Let's solve the problem step-by-step to determine the probability of getting at least 2 heads when tossing a coin 3 times.
### Step 1: Identify the sample space
The sample space for tossing a coin 3 times consists of all possible sequences of heads (H) and tails (T). There are [tex]\(2^3 = 8\)[/tex] possible outcomes, which are:
- HHH
- HHT
- HTH
- HTT
- THH
- THT
- TTH
- TTT
### Step 2: Identify the favorable outcomes
We are interested in outcomes with at least 2 heads. Let's list these outcomes and check how many heads each sequence contains:
- HHH (3 heads)
- HHT (2 heads)
- HTH (2 heads)
- THH (2 heads)
These are all the sequences with at least 2 heads.
### Step 3: Count the favorable outcomes
From the list above, we see there are 4 favorable outcomes:
- HHH
- HHT
- HTH
- THH
### Step 4: Calculate the probability
The probability [tex]\(P\)[/tex] of an event is given by the ratio of the number of favorable outcomes to the total number of outcomes. So, we have:
[tex]\[
P(\text{at least 2 heads}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{4}{8} = \frac{1}{2} = 0.5
\][/tex]
To express this probability as a percentage, we multiply by 100:
[tex]\[
P(\text{at least 2 heads}) \times 100 = 0.5 \times 100 = 50\%
\][/tex]
### Conclusion:
Therefore, the correct answer is 50%.
So, the probability of getting at least 2 heads when tossing a coin 3 times is 50%.
