Three coins are tossed simultaneously. Find the probability of getting
(i) at least 2 heads (ii) no heads

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:
\[
P(\text{no heads}) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{1}{8}
\]

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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.

Three coins are tossed simultaneously. Find the probability of getting (i) at least 2 heads (ii) no heads.

Discover how to calculate the probability of getting at least 2 heads or no heads when tossing 3 coins simultaneously. This problem covers fundamental concepts in probability and combinatorics, including the formula for calculating probabilities and the total number of outcomes. Suitable for students in Grades 6-8.

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