0039 countably innite sets in the previous two

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Unformatted text preview: e numbers can be arranged to get the sequence 10,11,12,13,1 (because, by tradition, the 1 card, or “ace,” can act as either the highest or lowest number). For example, the outcome {3D, 4H, 5H, 6S, 7D} is an element of STRAIGHT. Find P (FULL HOUSE) and P (STRAIGHT). Solution: The sample space is Ω = {A : A ⊂ C and |A| = 5}, and the number of possible outcomes is |Ω| = 552 . To select an outcome in FULL HOUSE, there are 13 ways to select a number for the three cards with the same number, and then 12 ways to select a number for the other two cards. Once the two numbers are selected, there are 4 ways to select 3 of the 4 suits 3 for the three cards with the same number, and 4 ways to select 2 of the 4 suits for the two cards 2 with the other number. Thus, P (FULL HOUSE) = = = 13 · 12 · 4 3 4 2 52 5 13 · 12 · 4 · 4 · 3 · 5 · 4 · 3 · 2 2 · 52 · 51 · 50 · 49 · 48 6 6 = ≈ 0.0014. 15 · 5 · 49 4165 To select an outcome in STRAIGHT, there are ten choices for the set of five integers on the five cards that can correspo...
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This note was uploaded on 02/09/2014 for the course ISYE 2027 taught by Professor Zahrn during the Spring '08 term at Georgia Tech.

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