and <math>m+n = \boxed{758}.</math> | |
== Video Solution by OmegaLearn == | == Video Solution by OmegaLearn == |
Latest revision as of 02:46, 31 July 2024
Problem
A deck of forty cards consists of four 's, four
's,..., and four
's. A matching pair (two cards with the same number) is removed from the deck. Given that these cards are not returned to the deck, let
be the probability that two randomly selected cards also form a pair, where
and
are relatively prime positive integers. Find
Solution 1
There are ways we can draw two cards from the reduced deck. The two cards will form a pair if both are one of the nine numbers that were not removed, which can happen in
ways, or if the two cards are the remaining two cards of the number that was removed, which can happen in
way. Thus, the answer is
, and
.
Solution 2
Instead of counting the cases and doing we can use probability directly.
For sake of simplicity, WLOG, assume that a pair of ones were removed from the deck of forty cards. We can split it into two cases:
Case 1: The pair is ones.
The probability that a one is chosen is The probability that a second one is chosen is
because one card was removed. Therefore, the probability that the pair is ones is
Case 2: The pair is
The probability that any other number is chosen is The probability that a number that is equal to this number of chosen (for example, if two was chosen originally then another two being chosen) is
Therefore, the probability that the pair is
is
Adding these two probabilities gives , and
Video Solution by OmegaLearn
https://youtu.be/mIJ8VMuuVvA?t=59
~ pi_is_3.14
See also
2000 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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