The problem in the image asks to "Find the probability of getting at least two heads in three throws of a coin."

Let's break this down:

1. A coin has two possible outcomes: heads (H) or tails (T), each with a probability of $\frac{1}{2}$.
2. There are three independent throws, meaning the number of outcomes can be listed as follows:
   • HHH
   • HHT
   • HTH
   • HTT
   • THH
   • THT
   • TTH
   • TTT
3. The total number of possible outcomes is $2^3 = 8$.

Now, we need to count how many of these outcomes result in at least two heads.

The outcomes that have at least two heads are:

• HHH
• HHT
• HTH
• THH

There are 4 favorable outcomes.

Thus, the probability of getting at least two heads is:

$P(\text{at least 2 heads}) = \frac{\text{favorable outcomes}}{\text{total outcomes}} = \frac{4}{8} = \frac{1}{2}$

Final Answer:

The probability of getting at least two heads in three throws of a coin is $\frac{1}{2}$.

Would you like more details or have any questions?

Here are 5 related questions:

1. What is the probability of getting exactly two heads in three coin flips?
2. How does the probability change if the coin is biased?
3. What is the probability of getting at least one head in three coin flips?
4. How would the probability change if the number of coin flips is increased to four?
5. What is the probability of getting exactly zero heads in three coin flips?

Tip: When solving probability problems, always identify the total number of possible outcomes and count favorable outcomes carefully.