14. Asymmetric coin. (From 9th grade, 2 points) Billy Bones has two coins - a gold one and a silver one. One of them is symmetric, and the other is not. It is unknown which coin is asymmetric, but it is known that the asymmetric coin lands heads with a probability of \( p = 0.6 \).

Billy Bones tossed the gold coin, and heads came up immediately. Then Billy Bones started tossing the silver coin, and heads came up only on the second toss. Find the probability that the asymmetric coin is the gold one.

![](https://cdn.mathpix.com/cropped/2024_05_06_aebb3441e4bbcfde777cg-10.jpg?height=588&width=577&top_left_y=1556&top_left_x=1342)
Solution. Let's introduce notations for the events:

$$
A=\{\text { the gold coin is biased }\},
$$

$B=\left\{\begin{array}{l}\text { when the gold coin is tossed, heads appear immediately, } \\ \text { and when the silver coin is tossed, heads appear on the second attempt. }\end{array}\right\}$

We need to find the conditional probability $\mathrm{P}(A \mid B)$. We will use Bayes' formula:

$$
\begin{gathered}
\mathrm{P}(A \mid B)=\frac{\mathrm{P}(B \mid A) \cdot \mathrm{P}(A)}{\mathrm{P}(B)}=\frac{\mathrm{P}(B \mid A) \cdot \mathrm{P}(A)}{\mathrm{P}(B \mid A) \cdot \mathrm{P}(A)+\mathrm{P}(B \mid \bar{A}) \cdot \mathrm{P}(\bar{A})}= \\
=\frac{p \cdot 0.5^{2}}{p \cdot 0.5^{2}+0.5 \cdot(1-p) p}=\frac{0.5}{0.5+(1-p)}=\frac{0.5}{0.5+0.4}=\frac{5}{9}
\end{gathered}
$$

Answer: $5 / 9$.