SOLUTION: In English: Eight cards are extracted without replacement from an ordinary deck. find the probability of getting exactly three ACEs or exactly three kings (or both). ocho car

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Question 681564: In English:
Eight cards are extracted without replacement from an ordinary deck. find the probability of getting exactly three ACEs or exactly three kings (or both).

ocho cartas se extraen sin reposición de una baraja ordinaria. encontrar la probabilidad de obtener exactamente tres ases o exactamente tres reyes (o ambos).
Answer by Edwin McCravy(20067) About Me  (Show Source):
You can put this solution on YOUR website!
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) = %28%22C%284%2C3%29%22%2A%22C%2848%2C5%29%22%29%2F%22C%2852%2C8%29%22 =.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) = %28%22C%284%2C3%29%22%2A%22C%284%2C3%29%22%2AC%2844%2C2%29%29%2F%22C%2852%2C8%29%22 =

.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) = %28%22C%284%2C3%29%22%2A%22C%2848%2C5%29%22%29%2F%22C%2852%2C8%29%22 = .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) = %28%22C%284%2C3%29%22%2A%22C%284%2C3%29%22%2AC%2844%2C2%29%29%2F%22C%2852%2C8%29%22 =  

.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