2022 AMC 12B Problems/Problem 12
Contents
Problem
Kayla rolls four fair -sided dice. What is the probability that at least one of the numbers Kayla rolls is greater than and at least two of the numbers she rolls are greater than ?
Solution 1 (Complementary Counting)
We will subtract from one the probability that the first condition is violated and the probability that only the second condition is violated, being careful not to double-count the probability that both conditions are violated.
For the first condition to be violated, all four dice must read or less, which happens with probability .
For the first condition to be met but the second condition to be violated, at least one of the dice must read greater than , but less than two of the dice can read greater than . Therefore, one of the four die must read or , while the remaining three dice must read or less, which happens with probability .
Therefore, the overall probability of meeting both conditions is .
Solution 2 (Direct and Complementary Counting)
There are either , , , , or dice that have values of more than . The probability of getting is , the probability of getting is , and the probability of getting or greater is .
It is obvious that the probability of getting at least two numbers greater than is if we have numbers greater than .
Let us calculate the probability of getting at least two numbers greater than if one die is greater than by using complementary counting. We already have one die that is greater than , and the probability that a die that is less than is greater than is . Thus, our probability is .
Finally, our total probability is
~mathboy100
Solution 3 (Complementary Counting and PIE)
We use Solution 1's idea of complementary counting. However, we will find the probability that the first condition is violated and the probability that the second condition is violated, then subtract the overlap (first and second condition both violated). We can do this because of PIE.
As in Solution 1, the probability that the is violated is
The probability that the is violated (regardless of the first condition) can be broken into two cases.
Then, there are four choices for which die is the one greater than 2, chance for whether it reads 3, 4, 5, or 6, and for the other three die that must read 1 or 2. The probability is then
Then, all four dice are either 1 or 2, so the probability of this is
Now, we look at the . In other words, we want to find the probability none of the rolls are greater than and less than two of the rolls are greater than . We split into the same two cases again.
There are still 4 choices for this die, and it must be 3 or 4, with probability . The other must read 1 or 2, with probability so this case yields
The first statement implies the second statement, so this case yields as from our previous Case 2.
Now we've got everything! Our answer is
~sirswagger21
Video Solution by mop 2024
https://youtu.be/ezGvZgBLe8k&t=171s
~r00tsOfUnity
Video Solution (Just 2 min!)
https://www.youtube.com/watch?v=m-rF2xwDLGM
- ~Education, the Study of Everything
Video Solution(1-16)
~~Hayabusa1
See Also
2022 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 11 |
Followed by Problem 13 |
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 12 Problems and Solutions |
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