Problem Analysis: Probability of Odd Tosses for First Head

The question asks for the probability that the number of tosses required to get the first head from a fair coin is an odd number (1, 3, 5, ...).

• A fair coin implies the probability of getting a head (H) is $p = 1/2$, and the probability of getting a tail (T) is $q = 1 - p = 1/2$.
• This scenario follows a geometric distribution, where the probability of the first success (head) occurring on the $k$-th trial is $P(X=k) = q^{k-1}p$.

Calculating the Probability

Let $P_{odd}$ be the probability that the number of tosses required is odd.

Consider the outcome of the first toss:

• Case 1: First toss is a Head (H). The probability of this is $p = 1/2$. The number of tosses is 1, which is odd. So, this case contributes $p$ to $P_{odd}$.
• Case 2: First toss is a Tail (T). The probability of this is $q = 1/2$. We have used one toss, and now we need the *remaining* tosses to result in the first head appearing on an *even* number of additional tosses. This means the total number of tosses will be $1 + (\text{even number})$, which is odd. The process essentially restarts after the first tail, but we require the subsequent process to finish in an even number of steps for the total to be odd. The probability of needing an even number of additional tosses is $1 - P_{odd}$. This case contributes $q \times (1 - P_{odd})$ to $P_{odd}$.

Combining these cases:

Substitute the values $p = 1/2$ and $q = 1/2$:

Now, solve for $P_{odd}$:

Conclusion

The probability that the number of required tosses is odd is $2/3$. This corresponds to Option C.