Difference between revisions of "2008 iTest Problems/Problem 35"
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==Problem== | ==Problem== | ||
− | Let <math>b</math> be the probability that the cards are from different suits. Compute <math>\lfloor1000b\rfloor</math>. | + | Let <math>b</math> be the probability that the cards are from different suits. Compute <math>\lfloor1000b\rfloor</math>. |
+ | |||
+ | '''Note: Two cards are drawn.''' | ||
==Solutions== | ==Solutions== |
Latest revision as of 14:44, 3 July 2018
Problem
Let be the probability that the cards are from different suits. Compute .
Note: Two cards are drawn.
Solutions
Solution 1
Use complementary counting to count the number of ways one can draw two cards with the same rank. There are ranks, and each rank has cards. That means the probability of getting two cards with the same rank is , so the probability of getting two cards with different ranks is . That means .
Solution 2
The first card can be any card, so the probability is . However, of the cards remaining, only of them have a different rank. Thus, the probability of getting two cards with different ranks is , so .
See Also
2008 iTest (Problems) | ||
Preceded by: Problem 34 |
Followed by: Problem 36 | |
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