Difference between revisions of "2022 AMC 12B Problems/Problem 12"
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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. | 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. |
Revision as of 22:00, 20 November 2022
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 (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
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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.