Difference between revisions of "2018 AMC 10B Problems/Problem 6"

Revision as of 15:30, 16 February 2018

A box contains $5$ chips, numbered $1$ , $2$ , $3$ , $4$ , and $5$ . Chips are drawn randomly one at a time without replacement until the sum of the values drawn exceeds $4$ . What is the probability that $3$ draws are required?

$\textbf{(A)} \frac{1}{15} \qquad \textbf{(B)} \frac{1}{10} \qquad \textbf{(C)} \frac{1}{6} \qquad \textbf{(D)} \frac{1}{5} \qquad \textbf{(E)} \frac{1}{4}$

Solution

Notice that the only two ways such that more than $2$ draws are required are $1,2$ , $1,3$ , $2,1$ , and $3,1$ . Notice that each of those cases has a $\frac{1}{5} \cdot \frac{1}{4}$ chance, so the answer is $\frac{1}{5} \cdot \frac{1}{4} \cdot 4 = \frac{1}{5}$ , or $\boxed{D}$ .

2018 AMC 10B (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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 ==See Also==
-{{AMC10 box|year=2018|ab=B|num-b=3|num-a=5}}
+{{AMC10 box|year=2018|ab=B|num-b=5|num-a=7}}
 {{MAA Notice}}

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Difference between revisions of "2018 AMC 10B Problems/Problem 6"

Revision as of 15:30, 16 February 2018

Solution

See Also