En español:
Ocho tarjetas se extraen sin reemplazo de una baraja ordinaria. hallar la probabilidad de obtener exactamente tres ases o Reyes exactamente (o ambos).
Hay 4 ases en una baraja normal.
Hay maneras de C(4,3) 3 Ases
Hay maneras de 5 tarjetas que no sean ases C(48,5)
Hay formas C(52,8) para obtener las 8 tarjetas
P(EXACTAMENTE 3 ASES) = =.0091014867
Hay 4 Reyes, por lo que la probabilidad de obtener exactamente 3 Reyes es el mismo:
P(EXACTAMENTE 3 REYES) =.0091014867
Pero también tenemos el número de as 3 y 3 Reyes,
Hay maneras de C(4,3) 3 Ases
Hay maneras de C(4,3) 3 Reyes
Hay maneras de C(44,2) 2 tarjetas que no sean ases ni Reyes
Hay formas C(52,8) para obtener las 8 tarjetas
P (exactamente 3 Ases y 3 Reyes) = =
.00000002126138057
Ahora utilizamos la ecuación:
P (EXACTAMENTE 3 ASES O REYES EXACTAMENTE 3 (O AMBOS)) =
P(EXACTAMENTE 3 ASES) + P (EXACTAMENTE 3 REYES) - P (EXACTAMENTE 3 ASES Y REYES EXACTAMENTE 3)
=.0091014867 +.0091014867 -.00000002126138057 =.0182029522
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In Engish:
Eight cards are extracted without replacement from an ordinary deck. find the probability of getting exactly three ACEs or exactly three kings (or both).
There are 4 ACES in an ordinary deck of cards.
There are C(4,3) ways to get 3 ACES
There are C(48,5) ways to get 5 cards which are not ACES
There are C(52,8) ways to get any 8 cards
P(exactly 3 ACES) = = .0091014867
There are 4 KINGS, so the probability of getting exactly 3 KINGS is the same:
P(exactly 3 KINGS) = .0091014867
But we must also get the number of 3 ACE and 3 KINGS,
There are C(4,3) ways to get 3 ACES
There are C(4,3) ways to get 3 KINGS
There are C(44,2) ways to get 2 cards which are neither ACES nor KINGS
There are C(52,8) ways to get any 8 cards
P(exactly 3 ACES and 3 KINGS) = =
.00000002126138057
Now we use the equation:
P(EXACTLY 3 ACES OR EXACTLY 3 KINGS (OR BOTH)) =
P(EXACTLY 3 ACES) + P(EXACTLY 3 KINGS) - P(EXACTLY 3 ACES AND EXACTLY 3 KINGS)
= .0091014867 + .0091014867 - .00000002126138057 = .0182029522
Edwin