Each toss of a fair coin has two possible outcomes: heads (H) or tails (T). Since the coin is tossed 9 times, the probability of getting tails in each individual toss is $\frac{1}{2}$.

To calculate the probability of getting all tails in 9 tosses, we raise the probability of getting tails in one toss to the power of 9:

$P(\text{all tails}) = \left( \frac{1}{2} \right)^9 = \frac{1}{512}$

Thus, the probability of getting all tails when a coin is tossed 9 times is:

$\frac{1}{512}$

As a decimal, this is:

$\frac{1}{512} \approx 0.001953$

So, the probability is approximately 0.001953, or exactly $\frac{1}{512}$.

Would you like more details on how this probability is derived?

Here are 5 related questions:

1. What is the probability of getting at least one head in 9 tosses?
2. What is the probability of getting exactly 5 tails in 9 tosses?
3. If a coin is tossed 10 times, what is the probability of getting all heads?
4. How would the probability change if the coin were biased with a 0.6 chance of tails?
5. What is the expected number of tails when tossing a coin 9 times?

Tip: The probability of independent events happening consecutively is the product of their individual probabilities.