Difference between revisions of "2008 iTest Problems/Problem 35"

Latest revision as of 14:44, 3 July 2018

Problem

Let $b$ be the probability that the cards are from different suits. Compute $\lfloor1000b\rfloor$ .

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 $4$ ranks, and each rank has $13$ cards. That means the probability of getting two cards with the same rank is $\tfrac{4 \cdot 13 \cdot 12}{52 \cdot 51} = \tfrac{12}{51}$ , so the probability of getting two cards with different ranks is $\tfrac{39}{51}$ . That means $\lfloor 1000b\rfloor = \boxed{764}$ .

Solution 2

The first card can be any card, so the probability is $1$ . However, of the $51$ cards remaining, only $3 \cdot 13 = 39$ of them have a different rank. Thus, the probability of getting two cards with different ranks is $1 \cdot \tfrac{39}{51} = \tfrac{39}{51}$ , so $\lfloor 1000b\rfloor = \boxed{764}$ .

2008 iTest (Problems)
Preceded by: Problem 34	Followed by: Problem 36
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@@ Line 1: / Line 1: @@
 ==Problem==
 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==

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