Difference between revisions of "2005 AMC 10B Problems/Problem 21"
Dairyqueenxd (talk | contribs) (→Solution 1 (where the order of drawing slips matters)) |
Dairyqueenxd (talk | contribs) (→Solution 1 (where the order of drawing slips matters)) |
||
Line 14: | Line 14: | ||
There are <math>{10 \choose 2} = 45</math> ways to determine which two numbers to pick for the second probability. There are <math>{4 \choose 2} = 6</math> ways to arrange the order which we draw the non-equal slips, and in each order there are <math>4 \times 3 \times 4 \times 3</math> ways to pick the slips, so <math>q = \frac{45 \cdot 6 \cdot 4^2 \cdot 3^2}{40 \cdot 39 \cdot 38 \cdot 37}</math>. | There are <math>{10 \choose 2} = 45</math> ways to determine which two numbers to pick for the second probability. There are <math>{4 \choose 2} = 6</math> ways to arrange the order which we draw the non-equal slips, and in each order there are <math>4 \times 3 \times 4 \times 3</math> ways to pick the slips, so <math>q = \frac{45 \cdot 6 \cdot 4^2 \cdot 3^2}{40 \cdot 39 \cdot 38 \cdot 37}</math>. | ||
− | Hence, the answer is <math>\frac{q}{p} = \frac{45 \cdot 6 \cdot 4^2 \cdot 3^2}{10\cdot 4!} = \boxed{\mathbf{(A)}162}</math>. | + | Hence, the answer is <math>\frac{q}{p} = \frac{45 \cdot 6 \cdot 4^2 \cdot 3^2}{10\cdot 4!} = \boxed{\mathbf{(A) }162}</math>. |
==Solution 2 (order does not matter)== | ==Solution 2 (order does not matter)== |
Revision as of 16:20, 16 December 2021
Contents
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
?
Video Solution
https://youtu.be/wopflrvUN2c?t=252
~ pi_is_3.14
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 (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, .
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.