2018 AMC 10B Problems/Problem 6
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
Notice that only the first two draws are important, it doesn't matter what number we get third because no matter what combination of numbers is picked, the sum will always be greater than 5. 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 .
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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