看到正面后是双头硬币的概率

设 $D$ 为选到双头硬币，$H$ 为出现正面。

$$P(D\mid H)=\frac{P(H\mid D)P(D)}{P(H)}.$$

其中：

• $P(D)=1/100$，$P(H\mid D)=1$；
• 选到公平硬币概率 $98/100$，其 $P(H)=1/2$；
• 选到双尾硬币概率 $1/100$，其 $P(H)=0$。

因此

$$P(D\mid H)=\frac{\frac{1}{100}}{\frac{1}{100}+\frac{98}{100}\cdot\frac{1}{2}}=\frac{1}{50}.$$

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Original Explanation

Let $D$ be the event that we select the double-headed coin and $H$ be the event that the coin showed heads. We want $\mathbb{P}[D \mid H]$. By Bayes' Rule, we have that $\mathbb{P}[D \mid H] = \dfrac{\mathbb{P}[H \mid D]\mathbb{P}[D]}{\mathbb{P}[H]}$. For the denominator, we should condition on the ways a head could be obtained. There are three types of coins: the fair coins, the double-headed coin, and the double-tailed coin. Let $F$ be the event of selecting a fair coin and $T$ be the event of selecting the double-tailed coin. We have that $\mathbb{P}[H \mid T] = 0$ since this coin has no heads on it. We know $\mathbb{P}[H \mid F] = \dfrac{1}{2}$ since it is fair and $\mathbb{P}[F] = \dfrac{98}{100}$ since 98 of the coins are fair. Lastly, $\mathbb{P}[H \mid D] = 1$ since both sides are heads and $\mathbb{P}[D] = \dfrac{1}{100}$ as there is only $1$ of them. Combining all of this,

$$\mathbb{P}[D \mid H] = \dfrac{\frac{1}{100} \cdot 1}{\frac{1}{100} \cdot 1 + \frac{98}{100} \cdot \frac{1}{2} + 0} = \dfrac{1}{50}$$