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The probability of a card higher than a five in any particular suit, when that suit is the only suit in the deck of 13, assuming the Ace is high, is 9 in 13, or about 0.6923. If the Ace is low, then the probability is 8 in 13, or about 0.6154.

If you draw from a full deck of 52 cards, and stipulate that you have to draw a club, then the probabilities become 9 in 52, or about 0.1731 (Ace high) or 8 in 52, or about 0.1538 (Ace low).

Q: What is the probability of a card higher than a five clubs appearing in a suit of clubs?

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Probability of 2 of clubs = 1/52 or 0.0192.

If only one card is dealt randomly from a deck of cards, the probability is 1/52.

There are 13 clubs (including the seven of clubs), and 3 other sevens (from the other suits). So that makes 16 cards that are clubs or sevens. Thus, the probability of drawing a club or a seven from a standard 52 card deck is 16/52 = 4/13 (about 31%).

the probablility of drawing a nine or clubs from all four randomly shuffled decks with 52 cards is 1 out of 7,311,616

In a standard 52 card deck, there are 13 clubs and 4 4s. However, the four of clubs is in both lists, so this represents only 16 distinct cards (the 13 clubs, and the 3 other 4s). The probability of drawing one of these 16 cards is 16/52, or 4/13.

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Probability of 2 of clubs = 1/52 or 0.0192.

If only one card is dealt randomly from a deck of cards, the probability is 1/52.

There are 13 clubs (including the seven of clubs), and 3 other sevens (from the other suits). So that makes 16 cards that are clubs or sevens. Thus, the probability of drawing a club or a seven from a standard 52 card deck is 16/52 = 4/13 (about 31%).

In an ordinary 52 card deck, there are 4 three's, and 13 clubs, one of which is the 3 of clubs, so there are 16 cards that are a three or a club. The probability of drawing a three or a club is 16 out of 52, or 4 out or 13, or 0.308.

If only one card is drawn, at random, then it is 1/52.

One in 52 - because there are 52 cars in a deck, and only one Queen of Clubs.

It is 5/52 for a single card, drawn randomly.

There are 12 face cards in a standard deck of 52 cards; the jacks, queens, and kings of spades, diamonds, clubs, and hearts. The probability, then, of drawing a face card is 12 in 52, or 3 in 13, or about 0.2308.

the probablility of drawing a nine or clubs from all four randomly shuffled decks with 52 cards is 1 out of 7,311,616

In a standard 52 card deck, there are 13 clubs and 4 4s. However, the four of clubs is in both lists, so this represents only 16 distinct cards (the 13 clubs, and the 3 other 4s). The probability of drawing one of these 16 cards is 16/52, or 4/13.

P(a^~a)==1 P(a&~a)==0 the line above is shorthand notation for an event that has a probability of 1, followed by an event that has a probability of 0. P(event) is an easy way to say the probability of "event". The "^" means "OR", the "~" means "NOT", and the "&" as you are probably familiar means "AND". So puting it together, "P(a^~a)" means the probability that an event "a" occurs OR that event "a" does not occur. So take an event "a", any event, like drawing an 8 of clubs out of a deck of cards. If you draw a random card out of a deck of cards, the probability that you will draw an 8 of clubs OR that you will not draw an 8 of clubs is 1. That means 100%. So when you draw a card you will either draw the 8 of clubs or not draw the 8 of clubs. It seems like an obvious statement to make but its a proof that becomes very important in proving less obvious theories. likewise, the second statement was "P(a&~a)", so the probability that event "a" occurs AND event "a" doesnt occur. Since the event has to either occur or not, it cant occur AND not occur, so the probability is 0.

Let's say there are 52 cards in a deckThere's one chance out of 52 to get the two of clubs from the first deckThe same chance with the second deckThe probability that the two events occur at the same time is the probability of the first multiplied by the probability of the second as the events are independentSo P = 1/52 x 1/52 = 1/2074