To determine the truth of the two statements regarding the probability of getting tails an odd number of times when tossing a fair coin, we will analyze each statement step by step.
### Step-by-Step Solution:
**Step 1: Analyze Statement I**
- We have a fair coin tossed 100 times.
- Let \( X \) be the number of tails obtained.
- The possible outcomes for \( X \) can be 0, 1, 2, ..., 100.
- We want to find the probability of \( X \) being odd, i.e., \( P(X \text{ is odd}) \).
**Step 2: Use the Binomial Distribution**
- The number of tails follows a binomial distribution: \( X \sim \text{Binomial}(n=100, p=0.5) \).
- The probability of getting exactly \( k \) tails in 100 tosses is given by:
\[
P(X = k) = \binom{100}{k} \left( \frac{1}{2} \right)^{100}
\]
- We need to sum this probability for all odd \( k \):
\[
P(X \text{ is odd}) = \sum_{k=1,3,5,\ldots}^{99} P(X = k) = \sum_{k=1,3,5,\ldots}^{99} \binom{100}{k} \left( \frac{1}{2} \right)^{100}
\]
**Step 3: Use the Binomial Theorem**
- By the binomial theorem:
\[
(p + q)^n = \sum_{k=0}^{n} \binom{n}{k} p^k q^{n-k}
\]
- For \( p = q = \frac{1}{2} \) and \( n = 100 \):
\[
\left( \frac{1}{2} + \frac{1}{2} \right)^{100} = 1 = \sum_{k=0}^{100} \binom{100}{k} \left( \frac{1}{2} \right)^{100}
\]
- The sum of probabilities for odd \( k \) and even \( k \) is equal, thus:
\[
P(X \text{ is odd}) + P(X \text{ is even}) = 1
\]
- Since there are equal numbers of odd and even outcomes:
\[
P(X \text{ is odd}) = \frac{1}{2}
\]
**Step 4: Conclusion for Statement I**
- Therefore, Statement I is true: The probability of getting tails an odd number of times when tossing a fair coin 100 times is indeed \( \frac{1}{2} \).
---
**Step 5: Analyze Statement II**
- Now consider a fair coin tossed 99 times.
- Again, let \( Y \) be the number of tails obtained.
- Similar to the previous case, \( Y \) follows a binomial distribution: \( Y \sim \text{Binomial}(n=99, p=0.5) \).
**Step 6: Calculate Probability for Odd Tails**
- We want to find \( P(Y \text{ is odd}) \):
\[
P(Y \text{ is odd}) = \sum_{k=1,3,5,\ldots}^{99} P(Y = k)
\]
- Using the same reasoning as before:
\[
P(Y \text{ is odd}) + P(Y \text{ is even}) = 1
\]
- Since there are equal numbers of odd and even outcomes:
\[
P(Y \text{ is odd}) = \frac{1}{2}
\]
**Step 7: Conclusion for Statement II**
- Therefore, Statement II is also true: The probability of getting tails an odd number of times when tossing a fair coin 99 times is \( \frac{1}{2} \).
---
### Final Conclusion
Both statements are true.
