Answer :
To solve the problem of finding the probability of drawing a four the first time and a diamond the second time from a standard 52-card deck with replacement, let's break it down step by step.
### Step 1: Find the probability of drawing a four the first time.
1. The total number of cards in a standard deck is 52.
2. There are four cards that are fours (one for each suit: hearts, diamonds, clubs, and spades).
The probability of drawing a four is therefore:
[tex]\[ P(\text{Drawing a four}) = \frac{\text{Number of fours}}{\text{Total number of cards}} = \frac{4}{52} = \frac{1}{13} \][/tex]
### Step 2: Find the probability of drawing a diamond the second time.
1. After replacing the first card and shuffling the deck, the total number of cards remains 52.
2. There are 13 diamonds in a deck (one for each rank: Ace, 2, 3, ..., 10, Jack, Queen, King).
The probability of drawing a diamond is:
[tex]\[ P(\text{Drawing a diamond}) = \frac{\text{Number of diamonds}}{\text{Total number of cards}} = \frac{13}{52} = \frac{1}{4} \][/tex]
### Step 3: Find the joint probability of both events happening.
Since both events (drawing a four the first time and drawing a diamond the second time) are independent, we can find the joint probability by multiplying their individual probabilities.
[tex]\[ P(\text{Drawing a four first AND a diamond second}) = P(\text{Drawing a four first}) \times P(\text{Drawing a diamond second}) \][/tex]
[tex]\[ = \left(\frac{1}{13}\right) \times \left(\frac{1}{4}\right) = \frac{1}{52} \][/tex]
Thus, the probability of drawing a four the first time and a diamond the second time is:
[tex]\[ \frac{1}{52} \][/tex]
This fraction is already in its simplest form. So the final answer is:
[tex]\[ \frac{1}{52} \][/tex]
### Step 1: Find the probability of drawing a four the first time.
1. The total number of cards in a standard deck is 52.
2. There are four cards that are fours (one for each suit: hearts, diamonds, clubs, and spades).
The probability of drawing a four is therefore:
[tex]\[ P(\text{Drawing a four}) = \frac{\text{Number of fours}}{\text{Total number of cards}} = \frac{4}{52} = \frac{1}{13} \][/tex]
### Step 2: Find the probability of drawing a diamond the second time.
1. After replacing the first card and shuffling the deck, the total number of cards remains 52.
2. There are 13 diamonds in a deck (one for each rank: Ace, 2, 3, ..., 10, Jack, Queen, King).
The probability of drawing a diamond is:
[tex]\[ P(\text{Drawing a diamond}) = \frac{\text{Number of diamonds}}{\text{Total number of cards}} = \frac{13}{52} = \frac{1}{4} \][/tex]
### Step 3: Find the joint probability of both events happening.
Since both events (drawing a four the first time and drawing a diamond the second time) are independent, we can find the joint probability by multiplying their individual probabilities.
[tex]\[ P(\text{Drawing a four first AND a diamond second}) = P(\text{Drawing a four first}) \times P(\text{Drawing a diamond second}) \][/tex]
[tex]\[ = \left(\frac{1}{13}\right) \times \left(\frac{1}{4}\right) = \frac{1}{52} \][/tex]
Thus, the probability of drawing a four the first time and a diamond the second time is:
[tex]\[ \frac{1}{52} \][/tex]
This fraction is already in its simplest form. So the final answer is:
[tex]\[ \frac{1}{52} \][/tex]
