2018 AMC 10B Problems/Problem 6: Difference between revisions
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== 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> | ||
Revision as of 15:33, 1 September 2021
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 | ||
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