A box containing the following three coins. I. A fair coin with head on one face ...

We have 3 coins:

🪙 A = HT
🪙 B = HH
🪙 C = TT

First pick

You pick a coin and get H.

So the first coin cannot be TT.

There are two possibilities:

Possibility 1: You picked HT
Remove it → remaining coins are:

$$HH,\ TT$$

Second pick:

• HH → H

• TT → T

So chance of H = 1/2.

---

Possibility 2: You picked HH
Remove it → remaining coins are:

$$HT,\ TT$$

Second pick:

• HT → H or T, each with probability 1/2

• TT → T

So chance of H = 1/4.

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Now, because you already got H on the first pick, HH was twice as likely as HT:

• HT chance = 1/3

• HH chance = 2/3

Therefore:

$$\frac13\times\frac12+\frac23\times\frac14$$

$$=\frac16+\frac16 =\boxed{\frac13}$$

Summary:

First H → remove that coin → depending on whether it was HT or HH, the remaining coins are different → combining those two cases gives $1/3$.

So the answer is $\boxed{1/3}$.