Understanding Probability with Card Drawing

This question asks for the probability of drawing three specific cards in a particular order from a standard deck of 52 cards without replacement. The key phrase here is "in order" and "without replacement". This means the order matters (permutation), and once a card is drawn, it is not put back into the deck.

Analyzing the Deck and the Draw

A standard deck has 52 cards. It contains:

• 4 Kings (one for each suit: hearts, diamonds, clubs, spades)
• 4 Queens (one for each suit)
• 4 Jacks (one for each suit)

We are drawing three cards in the specific sequence: King, then Queen, then Jack.

Step-by-Step Probability Calculation

We calculate the probability of each draw occurring in the specified order:

1. First Draw: Drawing a King

   There are 4 Kings in a deck of 52 cards.

   The probability of drawing a King first is:

   \(P(\text{King first}) = \frac{\text{Number of Kings}}{\text{Total number of cards}} = \frac{4}{52}\)

2. Second Draw: Drawing a Queen (after drawing a King)

   After drawing one King, there are now 51 cards left in the deck.

   There are still 4 Queens in the deck.

   The probability of drawing a Queen second, given a King was drawn first, is:

   \(P(\text{Queen second } | \text{ King first}) = \frac{\text{Number of Queens}}{\text{Remaining number of cards}} = \frac{4}{51}\)

3. Third Draw: Drawing a Jack (after drawing a King and a Queen)

   After drawing one King and one Queen, there are now 50 cards left in the deck.

   There are still 4 Jacks in the deck.

   The probability of drawing a Jack third, given a King and a Queen were drawn, is:

   \(P(\text{Jack third } | \text{ King first and Queen second}) = \frac{\text{Number of Jacks}}{\text{Remaining number of cards}} = \frac{4}{50}\)

Calculating the Combined Probability

To find the probability of all three events happening in this specific order, we multiply the probabilities of each step:

\(P(\text{King, then Queen, then Jack}) = P(\text{King first}) \times P(\text{Queen second } | \text{ King first}) \times P(\text{Jack third } | \text{ King first and Queen second})\)

\(P = \frac{4}{52} \times \frac{4}{51} \times \frac{4}{50}\)

Simplifying the Probability

Now, we perform the multiplication:

\(P = \frac{4 \times 4 \times 4}{52 \times 51 \times 50}\)

\(P = \frac{64}{132600}\)

We can simplify this fraction by dividing both the numerator and the denominator by their greatest common divisor. Both are divisible by 8.

• \(64 \div 8 = 8\)
• \(132600 \div 8 = 16575\)

So, the simplified probability is:

\(P = \frac{8}{16575}\)

Conclusion

The probability of drawing a King, a Queen, and a Jack in that specific order from a standard 52-card deck without replacement is \(\frac{8}{16575}\).

Revision Table: Probability of Drawing Specific Cards

Let's quickly review the key information for this probability problem:

• Total cards in a standard deck: 52
• Number of Kings, Queens, Jacks: 4 each
• Drawing is without replacement (card count decreases)
• Order matters (King, then Queen, then Jack)

Additional Information: Permutations vs. Combinations in Card Probability

This problem is about permutations because the order of drawing the cards matters (King first, then Queen, then Jack is different from Queen first, then King, then Jack). If the question had asked for the probability of drawing a King, a Queen, and a Jack in any order, we would use combinations or calculate the permutations for all possible orders and sum their probabilities.

• Permutations: Used when the order of selection is important. The number of permutations of selecting \(r\) items from a set of \(n\) items is \(P(n, r) = \frac{n!}{(n-r)!}\). In our case, for the denominators, we were implicitly using this idea for the sequence of remaining cards.
• Combinations: Used when the order of selection does not matter. The number of combinations of selecting \(r\) items from a set of \(n\) items is \(C(n, r) = \frac{n!}{r!(n-r)!}\).

Since the question specified "in order", we correctly used the sequential probability approach, which is equivalent to calculating a specific permutation's probability.