2018 AMC 10B Problems/Problem 6
Contents
Problem
A box contains chips, numbered , , , , and . Chips are drawn randomly one at a time without replacement until the sum of the values drawn exceeds . What is the probability that draws are required?
Solution 1
Notice that the only four ways such that draws are required are ; ; ; and . Notice that each of those cases has a chance, so the answer is , or .
Solution 2
We only have to analyze first two draws as that gives us insight on if third draw is necessary. Also, note that it is necessary to draw a in order to have 3 draws, otherwise will be attainable in two or less draws. So the probability of getting a is . It is necessary to pull either a or on the next draw and the probability of that is . But, the order of the draws can be switched so we get:
, or
By: Soccer_JAMS
Solution 3
We can use complementary probability. There is a chance of pulling either or . In both cases, there is a 100% chance that we need not pull a third number. There is a chance of pulling either or , for which there is a chance that we need not pull a third number, for this will only happen if is pulled next. Finally, if we pull a (for which the probability is ), there is a chance that we need not pull a third number, for this will happen if either or is pulled next.
Multiplying these fractions gives us the following expression:
Therefore, the complementary probability is so the answer is or .
Video Solution
~savannahsolver
Video Solution
https://youtu.be/wopflrvUN2c?t=20
~ pi_is_3.14
See Also
2018 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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.