Question 681564
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En español:
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Ocho tarjetas se extraen sin reemplazo de una baraja ordinaria. hallar la probabilidad de obtener exactamente tres ases o Reyes exactamente (o ambos).
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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) = {{{("C(4,3)"*"C(48,5)")/"C(52,8)"}}} =.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) = {{{("C(4,3)"*"C(4,3)"*C(44,2))/"C(52,8)"}}} =

.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:
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Eight cards are extracted without replacement from an ordinary deck. find the probability of getting exactly three ACEs or exactly three kings (or both).
<pre>
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) = {{{("C(4,3)"*"C(48,5)")/"C(52,8)"}}} = .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) = {{{("C(4,3)"*"C(4,3)"*C(44,2))/"C(52,8)"}}} =  

.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</pre>