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Sagot :
To determine the probability that Hiro pulls out a 3 first and then pulls out a 2 without replacement from the stack of cards, let's follow through the problem step-by-step.
1. Count the Total Number of Cards:
Hiro has a set of cards with the numbers [tex]\(1, 1, 2, 2, 3, 3, 3, 4\)[/tex]. This makes a total of 8 cards.
2. Count the Frequency of Each Number:
- There are 2 cards with the number 1.
- There are 2 cards with the number 2.
- There are 3 cards with the number 3.
- There is 1 card with the number 4.
3. Probability of Drawing a 3 First:
To find the probability of drawing a 3 first, we see that there are 3 cards with a 3 out of the total 8 cards. So, the probability is:
[tex]\[ P(\text{First draw is a 3}) = \frac{3}{8} \][/tex]
4. Update the Total Number of Cards After Drawing a 3:
After one card is drawn (which is a 3), the total number of remaining cards is 7. Out of these, the number of 3's is reduced by 1, so there are now 2 cards with a 3 left. The number of 2's remains unchanged, which is 2.
5. Probability of Drawing a 2 Second:
After having drawn a 3 first, the probability of drawing a 2 from the remaining 7 cards is calculated using the remaining cards:
[tex]\[ P(\text{Second draw is a 2}) = \frac{2}{7} \][/tex]
6. Combined Probability for Both Events:
To find the combined probability of both events happening consecutively (drawing a 3 first and then a 2), we multiply the probabilities of the two independent events:
[tex]\[ P(\text{3 first and then 2}) = P(\text{First draw is a 3}) \times P(\text{Second draw is a 2}) \][/tex]
[tex]\[ P(\text{3 first and then 2}) = \frac{3}{8} \times \frac{2}{7} = \frac{6}{56} \][/tex]
7. Simplify the Fraction:
Simplify the fraction [tex]\(\frac{6}{56}\)[/tex]:
[tex]\[ \frac{6}{56} = \frac{3}{28} \][/tex]
Thus, the probability that Hiro pulls out a 3 first and then pulls out a 2 without replacement is:
[tex]\[ \boxed{\frac{3}{28}} \][/tex]
Among the given options, the correct answer is [tex]\(\frac{3}{28}\)[/tex].
1. Count the Total Number of Cards:
Hiro has a set of cards with the numbers [tex]\(1, 1, 2, 2, 3, 3, 3, 4\)[/tex]. This makes a total of 8 cards.
2. Count the Frequency of Each Number:
- There are 2 cards with the number 1.
- There are 2 cards with the number 2.
- There are 3 cards with the number 3.
- There is 1 card with the number 4.
3. Probability of Drawing a 3 First:
To find the probability of drawing a 3 first, we see that there are 3 cards with a 3 out of the total 8 cards. So, the probability is:
[tex]\[ P(\text{First draw is a 3}) = \frac{3}{8} \][/tex]
4. Update the Total Number of Cards After Drawing a 3:
After one card is drawn (which is a 3), the total number of remaining cards is 7. Out of these, the number of 3's is reduced by 1, so there are now 2 cards with a 3 left. The number of 2's remains unchanged, which is 2.
5. Probability of Drawing a 2 Second:
After having drawn a 3 first, the probability of drawing a 2 from the remaining 7 cards is calculated using the remaining cards:
[tex]\[ P(\text{Second draw is a 2}) = \frac{2}{7} \][/tex]
6. Combined Probability for Both Events:
To find the combined probability of both events happening consecutively (drawing a 3 first and then a 2), we multiply the probabilities of the two independent events:
[tex]\[ P(\text{3 first and then 2}) = P(\text{First draw is a 3}) \times P(\text{Second draw is a 2}) \][/tex]
[tex]\[ P(\text{3 first and then 2}) = \frac{3}{8} \times \frac{2}{7} = \frac{6}{56} \][/tex]
7. Simplify the Fraction:
Simplify the fraction [tex]\(\frac{6}{56}\)[/tex]:
[tex]\[ \frac{6}{56} = \frac{3}{28} \][/tex]
Thus, the probability that Hiro pulls out a 3 first and then pulls out a 2 without replacement is:
[tex]\[ \boxed{\frac{3}{28}} \][/tex]
Among the given options, the correct answer is [tex]\(\frac{3}{28}\)[/tex].
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