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Let two cards be dealt successively, without replacement, from a standard 52-card deck. Find the probability of the event.

The first card is a ten, and the second is a face card.

The probability that the first card is a ten and the second is a face card is:
(Simplify your answer. Type an integer or a fraction.)


Sagot :

To determine the probability that the first card drawn from a deck of 52 cards is a ten and the second card drawn is a face card, we need to follow a few steps. Let's break it down:

1. Total Number of Cards in a Deck: A standard deck of cards contains 52 cards.

2. Number of Tens: There are 4 tens in the deck (one for each suit: hearts, diamonds, clubs, and spades).

3. Number of Face Cards: There are 12 face cards in the deck (Jack, Queen, and King of each suit, which makes 3 cards per suit multiplied by 4 suits).

4. First Card Being a Ten:
- Since there are 4 tens out of 52 cards, the probability of drawing a ten first is:
[tex]\[ \text{Probability of first card being a ten} = \frac{4}{52} = \frac{1}{13} \][/tex]

5. Second Card Being a Face Card (given the first was a ten):
- Once a ten is drawn, we are left with 51 cards.
- The number of face cards remains 12 since none have been removed yet.
- The probability of drawing a face card next is:
[tex]\[ \text{Probability of second card being a face card} = \frac{12}{51} \][/tex]

6. Combined Probability:
- To find the overall probability of both events happening in sequence (first drawing a ten and then a face card), we multiply the individual probabilities of these two independent events:
[tex]\[ \text{Combined Probability} = \left( \frac{1}{13} \right) \times \left( \frac{12}{51} \right) \][/tex]

Simplifying the fraction:
[tex]\[ \frac{1}{13} \times \frac{12}{51} = \frac{1 \cdot 12}{13 \cdot 51} = \frac{12}{663} = \frac{4}{221} \][/tex]

Therefore, the probability that the first card is a ten and the second is a face card is:
[tex]\[ \boxed{\frac{4}{221}} \][/tex]