Difference between revisions of "2017 AMC 8 Problems/Problem 10"
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− | Using complementary counting, <math>P_\textbf{4 is largest} = 1-P_\textbf{5 is largest} - P_\textbf{3 is largest} = 1- \frac{\dbinom{4}{2}}{\dbinom{5}{3}} - \frac{\dbinom{2}{2}}{\dbinom{5}{3}} = 1- \frac{6}{10} - \frac{1}{10} = \boxed{{\frac{3}{10} | + | Using complementary counting, <math>P_\textbf{4 is largest} = 1-P_\textbf{5 is largest} - P_\textbf{3 is largest} = 1- \frac{\dbinom{4}{2}}{\dbinom{5}{3}} - \frac{\dbinom{2}{2}}{\dbinom{5}{3}} = 1- \frac{6}{10} - \frac{1}{10} = \boxed{{{\textbf{(C)}\frac{3}{10}}}}</math> |
-mathfan2020 | -mathfan2020 | ||
Revision as of 14:32, 21 December 2022
Contents
Problem 10
A box contains five cards, numbered 1, 2, 3, 4, and 5. Three cards are selected randomly without replacement from the box. What is the probability that 4 is the largest value selected?
Solution 1 (combinations)
There are possible groups of cards that can be selected. If
is the largest card selected, then the other two cards must be either
,
, or
, for a total
groups of cards. Then the probability is just
Solution 2 (regular probability)
P (no 5)= *
*
=
this is the fraction of total cases with no fives.
p (no 4 and no 5)=
*
*
=
=
this is the intersection of no fours and no fives. Subtract fraction of no fours and no fives from no fives.
(C)
Solution 3 (Complementary Probability)
Using complementary counting,
-mathfan2020
Solution 4
Let's have three 'boxes'.
One of the boxes must be 4, so
Video Solutions
~savannahsolver
See Also:
2017 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.