# Problem Restatement A box contains: - **3 fair coins** (each gives Heads/Tails with probability \( \frac{1}{2} \)) - **2 double-headed coins** (always give Heads) A coin is **randomly selected** and **tossed 4 times**. All 4 tosses are **Heads**. **Question:** What is the probability the chosen coin is double-headed? --- ## Step 1: Define Events Let: - \( D \): Coin is **Double-Headed** - \( F \): Coin is **Fair** - \( H \): All 4 tosses result in Heads We want \( P(D|H) \): Probability the coin is double-headed given 4 Heads. --- ## Step 2: Use Bayes’ Theorem \[ P(D|H) = \frac{P(H|D) \cdot P(D)}{P(H)} \] ### Find Each Term #### 1. \( P(D) \): Probability coin is double-headed \[ P(D) = \frac{2}{5} \] There are 2 double-headed coins out of 5 total. #### 2. \( P(H|D) \): Probability of 4 Heads if coin is double-headed A double-headed coin **always** gives Heads: \[ P(H|D) = 1 \] #### 3. \( P(H) \): Total probability of 4 Heads By the law of total probability: \[ P(H) = P(H|D)P(D) + P(H|F)P(F) \] - \( P(H|F) \): Probability of 4 Heads with a fair coin Each toss: \( \frac{1}{2} \) for Heads. So, \[ P(H|F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] - \( P(F) = \frac{3}{5} \) So, \[ P(H) = (1) \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} = \frac{2}{5} + \frac{3}{80} \] Calculate: - \(\frac{2}{5} = \frac{32}{80}\) - \(\frac{32}{80} + \frac{3}{80} = \frac{35}{80}\) \[ P(H) = \frac{35}{80} \] --- ## Step 3: Compute \( P(D|H) \) Put it all together: \[ P(D|H) = \frac{P(H|D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{35}{80}} = \frac{2}{5} \times \frac{80}{35} = \frac{2 \times 80}{5 \times 35} = \frac{160}{175} = \frac{32}{35} \] --- ## **Final Answer** \[ \boxed{\frac{32}{35}} \] **If you get 4 heads in a row, the probability the coin is double-headed is \(\frac{32}{35}\) (about 91.4%).** --- ## **Summary Table** | Coin Type | Probability Chosen | Probability 4 Heads | Combined Probability for 4 Heads | |------------------|-------------------|---------------------|----------------------------------| | Double-headed | \( \frac{2}{5} \) | 1 | \( \frac{2}{5} \) | | Fair | \( \frac{3}{5} \) | \( \frac{1}{16} \) | \( \frac{3}{80} \) | | **Total** | | | \( \frac{35}{80} \) | --- **Key idea:** Even though there are more fair coins, getting 4 Heads in a row is so unlikely with a fair coin that it’s very strong evidence the coin is double-headed.

# Problem Overview A box contains: - **3 fair coins** (each gives Heads/Tails with probability \( \frac{1}{2} \)) - **2 double-headed coins** (always give Heads) A coin is **randomly selected** and **tossed 4 times**, resulting in **Heads** each time. **Question:** What is the probability the chosen coin is double-headed? --- ## Step 1: Define Events - **\( D \)**: Coin is **Double-Headed** - **\( F \)**: Coin is **Fair** - **\( H \)**: All 4 tosses result in Heads We need to find \( P(D|H) \): Probability the coin is double-headed given 4 Heads. --- ## Step 2: Apply Bayes’ Theorem Using Bayes' theorem: \[ P(D|H) = \frac{P(H|D) \cdot P(D)}{P(H)} \] ### Calculate Each Term 1. **\( P(D) \)**: Probability the coin is double-headed \[ P(D) = \frac{2}{5} \] 2. **\( P(H|D) \)**: Probability of 4 Heads if the coin is double-headed \[ P(H|D) = 1 \] 3. **\( P(H) \)**: Total probability of 4 Heads \[ P(H) = P(H|D)P(D) + P(H|F)P(F) \] - **\( P(H|F) \)**: Probability of 4 Heads with a fair coin \[ P(H|F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] - **\( P(F) \)**: Probability the coin is fair \[ P(F) = \frac{3}{5} \] Combining these: \[ P(H) = (1) \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Converting \( \frac{2}{5} \) to a common denominator: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} \] --- ## Step 3: Compute \( P(D|H) \) Putting it all together: \[ P(D|H) = \frac{P(H|D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{35}{80}} = \frac{2}{5} \times \frac{80}{35} = \frac{160}{175} = \frac{32}{35} \] --- ## **Final Answer** \[ \boxed{\frac{32}{35}} \] **Interpretation:** If all 4 tosses result in Heads, the probability that the chosen coin is double-headed is \(\frac{32}{35}\) (approximately 91.4%). --- ## Summary of Key Results - **Probability Chosen Coin is Double-Headed**: \( \frac{2}{5} \) - **Probability of 4 Heads if Double-Headed**: 1 - **Total Probability of 4 Heads**: \( \frac{35}{80} \) ### Conclusion Despite having more fair coins, the occurrence of 4 Heads in a row provides substantial evidence that the coin is likely double-headed.