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Kryger [21]
2 years ago
13

If you draw four cards at random from a standard deck of 52 cards, what is the probability that all 4 cards have distinct charac

ters (letters or numbers)
Mathematics
1 answer:
mezya [45]2 years ago
4 0

There are \binom{52}4 ways of drawing a 4-card hand, where

\dbinom nk = \dfrac{n!}{k!(n-k)!}

is the so-called binomial coefficient.

There are 13 different card values, of which we want the hand to represent 4 values, so there are \binom{13}4 ways of meeting this requirement.

For each card value, there are 4 choices of suit, of which we only pick 1, so there are \binom41 ways of picking a card of any given value. We draw 4 cards from the deck, so there are \binom41^4 possible hands in which each card has a different value.

Then there are \binom{13}4 \binom41^4 total hands in which all 4 cards have distinct values, and the probability of drawing such a hand is

\dfrac{\dbinom{13}4 \dbinom41^4}{\dbinom{52}4} = \boxed{\dfrac{2816}{4165}} \approx 0.6761

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3 years ago
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6 0
3 years ago
Examine the following table. Determine whether the rate of change is constant or variable. If it is constant, state the rate of
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5 0
2 years ago
In rectangle PQRS, PR = 18x – 28 and QS = x + 380. Find the value of x and the length of each diagonal.
chubhunter [2.5K]

Answer:

x = 24, PR = 404, QS = 404

Step-by-step explanation:

PR and QS are both diagonals of the rectangle. The diagonals of a rectangle bisect each other and are of equal length. Thus, we can say:

PR = QS

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Solving for x, we get:

18x - x = 380 + 28

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x = 408/17

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Thus, PR = 18(24) - 28 = 404  and  QS = 24 + 380 = 404

Third choice is right.

7 0
3 years ago
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