Answer :
To find the probability of drawing an ace and a 5 from an ordinary deck of 52 cards without replacement, we need to consider two scenarios:
1. Drawing an ace first and a 5 second.
2. Drawing a 5 first and an ace second.
Here is a detailed step-by-step solution:
### Case 1: Drawing an Ace First and a 5 Second
1. The total number of cards in the deck is 52.
2. There are 4 aces in the deck.
3. The probability of drawing an ace first is [tex]\(\frac{4}{52} = \frac{1}{13} \approx 0.07692307692307693 \)[/tex].
After drawing the ace:
4. There are now 51 cards left in the deck.
5. There are still 4 fives in the deck.
6. The probability of drawing a 5 second is [tex]\(\frac{4}{51} \approx 0.0784313725490196 \)[/tex].
The combined probability of drawing an ace first and a 5 second is:
[tex]\[ \left( \frac{1}{13} \right) \times \left( \frac{4}{51} \right) = \ 0.07692307692307693 \times 0.0784313725490196 \approx 0.006033182503770739 \][/tex]
### Case 2: Drawing a 5 First and an Ace Second
1. The total number of cards in the deck is 52.
2. There are 4 fives in the deck.
3. The probability of drawing a 5 first is [tex]\(\frac{4}{52} = \frac{1}{13} \approx 0.07692307692307693 \)[/tex].
After drawing the 5:
4. There are now 51 cards left in the deck.
5. There are still 4 aces in the deck.
6. The probability of drawing an ace second is [tex]\(\frac{4}{51} \approx 0.0784313725490196 \)[/tex].
The combined probability of drawing a 5 first and an ace second is:
[tex]\[ \left( \frac{1}{13} \right) \times \left( \frac{4}{51} \right) = 0.07692307692307693 \times 0.0784313725490196 \approx 0.006033182503770739 \][/tex]
### Total Probability
Since these two scenarios are mutually exclusive, we add the probabilities of both scenarios:
[tex]\[ 0.006033182503770739 + 0.006033182503770739 = 0.012066365007541479 \][/tex]
So, the probability of drawing an ace and a 5 in any order from an ordinary deck of 52 cards without replacement is approximately:
[tex]\[ 0.012066365007541479 \][/tex]
In fraction form:
[tex]\[ \boxed{\frac{12}{997}} \][/tex]
1. Drawing an ace first and a 5 second.
2. Drawing a 5 first and an ace second.
Here is a detailed step-by-step solution:
### Case 1: Drawing an Ace First and a 5 Second
1. The total number of cards in the deck is 52.
2. There are 4 aces in the deck.
3. The probability of drawing an ace first is [tex]\(\frac{4}{52} = \frac{1}{13} \approx 0.07692307692307693 \)[/tex].
After drawing the ace:
4. There are now 51 cards left in the deck.
5. There are still 4 fives in the deck.
6. The probability of drawing a 5 second is [tex]\(\frac{4}{51} \approx 0.0784313725490196 \)[/tex].
The combined probability of drawing an ace first and a 5 second is:
[tex]\[ \left( \frac{1}{13} \right) \times \left( \frac{4}{51} \right) = \ 0.07692307692307693 \times 0.0784313725490196 \approx 0.006033182503770739 \][/tex]
### Case 2: Drawing a 5 First and an Ace Second
1. The total number of cards in the deck is 52.
2. There are 4 fives in the deck.
3. The probability of drawing a 5 first is [tex]\(\frac{4}{52} = \frac{1}{13} \approx 0.07692307692307693 \)[/tex].
After drawing the 5:
4. There are now 51 cards left in the deck.
5. There are still 4 aces in the deck.
6. The probability of drawing an ace second is [tex]\(\frac{4}{51} \approx 0.0784313725490196 \)[/tex].
The combined probability of drawing a 5 first and an ace second is:
[tex]\[ \left( \frac{1}{13} \right) \times \left( \frac{4}{51} \right) = 0.07692307692307693 \times 0.0784313725490196 \approx 0.006033182503770739 \][/tex]
### Total Probability
Since these two scenarios are mutually exclusive, we add the probabilities of both scenarios:
[tex]\[ 0.006033182503770739 + 0.006033182503770739 = 0.012066365007541479 \][/tex]
So, the probability of drawing an ace and a 5 in any order from an ordinary deck of 52 cards without replacement is approximately:
[tex]\[ 0.012066365007541479 \][/tex]
In fraction form:
[tex]\[ \boxed{\frac{12}{997}} \][/tex]
