Difference between revisions of "2018 AMC 10B Problems/Problem 6"
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Jonathan Xu (pi_is_delicious_69420) | Jonathan Xu (pi_is_delicious_69420) | ||
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| + | == Solution 1.1 == | ||
| + | |||
| + | Since our favorable outcomes are <math>1,2</math>; <math>1,3</math>; <math>2,1</math>; and <math>3,1</math>, we have 4 favorable outcomes. We have 20 total outcomes. Probability will be <math>\frac{1}{5}</math>. Hence the answer is <math>\boxed{D}</math>. | ||
| + | |||
| + | ~east999 | ||
== Solution 2 == | == Solution 2 == | ||
| − | We only have to analyze first two draws as that gives us insight | + | 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 <math>1</math> to have 3 draws, otherwise <math>5</math> will be attainable in two or fewer 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> | ||
| − | + | ~Soccer_JAMS | |
| + | |||
| + | == Solution 3 (Inequalities) == | ||
| + | We know that the first <math>2</math> draws must be <math>\le{3}</math> leaving us with with the only option being that <math>1</math> and <math>2</math> are chosen in either order. This means there is a probability of <math>\frac{1}{\binom52} \cdot 2 = \frac{1}{5}</math>, or <math>\boxed {D}</math> | ||
| + | |||
| + | ~MC_ADe | ||
==Video Solution (HOW TO THINK CRITICALLY!!!)== | ==Video Solution (HOW TO THINK CRITICALLY!!!)== | ||
Latest revision as of 10:41, 5 November 2024
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 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 | |
| 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. 
