Difference between revisions of "2017 AMC 8 Problems/Problem 10"
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==Solution 2 (regular probability)== | ==Solution 2 (regular probability)== | ||
P (no 5)= <math>\frac{4}{5}</math> * <math>\frac{3}{4}</math> * <math>\frac{2}{3}</math> = <math>\frac{2}{5}</math>. This is the fraction of total cases with no fives. | P (no 5)= <math>\frac{4}{5}</math> * <math>\frac{3}{4}</math> * <math>\frac{2}{3}</math> = <math>\frac{2}{5}</math>. This is the fraction of total cases with no fives. | ||
− | p (no 4 and no 5)= <math>\frac{3}{5}</math> * <math>\frac{2}{4}</math> * <math>\frac{1}{3}</math> = <math>\frac{6}{60}</math> = <math>\frac{1}{10}</math>. This is the intersection of no fours and no fives. Subtract the fraction of no fours and no fives from that of no fives. <math>\frac{2}{5} - \frac{1}{10} = \frac{3}{10}</math>= \boxed | + | p (no 4 and no 5)= <math>\frac{3}{5}</math> * <math>\frac{2}{4}</math> * <math>\frac{1}{3}</math> = <math>\frac{6}{60}</math> = <math>\frac{1}{10}</math>. This is the intersection of no fours and no fives. Subtract the fraction of no fours and no fives from that of no fives. <math>\frac{2}{5} - \frac{1}{10} = \frac{3}{10}</math> = \boxed{\textbf{(C)}\frac{3}{10}}$. |
==Solution 3 (Complementary Probability)== | ==Solution 3 (Complementary Probability)== |
Revision as of 07:49, 5 March 2023
Contents
Problem 10
A box contains five cards, numbered 1, 2, 3, 4, and 5. Three cards are selected randomly without replacement from the box. What is the probability that 4 is the largest value selected?
Solution 1 (combinations)
There are possible groups of cards that can be selected. If is the largest card selected, then the other two cards must be either , , or , for a total groups of cards. Then, the probability is just .
Solution 2 (regular probability)
P (no 5)= * * = . This is the fraction of total cases with no fives. p (no 4 and no 5)= * * = = . This is the intersection of no fours and no fives. Subtract the fraction of no fours and no fives from that of no fives. = \boxed{\textbf{(C)}\frac{3}{10}}$.
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 |
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