Answer :
To find the probability of drawing two Kings in a row without replacement from a standard deck of 52 cards, we need to go through the problem step-by-step. Here is the detailed solution:
1. Total Number of Cards in a Deck: There are 52 cards in a standard deck.
2. Total Number of Kings in a Deck: There are 4 Kings in a standard deck.
3. Probability of Drawing the First King:
- When you draw the first card, there are 52 cards available.
- The probability that the first card drawn is a King is the number of Kings divided by the total number of cards.
- So, the probability of drawing the first King is [tex]\( \frac{4}{52} \)[/tex].
4. Probability of Drawing the Second King:
- After drawing the first King, you have 51 cards left in the deck.
- There are now 3 Kings remaining.
- The probability of drawing the second King is the number of remaining Kings divided by the remaining number of cards.
- So, the probability of drawing the second King after having already drawn a King is [tex]\( \frac{3}{51} \)[/tex].
5. Overall Probability:
- The overall probability of both events occurring (drawing the first King and then the second King) is the product of the two probabilities:
- [tex]\( \text{Overall Probability} = \left(\frac{4}{52}\right) \times \left(\frac{3}{51}\right) \)[/tex].
6. Calculating the Overall Probability:
- Calculating [tex]\( \frac{4}{52} \)[/tex]:
[tex]\[ \frac{4}{52} = 0.07692307692307693 \][/tex]
- Calculating [tex]\( \frac{3}{51} \)[/tex]:
[tex]\[ \frac{3}{51} = 0.058823529411764705 \][/tex]
- Multiplying both probabilities:
[tex]\[ 0.07692307692307693 \times 0.058823529411764705 = 0.004524886877828055 \][/tex]
7. Converting to Percentage:
- To express this probability as a percentage, multiply by 100:
[tex]\[ 0.004524886877828055 \times 100 = 0.4524886877828055 \% \][/tex]
So, the probability of drawing two Kings in a row without replacement is approximately [tex]\( 0.4525\% \)[/tex].
Considering the provided choices, the correct answer is not explicitly listed among them, but the exact detailed calculation aligns closely with:
[tex]\[ 0.004524886877828055 \text{ (which is approximately 0.4525%)} \][/tex]
1. Total Number of Cards in a Deck: There are 52 cards in a standard deck.
2. Total Number of Kings in a Deck: There are 4 Kings in a standard deck.
3. Probability of Drawing the First King:
- When you draw the first card, there are 52 cards available.
- The probability that the first card drawn is a King is the number of Kings divided by the total number of cards.
- So, the probability of drawing the first King is [tex]\( \frac{4}{52} \)[/tex].
4. Probability of Drawing the Second King:
- After drawing the first King, you have 51 cards left in the deck.
- There are now 3 Kings remaining.
- The probability of drawing the second King is the number of remaining Kings divided by the remaining number of cards.
- So, the probability of drawing the second King after having already drawn a King is [tex]\( \frac{3}{51} \)[/tex].
5. Overall Probability:
- The overall probability of both events occurring (drawing the first King and then the second King) is the product of the two probabilities:
- [tex]\( \text{Overall Probability} = \left(\frac{4}{52}\right) \times \left(\frac{3}{51}\right) \)[/tex].
6. Calculating the Overall Probability:
- Calculating [tex]\( \frac{4}{52} \)[/tex]:
[tex]\[ \frac{4}{52} = 0.07692307692307693 \][/tex]
- Calculating [tex]\( \frac{3}{51} \)[/tex]:
[tex]\[ \frac{3}{51} = 0.058823529411764705 \][/tex]
- Multiplying both probabilities:
[tex]\[ 0.07692307692307693 \times 0.058823529411764705 = 0.004524886877828055 \][/tex]
7. Converting to Percentage:
- To express this probability as a percentage, multiply by 100:
[tex]\[ 0.004524886877828055 \times 100 = 0.4524886877828055 \% \][/tex]
So, the probability of drawing two Kings in a row without replacement is approximately [tex]\( 0.4525\% \)[/tex].
Considering the provided choices, the correct answer is not explicitly listed among them, but the exact detailed calculation aligns closely with:
[tex]\[ 0.004524886877828055 \text{ (which is approximately 0.4525%)} \][/tex]