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

Latest revision as of 13:44, 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
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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 • 51 • 52 • 53 • 54 • 55 • 56 • 57 • 58 • 59 • 60 • 61 • 62 • 63 • 64 • 65 • 66 • 67 • 68 • 69 • 70 • 71 • 72 • 73 • 74 • 75 • 76 • 77 • 78 • 79 • 80 • 81 • 82 • 83 • 84 • 85 • 86 • 87 • 88 • 89 • 90 • 91 • 92 • 93 • 94 • 95 • 96 • 97 • 98 • 99 • 100

@@ 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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