A card is drawn from a well-shuffled deck of 52 cards. How do you find P(drawing an ace or a 9)?

Answer 1

#P("ace or 9") = 2/13#

There are two possible events that could happen. Either you draw an ace or a 9. There are four aces in a deck of #52# cards as well as #4# of #9#'s in the same deck. The conjunction "or" implies adding up the probabilities of the individual events . Thus,
#P("ace or 9") = 4/52 + 4/52 = 8/52#
Reduced to lowest terms, the answer is equivalent to #2/13#
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Answer 2

To find the probability of drawing an ace or a 9 from a well-shuffled deck of 52 cards, you add the probabilities of drawing each separately and then subtract the probability of drawing both an ace and a 9, since drawing both simultaneously is not possible:

P(drawing an ace or a 9) = P(ace) + P(9) - P(ace and 9)

There are 4 aces and 4 9s in a standard deck of cards:

P(ace) = 4/52 P(9) = 4/52

There are 4 cards that are both an ace and a 9 (the ace of hearts, diamonds, clubs, and spades):

P(ace and 9) = 4/52

Substituting the values:

P(drawing an ace or a 9) = (4/52) + (4/52) - (4/52)

P(drawing an ace or a 9) = 8/52

Simplify the fraction:

P(drawing an ace or a 9) = 2/13

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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