Question

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:

• A. \( \frac{19}{21} \)
• B. \( \frac{1}{3} \)
• C. \( \frac{1}{5} \)
• D. \( \frac{1}{6} \)

Hint:

When solving probability questions using Bayes' theorem, remember to first calculate the likelihood of the event, the prior probabilities, and the total probability of the event.

Answer 1

The Correct Option is A

Given 19 unbiased coins (each with a \( \frac{1}{2} \) probability of heads) and 1 biased coin (always heads), we aim to determine the probability that an unbiased coin was chosen, given that a head appeared. Let:- \( U \) denote the selection of an unbiased coin.- \( B \) denote the selection of the biased coin.- \( H \) denote the outcome of tossing a head.We seek \( P(U|H) \), the probability of having selected an unbiased coin given a head. Bayes' theorem states:\[P(U|H) = \frac{P(H|U) \cdot P(U)}{P(H)}.\]Step 1: \( P(H|U) \)The probability of obtaining a head from an unbiased coin is \( P(H|U) = \frac{1}{2} \).Step 2: \( P(U) \)The probability of selecting an unbiased coin is \( P(U) = \frac{19}{20} \).Step 3: \( P(H|B) \)The probability of obtaining a head from the biased coin is \( P(H|B) = 1 \).Step 4: \( P(B) \)The probability of selecting the biased coin is \( P(B) = \frac{1}{20} \).Step 5: \( P(H) \)The total probability of tossing a head is calculated as:\[P(H) = P(H|U) \cdot P(U) + P(H|B) \cdot P(B)\]\[P(H) = \frac{1}{2} \cdot \frac{19}{20} + 1 \cdot \frac{1}{20} = \frac{19}{40} + \frac{1}{20} = \frac{19 + 2}{40} = \frac{21}{40}.\]Step 6: Bayes' Theorem ApplicationApplying Bayes' theorem yields:\[P(U|H) = \frac{\frac{1}{2} \cdot \frac{19}{20}}{\frac{21}{40}} = \frac{\frac{19}{40}}{\frac{21}{40}} = \frac{19}{21}.\]The probability that an unbiased coin was selected, given that a head appeared, is \( \frac{19}{21} \).