When three coins are tossed simultaneously, the possible outcomes are as follows:

$\{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$

There are $2^3 = 8$ possible outcomes since each coin has two outcomes (head or tail) and there are three coins.

(i) Probability of getting at least 2 heads:

The outcomes where at least 2 heads appear are: $\{HHH, HHT, HTH, THH\}$ There are 4 favorable outcomes. Hence, the probability is: $P(\text{at least 2 heads}) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{4}{8} = \frac{1}{2}$

(ii) Probability of getting no heads:

The only outcome where no heads appear is: $\{TTT\}$ There is 1 favorable outcome. Hence, the probability is: $P(\text{no heads}) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{1}{8}$

---

Would you like further details on any part of this problem?

Here are 5 related questions:

1. What is the probability of getting exactly 1 head when tossing 3 coins?
2. How many outcomes have at least 1 head in a 3-coin toss?
3. If 4 coins are tossed, what is the probability of getting at least 2 heads?
4. What is the probability of getting all tails when tossing 4 coins?
5. How does the probability change if 5 coins are tossed simultaneously?

Tip: In probability, "at least" generally means you count all outcomes greater than or equal to a certain condition.