Answered

Discover a wealth of knowledge at Westonci.ca, where experts provide answers to your most pressing questions. Join our platform to get reliable answers to your questions from a knowledgeable community of experts. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.

If two cards are drawn without replacement from an ordinary deck, find the probability of an ace and a 5 being drawn.

The probability of an ace and a 5 being drawn is (Type an integer or a simplified fraction).

Sagot :

GinnyAnswer
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]
Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.