Task: there are 100 coins, one is dishonest (heads on both sides). When flipped, heads appear — find the probability that the coin is dishonest.

Let's denote:

• Total coins: 100
• Unfair coin: 1 (double-headed)
• Fair coins: 99

Given that a head has appeared when flipping a coin, we need to find the probability that the coin is unfair.

Define the events:

• H — the coin is unfair
• F — the coin is fair
• E — a head is flipped

From the problem:

• P(H) = 1/100
• P(F) = 99/100

The probability of flipping a head:

• P(E|H) = 1 (an unfair coin always shows heads)
• P(E|F) = 0.5 (a fair coin has a 50% chance of heads)

Using Bayes' theorem:

[ P(H|E) = \frac{P(E|H) \cdot P(H)}{P(E|H) \cdot P(H) + P(E|F) \cdot P(F)} = \frac{1 \cdot \frac{1}{100}}{1 \cdot \frac{1}{100} + 0.5 \cdot \frac{99}{100}} = \frac{\frac{1}{100}}{\frac{1}{100} + \frac{49.5}{100}} = \frac{1}{1 + 49.5} = \frac{1}{50.5} \approx 0.0198 ]

Answer: The probability that the coin is unfair given that a head was flipped is approximately 1.98%."}]}]},