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Sagot :
To determine the probability that Jude takes exactly one red counter when he picks 3 counters from a bag containing 3 red and 5 yellow counters, we need to consider the different orders in which he can pick the red counter. We will break this down into the three distinct cases and then sum the probabilities.
### Step-by-Step Solution:
1. Total Counters in the Bag:
- Red counters: \(3\)
- Yellow counters: \(5\)
- Total counters: \(3 + 5 = 8\)
2. Case 1: Red Counter First, Followed by Two Yellow Counters
- The probability of picking a red counter first:
[tex]\[ \frac{3}{8} \][/tex]
- After removing one red counter, the remaining counters:
- \(2\) red
- \(5\) yellow
\(\implies 7\) total counters
- The probability of picking a yellow counter next:
[tex]\[ \frac{5}{7} \][/tex]
- After removing one yellow counter, the remaining counters:
- \(2\) red
- \(4\) yellow
\(\implies 6\) total counters
- The probability of picking another yellow counter:
[tex]\[ \frac{4}{6} \][/tex]
- Thus, the total probability for this case is:
[tex]\[ \frac{3}{8} \times \frac{5}{7} \times \frac{4}{6} = \frac{60}{336} \][/tex]
3. Case 2: Red Counter Second, Surrounded by Two Yellow Counters
- The probability of picking a yellow counter first:
[tex]\[ \frac{5}{8} \][/tex]
- After removing one yellow counter, the remaining counters:
- \(3\) red
- \(4\) yellow
\(\implies 7\) total counters
- The probability of picking a red counter next:
[tex]\[ \frac{3}{7} \][/tex]
- After removing one red counter, the remaining counters:
- \(2\) red
- \(4\) yellow
\(\implies 6\) total counters
- The probability of picking a yellow counter:
[tex]\[ \frac{4}{6} \][/tex]
- Thus, the total probability for this case is:
[tex]\[ \frac{5}{8} \times \frac{3}{7} \times \frac{4}{6} = \frac{60}{336} \][/tex]
4. Case 3: Red Counter Third, Preceded by Two Yellow Counters
- The probability of picking a yellow counter first:
[tex]\[ \frac{5}{8} \][/tex]
- After removing one yellow counter, the remaining counters:
- \(3\) red
- \(4\) yellow
\(\implies 7\) total counters
- The probability of picking another yellow counter:
[tex]\[ \frac{4}{7} \][/tex]
- After removing one yellow counter, the remaining counters:
- \(3\) red
- \(3\) yellow
\(\implies 6\) total counters
- The probability of picking a red counter:
[tex]\[ \frac{3}{6} \][/tex]
- Thus, the total probability for this case is:
[tex]\[ \frac{5}{8} \times \frac{4}{7} \times \frac{3}{6} = \frac{60}{336} \][/tex]
5. Total Probability of Exactly One Red Counter
- Summing the probabilities of the three cases:
[tex]\[ \frac{60}{336} + \frac{60}{336} + \frac{60}{336} = \frac{180}{336} \][/tex]
Simplifying this result:
[tex]\[ \frac{180}{336} = \frac{15}{28} \][/tex]
- Converting to a decimal (optional for better understanding):
[tex]\[ \frac{15}{28} \approx 0.5357 \][/tex]
Thus, the probability that Jude will pick exactly one red counter when he picks 3 counters from the bag is [tex]\(\frac{15}{28}\)[/tex] or approximately [tex]\(0.5357\)[/tex], which can also be expressed as about 53.57%.
### Step-by-Step Solution:
1. Total Counters in the Bag:
- Red counters: \(3\)
- Yellow counters: \(5\)
- Total counters: \(3 + 5 = 8\)
2. Case 1: Red Counter First, Followed by Two Yellow Counters
- The probability of picking a red counter first:
[tex]\[ \frac{3}{8} \][/tex]
- After removing one red counter, the remaining counters:
- \(2\) red
- \(5\) yellow
\(\implies 7\) total counters
- The probability of picking a yellow counter next:
[tex]\[ \frac{5}{7} \][/tex]
- After removing one yellow counter, the remaining counters:
- \(2\) red
- \(4\) yellow
\(\implies 6\) total counters
- The probability of picking another yellow counter:
[tex]\[ \frac{4}{6} \][/tex]
- Thus, the total probability for this case is:
[tex]\[ \frac{3}{8} \times \frac{5}{7} \times \frac{4}{6} = \frac{60}{336} \][/tex]
3. Case 2: Red Counter Second, Surrounded by Two Yellow Counters
- The probability of picking a yellow counter first:
[tex]\[ \frac{5}{8} \][/tex]
- After removing one yellow counter, the remaining counters:
- \(3\) red
- \(4\) yellow
\(\implies 7\) total counters
- The probability of picking a red counter next:
[tex]\[ \frac{3}{7} \][/tex]
- After removing one red counter, the remaining counters:
- \(2\) red
- \(4\) yellow
\(\implies 6\) total counters
- The probability of picking a yellow counter:
[tex]\[ \frac{4}{6} \][/tex]
- Thus, the total probability for this case is:
[tex]\[ \frac{5}{8} \times \frac{3}{7} \times \frac{4}{6} = \frac{60}{336} \][/tex]
4. Case 3: Red Counter Third, Preceded by Two Yellow Counters
- The probability of picking a yellow counter first:
[tex]\[ \frac{5}{8} \][/tex]
- After removing one yellow counter, the remaining counters:
- \(3\) red
- \(4\) yellow
\(\implies 7\) total counters
- The probability of picking another yellow counter:
[tex]\[ \frac{4}{7} \][/tex]
- After removing one yellow counter, the remaining counters:
- \(3\) red
- \(3\) yellow
\(\implies 6\) total counters
- The probability of picking a red counter:
[tex]\[ \frac{3}{6} \][/tex]
- Thus, the total probability for this case is:
[tex]\[ \frac{5}{8} \times \frac{4}{7} \times \frac{3}{6} = \frac{60}{336} \][/tex]
5. Total Probability of Exactly One Red Counter
- Summing the probabilities of the three cases:
[tex]\[ \frac{60}{336} + \frac{60}{336} + \frac{60}{336} = \frac{180}{336} \][/tex]
Simplifying this result:
[tex]\[ \frac{180}{336} = \frac{15}{28} \][/tex]
- Converting to a decimal (optional for better understanding):
[tex]\[ \frac{15}{28} \approx 0.5357 \][/tex]
Thus, the probability that Jude will pick exactly one red counter when he picks 3 counters from the bag is [tex]\(\frac{15}{28}\)[/tex] or approximately [tex]\(0.5357\)[/tex], which can also be expressed as about 53.57%.
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