Question

A box contains 3 coins:

Coin A: Fair coin (head probability) = $\frac{1}{2}$.

Coin B: Biased coin with head probability = $\frac{3}{4}$.

Coin C: Biased coin with head probability = $\frac{1}{4}$.

A coin is chosen at random and tossed twice. It is observed that both tosses are heads. Then the probability that the chosen coin was coin B is equal to

• A. $\frac{9}{14}$
• B. $\frac{1}{3}$
• C. $\frac{7}{24}$
• D. $\frac{9}{16}$

Answer 1

Answer: (A)

$\frac{9}{14}$

Answer 2

1. Probabilities of choosing coins: Since a coin is chosen at random, the prior probabilities are equal: $P(C_A) = P(C_B) = P(C_C) = \frac{1}{3}$.
2. Probabilities of observing two heads ($HH$) given the chosen coin:
   • Coin A (Fair): $P(HH | C_A) = (\frac{1}{2})^2 = \frac{1}{4}$.
   • Coin B (Biased): $P(HH | C_B) = (\frac{3}{4})^2 = \frac{9}{16}$.
   • Coin C (Biased): $P(HH | C_C) = (\frac{1}{4})^2 = \frac{1}{16}$.
3. Apply Bayes' Theorem: We want to find the posterior probability $P(C_B | HH)$. Bayes' Theorem states: $P(C_B | HH) = \frac{P(HH | C_B) P(C_B)}{P(HH)}$ The probability of observing two heads, $P(HH)$, is calculated using the law of total probability: $P(HH) = P(HH | C_A)P(C_A) + P(HH | C_B)P(C_B) + P(HH | C_C)P(C_C)$
4. Substitute values and calculate: $P(HH) = \left(\frac{1}{4}\right)\left(\frac{1}{3}\right) + \left(\frac{9}{16}\right)\left(\frac{1}{3}\right) + \left(\frac{1}{16}\right)\left(\frac{1}{3}\right)$ $P(HH) = \frac{1}{3} \left(\frac{1}{4} + \frac{9}{16} + \frac{1}{16}\right) = \frac{1}{3} \left(\frac{4}{16} + \frac{9}{16} + \frac{1}{16}\right) = \frac{1}{3} \left(\frac{14}{16}\right) = \frac{1}{3} \left(\frac{7}{8}\right) = \frac{7}{24}$ Now, substitute this into Bayes' Theorem: $P(C_B | HH) = \frac{\left(\frac{9}{16}\right)\left(\frac{1}{3}\right)}{\frac{7}{24}} = \frac{\frac{9}{48}}{\frac{7}{24}} = \frac{9}{48} \times \frac{24}{7} = \frac{9}{2 \times 7} = \frac{9}{14}$