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alina1380 [7]
3 years ago
10

A card is chosen from a standard deck of cards. What is the probability that the card is a club, given that the card is black?

Mathematics
2 answers:
Misha Larkins [42]3 years ago
7 0
Answer : 13/52

There are 52 cards in a standard deck

13 of those cards are clubs, and they are all black.

13/52 will be a black club


leonid [27]3 years ago
5 0
From a standard deck of cards, one card is drawn. What is the probability that the card is black and a jack? P(Black and Jack) P(Black) = 26/52 or ½ , P(Jack) is 4/52 or 1/13 so P(Black and Jack) = ½ * 1/13 = 1/26 A standard deck of cards is shuffled and one card is drawn. Find the probability that the card is a queen or an ace. P(Q or A) = P(Q) = 4/52 or 1/13 + P(A) = 4/52 or 1/13 = 1/13 + 1/13 = 2/13 WITHOUT REPLACEMENT: If you draw two cards from the deck without replacement, what is the probability that they will both be aces? P(AA) = (4/52)(3/51) = 1/221. 1 WITHOUT REPLACEMENT: What is the probability that the second card will be an ace if the first card is a king? P(A|K) = 4/51 since there are four aces in the deck but only 51 cards left after the king has been removed. WITH REPLACEMENT: Find the probability of drawing three queens in a row, with replacement. We pick a card, write down what it is, then put it back in the deck and draw again. To find the P(QQQ), we find the probability of drawing the first queen which is 4/52. The probability of drawing the second queen is also 4/52 and the third is 4/52. We multiply these three individual probabilities together to get P(QQQ) = P(Q)P(Q)P(Q) = (4/52)(4/52)(4/52) = .00004 which is very small but not impossible. Probability of getting a royal flush = P(10 and Jack and Queen and King and Ace of the same suit) What's the probability of being dealt a royal flush in a five card hand from a standard deck of cards? (Note: A royal flush is a 10, Jack, Queen, King, and Ace of the same suit. A standard deck has 4 suits, each with 13 distinct cards, including these five above.) (NB: The order in which the cards are dealt is unimportant, and you keep each card as it is dealt -- it's not returned to the deck.) The probability of drawing any card which could fit into some royal flush is 5/13. Once that card is taken from the pack, there are 4 possible cards which are useful for making a royal flush with that first card, and there are 51 cards left in the pack. therefore the probability of drawing a useful second card (given that the first one was useful) is 4/51. By similar logic you can calculate the probabilities of drawing useful cards for the other three. The probability of the royal flush is therefore the product of these numbers, or 5/13 * 4/51 * 3/50 * 2/49 * 1/48 = .00000154
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Illusion [34]
Solve for y
eliminate x
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-15x-30y=-15
<u>15x-3y=81 +</u>
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7 0
3 years ago
When Eric multiplied two binomials together, his result was a trinomial. An example is
leva [86]

Answer:

(x + 4)(x - 4)

Step-by-step explanation:

There are actually quite a lot of pairs of binomials the disproves Eric's conclusion, but they all model after the same special product: a^2 - b^2.

The special product a^2 - b^2 can be factored into (a + b)(a - b) and for all real a and b, it will come out as a binomial.

Here is an example:

(x + 4)(x - 4)

We can use the distributive property to get:

x^2 - 4x + 4x - 16

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x^2 - 16

This would disprove Eric's conclusion.

6 0
3 years ago
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seraphim [82]

Answer:

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5 0
2 years ago
P is located at -23 and q is located at 18. what is the best distance between the points
Rina8888 [55]
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5 0
3 years ago
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2 is subtracted from a number, and then the the difference is multiplied by 5. The result is 30
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