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Difference between revisions of "2017 AMC 8 Problems/Problem 10"

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==Solution 4==
 
==Solution 4==
 
Let's have three 'boxes'.
 
Let's have three 'boxes'.
−
One of the boxes must be 4, so 3C1 x 3 x 2/5 x 4 x 3 = 3/10
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One of the boxes must be 4, so <math>\frac{\binom{3}{1} x 3 x 2}{5 x 4 x 3} = \frac{3}{10}</math>
  
 
==Video Solutions==
 
==Video Solutions==

Revision as of 13:06, 11 December 2021

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?

$\textbf{(A) }\frac{1}{10}\qquad\textbf{(B) }\frac{1}{5}\qquad\textbf{(C) }\frac{3}{10}\qquad\textbf{(D) }\frac{2}{5}\qquad\textbf{(E) }\frac{1}{2}$

Solution

There are $\binom{5}{3}$ possible groups of cards that can be selected. If $4$ is the largest card selected, then the other two cards must be either $1$, $2$, or $3$, for a total $\binom{3}{2}$ groups of cards. Then the probability is just ${\frac{{\dbinom{3}{2}}}{{\dbinom{5}{3}}}} = \boxed{{\textbf{(C) }} {\frac{3}{10}}}$

Solution 2 (regular probability)

P (no 5)= $\frac{4}{5}$ * $\frac{3}{4}$ * $\frac{2}{3}$ = $\frac{2}{5}$ this is the fraction of total cases with no fives. p (no 4 and no 5)= $\frac{3}{5}$ * $\frac{2}{4}$ * $\frac{1}{3}$ = $\frac{6}{60}$ = $\frac{1}{10}$ this is the intersection of no fours and no fives. Subtract fraction of no fours and no fives from no fives. $\frac{2}{5} - \frac{1}{10} = \frac{3}{10}$ (C)


Solution 3 (Complementary Probability)

Using complementary counting, $P_\textbf{4 is largest} = 1-P_\textbf{5 is largest} - P_\textbf{3 is largest} = 1- \frac{\dbinom{4}{2}}{\dbinom{5}{3}} - \frac{\dbinom{2}{2}}{\dbinom{5}{3}} = 1- \frac{6}{10} - \frac{1}{10} = \boxed{{\frac{3}{10}}{\textbf{(C)}}}$ -mathfan2020

Solution 4

Let's have three 'boxes'. One of the boxes must be 4, so $\frac{\binom{3}{1} x 3 x 2}{5 x 4 x 3} = \frac{3}{10}$

Video Solutions

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

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