Difference between revisions of "2005 AMC 10B Problems/Problem 21"
Dairyqueenxd (talk | contribs) (→Solution 2 (order does not matter)) |
Firebolt360 (talk | contribs) |
||
Line 3: | Line 3: | ||
<math>\textbf{(A) } 162 \qquad \textbf{(B) } 180 \qquad \textbf{(C) } 324 \qquad \textbf{(D) } 360 \qquad \textbf{(E) } 720 </math> | <math>\textbf{(A) } 162 \qquad \textbf{(B) } 180 \qquad \textbf{(C) } 324 \qquad \textbf{(D) } 360 \qquad \textbf{(E) } 720 </math> | ||
− | |||
− | |||
− | |||
− | |||
− | |||
== Solution 1 (where the order of drawing slips matters) == | == Solution 1 (where the order of drawing slips matters) == | ||
Line 27: | Line 22: | ||
Hence, <math>\frac{q}{p}=\frac{45\cdot6\cdot6}{10}=9\cdot6\cdot3=\boxed{\textbf{(A) }162}</math>. | Hence, <math>\frac{q}{p}=\frac{45\cdot6\cdot6}{10}=9\cdot6\cdot3=\boxed{\textbf{(A) }162}</math>. | ||
+ | |||
+ | == Video Solution == | ||
+ | https://youtu.be/wopflrvUN2c?t=252 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
==See Also== | ==See Also== |
Revision as of 23:34, 26 October 2022
Contents
[hide]Problem
Forty slips are placed into a hat, each bearing a number , , , , , , , , , or , with each number entered on four slips. Four slips are drawn from the hat at random and without replacement. Let be the probability that all four slips bear the same number. Let be the probability that two of the slips bear a number and the other two bear a number . What is the value of ?
Solution 1 (where the order of drawing slips matters)
There are ways to determine which number to pick. There are ways to then draw those four slips with that number, and total ways to draw four slips. Thus .
There are ways to determine which two numbers to pick for the second probability. There are ways to arrange the order which we draw the non-equal slips, and in each order there are ways to pick the slips, so .
Hence, the answer is .
Solution 2 (where the order does not matter)
For probability , there are ways to choose the number you want to show up times.
Hence, the probability is .
For probability , there are ways to choose the numbers you want to show up twice. There are ways to pick which slips you want out of the of each.
Hence, the probability is
Hence, .
Video Solution
https://youtu.be/wopflrvUN2c?t=252
~ pi_is_3.14
See Also
2005 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.