unicornjazz666
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You have also set up a card game in which a player picks a card from a standard deck of 52 cards. The player wins if these two events occur together: E1, in which the card drawn is a black card, and E2, in which the card drawn is a numbered card, 2 through 10.What is the probability of getting a black card and a numbered card? Calculate the probabilities P(E1) and P(E2) as fractions.

Respuesta :

remotecontrolbrain

First, let's count:

there are 26 possible outcomes for E1 (black card)

there are 4x9 = 36 possible outcomes for E2, to pick a numbered card (any color)

there are 2x9 =18 possible outcomes for E1 (black) AND E2 (numbered, spade + clower)

the probability of E1 AND E2 is the ratio of the count of possible outcomes for E1 + E2 and the count of all possible outcomes (52 choices to pick a card from the deck):

P(E1 and E2) = 18/52 (34.6%)

And as asked:

P(E1) = 26/52 = 1/2 (50%)

P(E2) = 36/52 = 9/13 (69.2%)

omgitzgali431

[tex]n (S) = 5\\ n (E_1) = 2 x 13 = 26\\P (E_1) \frac{n (E_1)}{n (S)} = \frac {26}{52} = \frac{1}{2} \\\\ n (E_2) 4 x 9 = 36\\P (E_2)\frac{n (E_2)}{n (S)}=\frac{36}{52}= \frac{9}{13}[/tex]

EDMENTUM / PLATO ANSWER!!!!!!!!

CAN JUST WRITE:

[tex]P (E_1) \frac{n (E_1)}{n (S)} = \frac {26}{52} = \frac{1}{2}[/tex] =  (50%)

[tex]P (E_2)\frac{n (E_2)}{n (S)}=\frac{36}{52}= \frac{9}{13}[/tex] =  (69.2%) :)))))

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