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 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 | |
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− | |||
+ | == 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!!!)== | |
+ | https://youtu.be/gtsDrM16J9U | ||
− | + | ~Education, the Study of Everything | |
==Video Solution== | ==Video Solution== | ||
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~savannahsolver | ~savannahsolver | ||
+ | |||
+ | == Video Solution == | ||
+ | https://youtu.be/wopflrvUN2c?t=20 | ||
==See Also== | ==See Also== |
Latest revision as of 00:15, 11 November 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 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.