Difference between revisions of "2018 AMC 10B Problems/Problem 6"
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Notice that the only four ways such that <math>3</math> draws are required are <math>1,2</math>; <math>1,3</math>; <math>2,1</math>; and <math>3,1</math>. Notice that each of those cases has a <math>\frac{1}{5} \cdot \frac{1}{4}</math> chance, so the answer is <math>\frac{1}{5} \cdot \frac{1}{4} \cdot 4 = \frac{1}{5}</math>, or <math>\boxed{D}</math>. | Notice that the only four ways such that <math>3</math> draws are required are <math>1,2</math>; <math>1,3</math>; <math>2,1</math>; and <math>3,1</math>. Notice that each of those cases has a <math>\frac{1}{5} \cdot \frac{1}{4}</math> chance, so the answer is <math>\frac{1}{5} \cdot \frac{1}{4} \cdot 4 = \frac{1}{5}</math>, or <math>\boxed{D}</math>. | ||
+ | |||
+ | Jonathan Xu (pi_is_delicious_69420) | ||
== Solution 2 == | == 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 <math>1</math> in order to have 3 draws, otherwise <math>5</math> will be attainable in two or less draws. So the probability of getting a <math>1</math> is <math>\frac{1}{5}</math>. It is necessary to pull either a <math>2</math> or <math>3</math> on the next draw and the probability of that is <math>\frac{1}{2}</math>. But, the order of the draws can be switched so we get: | |
<math>\frac{1}{5} \cdot \frac{1}{2} \cdot 2 = \frac{1}{5}</math>, or <math>\boxed {D}</math> | <math>\frac{1}{5} \cdot \frac{1}{2} \cdot 2 = \frac{1}{5}</math>, or <math>\boxed {D}</math> | ||
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By: Soccer_JAMS | By: Soccer_JAMS | ||
− | == Solution | + | ==Video Solution (HOW TO THINK CRITICALLY!!!)== |
+ | https://youtu.be/gtsDrM16J9U | ||
− | + | ~Education, the Study of Everything | |
− | + | ==Video Solution== | |
+ | https://youtu.be/ctQ3VbKAFBg | ||
− | + | ~savannahsolver | |
− | + | == Video Solution == | |
+ | https://youtu.be/wopflrvUN2c?t=20 | ||
==See Also== | ==See Also== |
Latest revision as of 14:22, 7 July 2023
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 .
Jonathan Xu (pi_is_delicious_69420)
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
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 | |
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.