2018 AMC 10B Problems/Problem 6
Contents
[hide]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 .
Jonathan Xu (pi_is_delicious_69420)
Solution 1.1
Since our favorable outcomes are ; ; ; and , we have 4 favorable outcomes. We have 20 total outcomes. Probability will be . Hence the answer is .
~east999
Solution 2
We only have to analyze the first two draws as that gives us insight into if a third draw is necessary. Also, note that it is necessary to draw a to have 3 draws, otherwise will be attainable in two or fewer 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
~Soccer_JAMS
Solution 3 (Inequalities)
We know that the first draws must be leaving us with with the only option being that and are chosen in either order. This means there is a probability of , or
~MC_ADe
Video Solution (HOW TO THINK CRITICALLY!!!)
~Education, the Study of Everything
Video Solution
~savannahsolver
Video Solution
https://youtu.be/wopflrvUN2c?t=20
See Also
2018 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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