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Sagot :
Let's walk through the steps to solve both parts of the question about drawing two jacks from a standard deck of 52 cards.
### Part (a): Drawing with Replacement
In this scenario, each time a card is drawn, it is placed back into the deck, and the deck is then reshuffled. This means that the probability of drawing a jack remains the same for both draws.
1. Determine the probability of drawing a jack in the first draw:
- There are 4 jacks in a standard deck of 52 cards.
- Therefore, the probability of drawing a jack is [tex]\( \frac{4}{52} \)[/tex].
2. Determine the probability of drawing a jack in the second draw after replacement:
- Because the card is replaced and the deck is reshuffled, the probability remains [tex]\( \frac{4}{52} \)[/tex].
3. Calculate the combined probability:
- Since these are independent events (due to replacement), the combined probability is the product of the individual probabilities:
[tex]\[ P(\text{Both Jacks}) = \left( \frac{4}{52} \right) \times \left( \frac{4}{52} \right) = \left( \frac{1}{13} \right) \times \left( \frac{1}{13} \right) = \frac{1}{169} \][/tex]
4. Convert to a decimal and round to the nearest ten-thousandth:
- [tex]\[ \frac{1}{169} \approx 0.0059 \][/tex]
So, the probability that both cards drawn are jacks with replacement is 0.0059.
### Part (b): Drawing without Replacement
In this scenario, once a card is drawn, it is not placed back into the deck, which affects the probabilities for the subsequent draws.
1. Determine the probability of drawing a jack in the first draw:
- There are still 4 jacks in a deck of 52 cards.
- Therefore, the probability of drawing a jack is [tex]\( \frac{4}{52} \)[/tex].
2. Determine the probability of drawing a jack in the second draw without replacement:
- After one jack is drawn, only 3 jacks are left in a reduced deck of 51 cards.
- So, the probability of drawing a jack in the second draw is [tex]\( \frac{3}{51} \)[/tex].
3. Calculate the combined probability:
- Since these are dependent events (due to no replacement), the combined probability is the product of the individual probabilities:
[tex]\[ P(\text{Both Jacks}) = \left( \frac{4}{52} \right) \times \left( \frac{3}{51} \right) = \left( \frac{1}{13} \right) \times \left( \frac{1}{17} \right) = \frac{1}{221} \][/tex]
4. Convert to a decimal and round to the nearest ten-thousandth:
- [tex]\[ \frac{1}{221} \approx 0.0045 \][/tex]
So, the probability that both cards drawn are jacks without replacement is 0.0045.
In summary:
- (a) The probability that both cards are jacks with replacement is 0.0059.
- (b) The probability that both cards are jacks without replacement is 0.0045.
### Part (a): Drawing with Replacement
In this scenario, each time a card is drawn, it is placed back into the deck, and the deck is then reshuffled. This means that the probability of drawing a jack remains the same for both draws.
1. Determine the probability of drawing a jack in the first draw:
- There are 4 jacks in a standard deck of 52 cards.
- Therefore, the probability of drawing a jack is [tex]\( \frac{4}{52} \)[/tex].
2. Determine the probability of drawing a jack in the second draw after replacement:
- Because the card is replaced and the deck is reshuffled, the probability remains [tex]\( \frac{4}{52} \)[/tex].
3. Calculate the combined probability:
- Since these are independent events (due to replacement), the combined probability is the product of the individual probabilities:
[tex]\[ P(\text{Both Jacks}) = \left( \frac{4}{52} \right) \times \left( \frac{4}{52} \right) = \left( \frac{1}{13} \right) \times \left( \frac{1}{13} \right) = \frac{1}{169} \][/tex]
4. Convert to a decimal and round to the nearest ten-thousandth:
- [tex]\[ \frac{1}{169} \approx 0.0059 \][/tex]
So, the probability that both cards drawn are jacks with replacement is 0.0059.
### Part (b): Drawing without Replacement
In this scenario, once a card is drawn, it is not placed back into the deck, which affects the probabilities for the subsequent draws.
1. Determine the probability of drawing a jack in the first draw:
- There are still 4 jacks in a deck of 52 cards.
- Therefore, the probability of drawing a jack is [tex]\( \frac{4}{52} \)[/tex].
2. Determine the probability of drawing a jack in the second draw without replacement:
- After one jack is drawn, only 3 jacks are left in a reduced deck of 51 cards.
- So, the probability of drawing a jack in the second draw is [tex]\( \frac{3}{51} \)[/tex].
3. Calculate the combined probability:
- Since these are dependent events (due to no replacement), the combined probability is the product of the individual probabilities:
[tex]\[ P(\text{Both Jacks}) = \left( \frac{4}{52} \right) \times \left( \frac{3}{51} \right) = \left( \frac{1}{13} \right) \times \left( \frac{1}{17} \right) = \frac{1}{221} \][/tex]
4. Convert to a decimal and round to the nearest ten-thousandth:
- [tex]\[ \frac{1}{221} \approx 0.0045 \][/tex]
So, the probability that both cards drawn are jacks without replacement is 0.0045.
In summary:
- (a) The probability that both cards are jacks with replacement is 0.0059.
- (b) The probability that both cards are jacks without replacement is 0.0045.
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