2008 iTest Problems/Problem 34

Problem

While entertaining his younger sister Alexis, Michael drew two different cards from an ordinary deck of playing cards. Let a be the probability that the cards are of different ranks. Compute $\lfloor 1000a\rfloor$.

Solutions

Solution 1

Use complementary counting to count the number of ways one can draw two cards with the same rank. There are $13$ ranks, and each rank has $4$ cards. That means the probability of getting two cards with the same rank is $\tfrac{13 \cdot 4 \cdot 3}{52 \cdot 51} = \tfrac{3}{51}$, so the probability of getting two cards with different ranks is $\tfrac{48}{51}$. That means $\lfloor 1000a\rfloor = \boxed{941}$.

Solution 2

The first card can be any card, so the probability is $1$. However, of the $51$ cards remaining, only $12 \cdot 4 = 48$ of them have a different rank. Thus, the probability of getting two cards with different ranks is $1 \cdot \tfrac{48}{51} = \tfrac{48}{51}$, so $\lfloor 1000a\rfloor = \boxed{941}$.

See Also

2008 iTest (Problems)
Preceded by:
Problem 33
Followed by:
Problem 35
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