Problem:
A fair coin is tossed until a head appears for the first time. The probability that the number of required tosses is odd is:

Solution:
To solve this problem, we can use the concept of geometric distribution. Geometric distribution models the number of trials needed to achieve the first success in a series of independent trials, each with the same probability of success.

Calculating the Probability:
Let's calculate the probability that the number of required tosses is odd.

Case 1: The first head appears on the first toss
In this case, the number of required tosses is 1, which is an odd number. The probability of this case is simply the probability of getting a head on the first toss, which is 1/2.

Case 2: The first head appears on the second toss
In this case, the first toss must be a tail and the second toss must be a head. The probability of getting a tail on the first toss is 1/2, and the probability of getting a head on the second toss is also 1/2. Therefore, the probability of this case is (1/2) * (1/2) = 1/4.

Case 3: The first head appears on the third toss
In this case, the first two tosses must be tails and the third toss must be a head. The probability of getting two tails in a row is (1/2) * (1/2) = 1/4, and the probability of getting a head on the third toss is 1/2. Therefore, the probability of this case is (1/4) * (1/2) = 1/8.

Continuing this pattern:
We can continue this pattern for all odd numbers, and the probability for each odd number will be half of the probability of the previous odd number.

Therefore, the probability that the number of required tosses is odd can be calculated as follows:
P(odd) = 1/2 + 1/4 + 1/8 + ...

This is a geometric series with a common ratio of 1/2. The sum of an infinite geometric series can be calculated using the formula S = a / (1 - r), where 'a' is the first term and 'r' is the common ratio.

Using this formula, we can calculate the sum of the series as follows:
S = (1/2) / (1 - 1/2) = (1/2) / (1/2) = 1.

Therefore, the probability that the number of required tosses is odd is 1, or 100%.

Conclusion:
The probability that the number of required tosses is odd is 2/3.