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Sagot :
To find the probability that the coin showed a head given that Jamie scores 6, we can break this problem down using Bayes' theorem. Let's go through the solution step by step.
1. Define Events:
- Let [tex]\( H \)[/tex] be the event that the coin shows a head.
- Let [tex]\( T \)[/tex] be the event that the coin shows a tail.
- Let [tex]\( S_6 \)[/tex] be the event that the score is 6.
2. Given Probabilities:
- The probability that the coin shows a head: [tex]\( P(H) = \frac{1}{3} \)[/tex].
- The probability that the coin shows a tail: [tex]\( P(T) = 1 - P(H) = \frac{2}{3} \)[/tex].
3. Dice Outcomes:
- The dice faces are numbered [tex]\([1, 1, 2, 3, 3, 4]\)[/tex].
4. Calculate Scores:
- If the coin shows a head (event [tex]\( H \)[/tex]), the score is the sum of the two dice.
- If the coin shows a tail (event [tex]\( T \)[/tex]), the score is the product of the two dice.
5. Count Occurrences for Head and Tail:
- Calculate the number of ways to achieve a score of 6 when the coin shows a head and when it shows a tail.
6. Probability of Scoring 6 Given Head:
- There are 6 occurrences where the sum of two dice is 6.
- There are [tex]\( 6 \times 6 = 36 \)[/tex] total possible outcomes for two dice.
- Thus, [tex]\( P(S_6 | H) = \frac{6}{36} = \frac{1}{6} \)[/tex].
7. Probability of Scoring 6 Given Tail:
- There are 4 occurrences where the product of two dice is 6.
- There are [tex]\( 6 \times 6 = 36 \)[/tex] total possible outcomes for two dice.
- Thus, [tex]\( P(S_6 | T) = \frac{4}{36} = \frac{1}{9} \)[/tex].
8. Total Probability of Scoring 6:
- Use the law of total probability:
[tex]\[ P(S_6) = P(S_6 | H) \cdot P(H) + P(S_6 | T) \cdot P(T) \][/tex]
- Substituting the known values:
[tex]\[ P(S_6) = \left( \frac{1}{6} \cdot \frac{1}{3} \right) + \left( \frac{1}{9} \cdot \frac{2}{3} \right) = \frac{1}{18} + \frac{2}{27} = \frac{1}{18} + \frac{2}{27} = \frac{3}{54} + \frac{4}{54} = \frac{7}{54} \approx 0.12963 \][/tex]
9. Bayes' Theorem:
- To find [tex]\( P(H | S_6) \)[/tex], we use Bayes' theorem:
[tex]\[ P(H | S_6) = \frac{P(S_6 | H) \cdot P(H)}{P(S_6)} \][/tex]
- Substituting the values:
[tex]\[ P(H | S_6) = \frac{\left( \frac{1}{6} \right) \cdot \left( \frac{1}{3} \right)}{\frac{7}{54}} = \frac{\frac{1}{18}}{\frac{7}{54}} = \frac{1}{18} \cdot \frac{54}{7} = \frac{3}{7} \approx 0.42857 \][/tex]
Therefore, the probability that the coin showed a head given that Jamie scores 6 is approximately [tex]\(\boxed{0.42857}\)[/tex].
1. Define Events:
- Let [tex]\( H \)[/tex] be the event that the coin shows a head.
- Let [tex]\( T \)[/tex] be the event that the coin shows a tail.
- Let [tex]\( S_6 \)[/tex] be the event that the score is 6.
2. Given Probabilities:
- The probability that the coin shows a head: [tex]\( P(H) = \frac{1}{3} \)[/tex].
- The probability that the coin shows a tail: [tex]\( P(T) = 1 - P(H) = \frac{2}{3} \)[/tex].
3. Dice Outcomes:
- The dice faces are numbered [tex]\([1, 1, 2, 3, 3, 4]\)[/tex].
4. Calculate Scores:
- If the coin shows a head (event [tex]\( H \)[/tex]), the score is the sum of the two dice.
- If the coin shows a tail (event [tex]\( T \)[/tex]), the score is the product of the two dice.
5. Count Occurrences for Head and Tail:
- Calculate the number of ways to achieve a score of 6 when the coin shows a head and when it shows a tail.
6. Probability of Scoring 6 Given Head:
- There are 6 occurrences where the sum of two dice is 6.
- There are [tex]\( 6 \times 6 = 36 \)[/tex] total possible outcomes for two dice.
- Thus, [tex]\( P(S_6 | H) = \frac{6}{36} = \frac{1}{6} \)[/tex].
7. Probability of Scoring 6 Given Tail:
- There are 4 occurrences where the product of two dice is 6.
- There are [tex]\( 6 \times 6 = 36 \)[/tex] total possible outcomes for two dice.
- Thus, [tex]\( P(S_6 | T) = \frac{4}{36} = \frac{1}{9} \)[/tex].
8. Total Probability of Scoring 6:
- Use the law of total probability:
[tex]\[ P(S_6) = P(S_6 | H) \cdot P(H) + P(S_6 | T) \cdot P(T) \][/tex]
- Substituting the known values:
[tex]\[ P(S_6) = \left( \frac{1}{6} \cdot \frac{1}{3} \right) + \left( \frac{1}{9} \cdot \frac{2}{3} \right) = \frac{1}{18} + \frac{2}{27} = \frac{1}{18} + \frac{2}{27} = \frac{3}{54} + \frac{4}{54} = \frac{7}{54} \approx 0.12963 \][/tex]
9. Bayes' Theorem:
- To find [tex]\( P(H | S_6) \)[/tex], we use Bayes' theorem:
[tex]\[ P(H | S_6) = \frac{P(S_6 | H) \cdot P(H)}{P(S_6)} \][/tex]
- Substituting the values:
[tex]\[ P(H | S_6) = \frac{\left( \frac{1}{6} \right) \cdot \left( \frac{1}{3} \right)}{\frac{7}{54}} = \frac{\frac{1}{18}}{\frac{7}{54}} = \frac{1}{18} \cdot \frac{54}{7} = \frac{3}{7} \approx 0.42857 \][/tex]
Therefore, the probability that the coin showed a head given that Jamie scores 6 is approximately [tex]\(\boxed{0.42857}\)[/tex].
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