Difference between revisions of "2021 AMC 10B Problems/Problem 18"
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~Jske25 | ~Jske25 | ||
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==Solution 7== | ==Solution 7== |
Revision as of 22:18, 26 April 2021
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3 (Quicksolve)
- 5 Solution 4
- 6 Solution 5
- 7 Solution 6 (same as solution 1 but with a little more of explanation)
- 8 Solution 6 (Infinite Geometric Sequence Method)
- 9 Solution 7
- 10 Video Solution by hurdler (complementary probability)
- 11 Video Solution by TheBeautyofMath
- 12 See Also
Problem
A fair -sided die is repeatedly rolled until an odd number appears. What is the probability that every even number appears at least once before the first occurrence of an odd number?
Solution 1
There is a chance that the first number we choose is even.
There is a chance that the next number that is distinct from the first is even.
There is a chance that the next number distinct from the first two is even.
, so the answer is
~Tucker
Solution 2
Every set of three numbers chosen from has an equal chance of being the first 3 distinct numbers rolled.
Therefore, the probability that the first 3 distinct numbers are is
~kingofpineapplz
Solution 3 (Quicksolve)
Note that the problem is basically asking us to find the probability that in some permutation of that we get the three even numbers in the first three spots.
There are ways to order the
numbers and
ways to order the evens in the first three spots and the odds in the next three spots.
Therefore the probability is .
--abhinavg0627
Solution 4
Let denote the probability that the first odd number appears on roll
and all our conditions are met. We now proceed with complementary counting.
For , it's impossible to have all
evens appear before an odd. Note that for
since there's a
chance that the first odd appears on roll
(disregarding the other conditions) and the other term is subtracting the probability that less than
of the evens show up before the first odd roll. Simplifying, we arrive at
Summing for all , we get our answer of
~ike.chen
Solution 5
Let be that probability that the condition in the problem is satisfied given that we need
more distinct even numbers. Then,
since there is a
probability that we will roll an even number we already have rolled and will be in the same position again. Solving, we find that
.
We can apply the same concept for and
. We find that
and so
. Also,
so
. Since the problem is asking for
, our answer is
. -BorealBear
Solution 6 (same as solution 1 but with a little more of explanation)
The probability of choosing an even number on the first turn is , now since you already chose that number, it is irrelevant to the problem now, so, if you chose the number again, it doesn't really matter to our problem anymore. Now, with our remaining
numbers, the probability of choosing another even number is
, and again, after you have chosen that number, it is out of our problem. Now, you just have
numbers left and the probability of choosing the last even number is
, so the answer is
.
~math31415926535
Solution 6 (Infinite Geometric Sequence Method)
Let's say that our even integers are found in the first numbers where n must be greater than or equal to
. Then, we can form an argument based on this. There are
total ways to assign even numbers being
2
4
6
to each space. Furthermore, we must subtract the cases where we have only a single even integer present in the total
spaces and the case where there are only
distinct even integers present. There is
way we can have a single even integer in the entire
spaces, therefore giving us just
option for each of the three integers, so we have
total cases for this. Moreover, the amount of cases with just
distinct even integers is
but subtract the cases where all of the n spaces is either a single integer giving us
, but we multiply by
because of the ways to choose
distinct even integers that are used in the sequence of
. Finally, we have
note: we must divide all of this by
for probability. Additionally, over the entire summation, we multiply by
because of the
probability of selecting an odd at the end of all the evens. Therefore, if you compute this using infinite geometric sequences, you get
.
~Jske25
https://www.youtube.com/watch?v=T-kdDFeHzns&t=17s
Solution 7
The question basically asks for the probability that ,
, and
are all rolled before
,
or
. This probability is equal to the probability that any 3 number combination is rolled before the remaining numbers are rolled (for example, rolling
and
before
,
and
). There are
combinations possible and summing them all up must result in 1 so the probability of the specific combination
being chosen is
~AwesomeK
Video Solution by hurdler (complementary probability)
https://www.youtube.com/watch?v=k2Jy4ni9tK8
Video Solution by TheBeautyofMath
https://youtu.be/FV9AnyERgJQ?t=480
~IceMatrix
See Also
2021 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.