If a box contains 19 unbiased coins and 1 biased coin with both faces heads, and a coin is randomly chosen from this box and tossed. If the head appears, then the probability that the unbiased coin was selected is:
Show Hint
- \( \frac{19}{21} \)
- \( \frac{1}{3} \)
- \( \frac{1}{5} \)
- \( \frac{1}{6} \)
The Correct Option is A
Solution and Explanation
We are given 19 unbiased coins and 1 biased coin. The unbiased coin has a probability of \( \frac{1}{2} \) of showing heads, while the biased coin always shows heads. We are asked to find the probability that an unbiased coin was selected given that the head appeared on the toss. Let: - \( U \) be the event that an unbiased coin was selected. - \( B \) be the event that the biased coin was selected. - \( H \) be the event that a head is tossed. We are looking for \( P(U|H) \), the probability that an unbiased coin was selected given that the head appeared. By Bayes' theorem: \[ P(U|H) = \frac{P(H|U) \cdot P(U)}{P(H)}. \] ### Step 1: Calculate \( P(H|U) \) The probability of getting heads given that an unbiased coin was selected is \( P(H|U) = \frac{1}{2} \). ### Step 2: Calculate \( P(U) \) The probability of selecting an unbiased coin is \( P(U) = \frac{19}{20} \). ### Step 3: Calculate \( P(H|B) \) The probability of getting heads given that the biased coin was selected is \( P(H|B) = 1 \). ### Step 4: Calculate \( P(B) \) The probability of selecting the biased coin is \( P(B) = \frac{1}{20} \). ### Step 5: Calculate \( P(H) \) The total probability of getting heads is: \[ P(H) = P(H|U) \cdot P(U) + P(H|B) \cdot P(B) \] \[ P(H) = \frac{1}{2} \cdot \frac{19}{20} + 1 \cdot \frac{1}{20} = \frac{19}{40} + \frac{1}{20} = \frac{19 + 2}{40} = \frac{21}{40}. \] ### Step 6: Apply Bayes' theorem Now, applying Bayes' theorem: \[ P(U|H) = \frac{\frac{1}{2} \cdot \frac{19}{20}}{\frac{21}{40}} = \frac{\frac{19}{40}}{\frac{21}{40}} = \frac{19}{21}. \] Thus, the correct answer is \( \frac{19}{21} \).
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