2017 AMC 8 Problems/Problem 10
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[hide]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 = \boxed{{\textbf{(C) }} {\frac{3}{10}}}$.
Solution 2 (regular probability)
P (no 5)= * * = . This is the fraction of total cases with no fives. p (no 4 and no 5)= * * = = . This is the intersection of no fours and no fives. Subtract the fraction of no fours and no fives from that of no fives. = \boxed{{\textbf{(C) }} {\frac{3}{10}}}$.
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) | ||
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Followed by Problem 11 | |
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