Question

A box contains 5 coins: 4 regular coins and 1 fake coin. When a regular coin is tossed, the probability \( P({head}) = 0.5 \), and for a fake coin, \( P({head}) = 1 \). You pick a coin at random and toss it twice, and get two heads. The probability that the coin you have chosen is the fake coin is __________.

Hint:

Bayes' Theorem allows us to update the probability of an event based on new evidence. In this case, it helps us find the probability of having chosen the fake coin, given the outcome of two heads.

Answer 1

Correct Answer: 0.49

Problem Analysis:

We need to determine the probability that the coin is fake based on the evidence of two heads. We can compare the relative "frequency" of getting two heads from the fake coin versus the regular coins.

---

Step 1: Calculate the "Weight" of the Fake Coin Outcome

Suppose we perform this experiment 100 times. In 20 instances ($1/5$), we pick the fake coin. Since it is two-headed, it always produces two heads. Successful "Two Head" outcomes from fake coin: $1/5 \times 1 = \mathbf{1/5}$

---

Step 2: Calculate the "Weight" of the Regular Coin Outcomes

In 80 instances ($4/5$), we pick a regular coin. The probability of a regular coin yielding two heads in a row is $(1/2)^2 = 1/4$. Successful "Two Head" outcomes from regular coins: $4/5 \times 1/4 = \mathbf{1/5}$

---

Step 3: Compare Relative Frequencies

The total "pool" of ways to get two heads is the sum of the weights from both paths: $\text{Total weight} = \frac{1}{5} \text{ (from fake)} + \frac{1}{5} \text{ (from regular)} = \frac{2}{5}$

The probability that the coin is fake is simply the fake coin's contribution divided by the total pool: $P(\text{Fake} \mid \text{2 Heads}) = \frac{\text{Fake Weight}}{\text{Total Weight}} = \frac{1/5}{2/5}$

---

Final Answer:

$P(\text{Fake} \mid \text{2 Heads}) = \frac{1}{2} = \mathbf{0.50}$