Answer :
To find the probability of picking a 7 and then picking a 6 from a standard deck of playing cards, where you replace the first card before drawing the second card, we need to consider the process step by step. Please note that a standard deck of cards contains 52 cards, with 4 cards for each rank (Ace through King), which means there are 4 sevens and 4 sixes.
First, we calculate the probability of drawing a 7:
1. There are 4 sevens in the deck, and a total of 52 cards. Therefore, the probability of drawing a 7 on your first draw is:
[tex]\( P(\text{7 first}) = \frac{4}{52} \)[/tex]
Simplify this by dividing both the numerator and the denominator by 4 to get:
[tex]\( P(\text{7 first}) = \frac{1}{13} \)[/tex]
Since we are replacing the seven back into the deck, the composition of the deck returns to its initial state of 52 cards.
Next, we calculate the probability of drawing a 6 after having put the 7 back:
2. Similarly, there are 4 sixes in the deck and the total remains 52 cards. Thus, the probability of drawing a 6 after replacing the 7 is:
[tex]\( P(\text{6 second}) = \frac{4}{52} \)[/tex]
Simplify this to get:
[tex]\( P(\text{6 second}) = \frac{1}{13} \)[/tex]
To find the combined probability of both independent events occurring (drawing a 7 first, and then a 6 after replacing the first card), we multiply the probabilities of the two individual events:
[tex]\( P(\text{7 first and 6 second}) = P(\text{7 first}) \times P(\text{6 second}) \)[/tex]
Substitute the simplified probabilities into the equation:
[tex]\( P(\text{7 first and 6 second}) = \frac{1}{13} \times \frac{1}{13} \)[/tex]
Multiply the fractions:
[tex]\( P(\text{7 first and 6 second}) = \frac{1 \times 1}{13 \times 13} \)[/tex]
[tex]\( P(\text{7 first and 6 second}) = \frac{1}{169} \)[/tex]
The probability of picking a 7 and then picking a 6, with replacement, is [tex]\(\frac{1}{169}\)[/tex].
First, we calculate the probability of drawing a 7:
1. There are 4 sevens in the deck, and a total of 52 cards. Therefore, the probability of drawing a 7 on your first draw is:
[tex]\( P(\text{7 first}) = \frac{4}{52} \)[/tex]
Simplify this by dividing both the numerator and the denominator by 4 to get:
[tex]\( P(\text{7 first}) = \frac{1}{13} \)[/tex]
Since we are replacing the seven back into the deck, the composition of the deck returns to its initial state of 52 cards.
Next, we calculate the probability of drawing a 6 after having put the 7 back:
2. Similarly, there are 4 sixes in the deck and the total remains 52 cards. Thus, the probability of drawing a 6 after replacing the 7 is:
[tex]\( P(\text{6 second}) = \frac{4}{52} \)[/tex]
Simplify this to get:
[tex]\( P(\text{6 second}) = \frac{1}{13} \)[/tex]
To find the combined probability of both independent events occurring (drawing a 7 first, and then a 6 after replacing the first card), we multiply the probabilities of the two individual events:
[tex]\( P(\text{7 first and 6 second}) = P(\text{7 first}) \times P(\text{6 second}) \)[/tex]
Substitute the simplified probabilities into the equation:
[tex]\( P(\text{7 first and 6 second}) = \frac{1}{13} \times \frac{1}{13} \)[/tex]
Multiply the fractions:
[tex]\( P(\text{7 first and 6 second}) = \frac{1 \times 1}{13 \times 13} \)[/tex]
[tex]\( P(\text{7 first and 6 second}) = \frac{1}{169} \)[/tex]
The probability of picking a 7 and then picking a 6, with replacement, is [tex]\(\frac{1}{169}\)[/tex].
