Cap_2_p_146

GRUPO areyesg -> 1 punto

=Capítulo 2, Problema 146=

In a game of poker find the probability of getting a) a royal flush, which consists of the ten, jack, queen, king and aces of a single suit; b) a full house, which consists of 3 cards of one face value and 2 of the another ( such as 3 tens and 2 jacks, etc,); c) all diferent cards, but of a single suit; d) 4 aces.

En un juego de póker encontrar la probabilidad de:


 * A)** Una Escalera Real, que consiste en: 10, Jack, Reina, Rey y un As de un solo Palo:

math P(\text{escalera real}) = \frac{\binom{13}{5}}{\binom{52}{5}} math

math \binom{13}{5}=\frac{13!}{(13-5)!*5!}=\frac{6227020800}{4838400}=1287 math

math \binom{52}{5}=\frac{52!}{(52-5)!*5!}=\frac{80658*10^{67}}{3.1034*10^{61}}=2598960 math

math P(\text{escalera real}) = \frac{1287}{2598960}=4.9519808\times 10^{-4}=0.00049519 math


 * B)** Una casa llena que consiste de tres cartas de un valor nominal y dos de otro (como 3 de diez y 2 de Jack, etc.)

math P(\text{full}) = \frac{\left (13\times\binom{4}{3} \right ) \times \left (12\times\binom{4}{2} \right )}{\binom{52}{5}} math

math \binom{4}{3}!=\frac{4!}{(4-3)!\times3!}=\frac{24}{6}=4 math

math \binom{4}{2}!=\frac{4!}{(4-2)!\times2!}=\frac{24}{4}=6 math

math \binom{52}{5}!=\frac{52!}{(52-5)!\times5!}=\frac{80658*10^{67}}{3.1034*10^{61}}=2598960 math

math P(\text{full}) = \frac{\left (13\times4 \right )\times \left (12\times6 \right )}{2598960}=\frac{3744}{2598960}=0.00144 math


 * C)** Todas las cartas diferentes de un solo palo

math P(\text{todas diferentes}) = \frac{4*\binom{13}{5}!}{\binom{52}{5}!}=\frac{4*1287}{2598960}=0.00198 math

math \\ P(\text{cuatro ases}) = \binom{5}{4} P(A1 \cap A2 \cap A3 \cap A4) = \\ P(\text{cuatro ases}) = \binom{5}{4} P(A1|A2 \cap A3 \cap A4) P(A2|A3 \cap A4) P( A3|A4)P(A4) = \\ P(\text{cuatro ases}) = \frac{5!}{(5-4)!4!} \frac{1}{49} \times \frac{2}{50} \times \frac{3}{51} \times \frac{4}{52} = 1.8469\times 10^{-5} math
 * D)** Cuatro ases:

Forma alternativa: math P(\text{cuatro ases}) = \frac{\binom{4}{4}\binom{48}{1}}{\binom{52}{5}} math

math \binom{4}{4}=\frac{4!}{(4-4)!*4!}=\frac{24}{24}=1 math

math \binom{48}{1}=\frac{48!}{(48-1)!*1!}=\frac{1.2414*10^{61}}{2.5862*10^{59}}=48 math

math P(\text{cuatro ases}) = \frac{\binom{4}{4}!\binom{48}{1}}{\binom{52}{5}}=\frac{1*48}{2598960} = 1.8469\times 10^{-5} math

Solucionado Por:
 * Diana Ramirez
 * David Osorio
 * Alexandra Reyes