To find the probability of getting 0 heads when three coins are tossed, we can use the concept of binomial probability.

In this scenario:

• Each coin has 2 possible outcomes: head (H) or tail (T).
• There are a total of $2^3 = 8$ possible outcomes when three coins are tossed.
• We define success as getting a head.

We need to calculate the probability of getting exactly 0 heads (i.e., all tails).

First, let's enumerate the outcomes:

• HHH
• HHT
• HTH
• HTT
• THH
• THT
• TTH
• TTT

Only one of these outcomes results in 0 heads: TTT.

Now, calculate the probability:

$P(\text{0 heads}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{1}{8}$

Convert this fraction to a decimal and round to three decimal places:

$\frac{1}{8} = 0.125$

Thus, the probability of getting 0 heads when three coins are tossed is:

$\boxed{0.125}$

Would you like further details or have any questions? Here are some related questions for deeper understanding:

1. What is the probability of getting exactly 1 head in three coin tosses?
2. What is the probability of getting exactly 2 heads in three coin tosses?
3. What is the probability of getting exactly 3 heads in three coin tosses?
4. How would the probability change if we tossed 4 coins instead of 3?
5. What is the probability of getting at least 1 head in three coin tosses?
6. How many possible outcomes result in getting at least 2 heads in three coin tosses?
7. What is the expected number of heads in three coin tosses?
8. How would you calculate the probability of getting exactly 0 heads using the binomial formula?

Tip: When dealing with probabilities of multiple events, visualizing all possible outcomes using a tree diagram or a probability table can greatly aid in understanding and solving the problem.