tyleighdayton
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For the experiment of drawing a single card from a standard 52-card deck, find (a) the probability of the following event, and (b) the
odds in favor of the following event.
Neither a spade nor an ace
(a) The probability that the card is neither a spade nor an ace is ☐ .
(Simplify your answer.)
(b) The odds, in simplified form, in favor of the event of the card being neither a spade nor an ace, are
to

Respuesta :

A. Since there are 13 spades in a deck and 3 aces that aren't spades, the probability of drawing neither is 36/52.  If you add 13 and 3 you get 16.  52 minus 16 is 36.  In its simplest form, you would reduce it to 9/13.

B. The odds in favor would be 16/52 because it is the opposide.  In the simplest form, you would reduce this by dividing 16 by 4 with is 4 and 52 by 4 which is 13 giving you the final answer of 4/13.

angelenguillo059

【Answer】: (a) 36/52 (b) 36:16

【Explanation】: (a) In a standard 52-card deck, there are 13 cards of each suit (hearts, diamonds, clubs, and spades). There are 4 aces in the deck (one for each suit). The event of drawing neither a spade nor an ace can be calculated by subtracting the number of spades and aces from the total number of cards. There are 13 spades and 4 aces, but we must be careful not to double-count the ace of spades. So, the total number of cards that are either spades or aces is 13 (spades) + 4 (aces) - 1 (ace of spades) = 16. Therefore, the probability of drawing neither a spade nor an ace is 1 - (16/52) = 36/52.

(b) The odds in favor of an event are defined as the ratio of the probability of the event happening to the probability of the event not happening. In this case, the probability of the event happening is 36/52 and the probability of the event not happening is 16/52. Therefore, the odds in favor of the event are 36:16.

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