This problem involves understanding conditional probability and applying Bayes' theorem in a multi-step experiment. We need to calculate the probability of getting a head in the second toss, given that the first toss resulted in a head.

Coins Description and Initial Probabilities

First, let's identify the types of coins in the box and their initial probabilities of being picked.

• Coin I: A fair coin (Head on one face, Tail on the other). Let's denote this as \(C_F\).
• Coin II: A two-headed coin (Heads on both faces). Let's denote this as \(C_{HH}\).
• Coin III: A two-tailed coin (Tails on both faces). Let's denote this as \(C_{TT}\).

There are three coins in total. When a coin is picked randomly from the box, the probability of picking any specific coin is:

• \(P(C_F) = \frac{1}{3}\)
• \(P(C_{HH}) = \frac{1}{3}\)
• \(P(C_{TT}) = \frac{1}{3}\)

First Toss Resulting in a Head

Let \(H_1\) be the event that the first toss results in a Head. We need to find the probability of \(H_1\). This depends on which coin was picked for the first toss.

• If \(C_F\) is picked, \(P(H_1 | C_F) = \frac{1}{2}\) (probability of head with a fair coin).
• If \(C_{HH}\) is picked, \(P(H_1 | C_{HH}) = 1\) (probability of head with a two-headed coin).
• If \(C_{TT}\) is picked, \(P(H_1 | C_{TT}) = 0\) (probability of head with a two-tailed coin).

Using the Law of Total Probability, the probability of getting a Head in the first toss is:

\(P(H_1) = P(H_1 | C_F)P(C_F) + P(H_1 | C_{HH})P(C_{HH}) + P(H_1 | C_{TT})P(C_{TT})\)

\(P(H_1) = \left(\frac{1}{2}\right)\left(\frac{1}{3}\right) + (1)\left(\frac{1}{3}\right) + (0)\left(\frac{1}{3}\right)\)

\(P(H_1) = \frac{1}{6} + \frac{1}{3} + 0 = \frac{1}{6} + \frac{2}{6} = \frac{3}{6} = \frac{1}{2}\)

So, the probability of the first toss being a head is \(\frac{1}{2}\).

Conditional Probability of the First Coin Picked Given \(H_1\)

Since we know the first toss resulted in a head (\(H_1\)), we can update our probabilities for which coin was initially picked using Bayes' Theorem. This is crucial because the identity of the first coin affects which coins are left in the box for the second draw.

• Probability that the first coin was \(C_F\) given \(H_1\): \(P(C_F | H_1) = \frac{P(H_1 | C_F)P(C_F)}{P(H_1)} = \frac{\left(\frac{1}{2}\right)\left(\frac{1}{3}\right)}{\frac{1}{2}} = \frac{\frac{1}{6}}{\frac{1}{2}} = \frac{1}{6} \times 2 = \frac{2}{6} = \frac{1}{3}\)
• Probability that the first coin was \(C_{HH}\) given \(H_1\): \(P(C_{HH} | H_1) = \frac{P(H_1 | C_{HH})P(C_{HH})}{P(H_1)} = \frac{(1)\left(\frac{1}{3}\right)}{\frac{1}{2}} = \frac{\frac{1}{3}}{\frac{1}{2}} = \frac{1}{3} \times 2 = \frac{2}{3}\)
• Probability that the first coin was \(C_{TT}\) given \(H_1\): \(P(C_{TT} | H_1) = \frac{P(H_1 | C_{TT})P(C_{TT})}{P(H_1)} = \frac{(0)\left(\frac{1}{3}\right)}{\frac{1}{2}} = 0\) (This makes sense, as a two-tailed coin cannot produce a head).

These probabilities sum to \( \frac{1}{3} + \frac{2}{3} + 0 = 1 \), as expected.

Probability of Getting a Head in the Second Toss (\(H_2\)) Given \(H_1\)

We need to find \(P(H_2 | H_1)\). This depends on which coin was chosen first. We will consider two cases based on the probabilities calculated above:

Case 1: The first coin picked was the Fair Coin (\(C_F\))

This occurs with probability \(P(C_F | H_1) = \frac{1}{3}\). If \(C_F\) was picked first, the remaining coins in the box are \(C_{HH}\) (two-headed) and \(C_{TT}\) (two-tailed).

A coin is then picked randomly from these two remaining coins and tossed.

• The probability of picking \(C_{HH}\) from the remaining two is \(\frac{1}{2}\). If \(C_{HH}\) is tossed, the probability of getting a Head is \(1\).
• The probability of picking \(C_{TT}\) from the remaining two is \(\frac{1}{2}\). If \(C_{TT}\) is tossed, the probability of getting a Head is \(0\).

So, the probability of getting a head in the second toss, given the first coin was \(C_F\), is:

\(P(H_2 | H_1, C_F \text{ first}) = \left(\frac{1}{2}\right)(1) + \left(\frac{1}{2}\right)(0) = \frac{1}{2}\)

Case 2: The first coin picked was the Two-Headed Coin (\(C_{HH}\))

This occurs with probability \(P(C_{HH} | H_1) = \frac{2}{3}\). If \(C_{HH}\) was picked first, the remaining coins in the box are \(C_F\) (fair) and \(C_{TT}\) (two-tailed).

A coin is then picked randomly from these two remaining coins and tossed.

• The probability of picking \(C_F\) from the remaining two is \(\frac{1}{2}\). If \(C_F\) is tossed, the probability of getting a Head is \(\frac{1}{2}\).
• The probability of picking \(C_{TT}\) from the remaining two is \(\frac{1}{2}\). If \(C_{TT}\) is tossed, the probability of getting a Head is \(0\).

So, the probability of getting a head in the second toss, given the first coin was \(C_{HH}\), is:

\(P(H_2 | H_1, C_{HH} \text{ first}) = \left(\frac{1}{2}\right)\left(\frac{1}{2}\right) + \left(\frac{1}{2}\right)(0) = \frac{1}{4}\)

Final Calculation of \(P(H_2 | H_1)\)

Now, we combine the probabilities from Case 1 and Case 2, weighted by the updated probabilities of which coin was picked first:

\(P(H_2 | H_1) = P(H_2 | H_1, C_F \text{ first})P(C_F | H_1) + P(H_2 | H_1, C_{HH} \text{ first})P(C_{HH} | H_1)\)

\(P(H_2 | H_1) = \left(\frac{1}{2}\right)\left(\frac{1}{3}\right) + \left(\frac{1}{4}\right)\left(\frac{2}{3}\right)\)

\(P(H_2 | H_1) = \frac{1}{6} + \frac{2}{12}\)

\(P(H_2 | H_1) = \frac{1}{6} + \frac{1}{6}\)

\(P(H_2 | H_1) = \frac{2}{6} = \frac{1}{3}\)

Therefore, the probability of getting a head in the second toss, given that the first toss resulted in a head, is \(\frac{1}{3}\).