To calculate the probability of obtaining exactly two heads when tossing an unbiased coin three times, follow these steps:

Step 1: Identify all possible outcomes

When you toss a coin 3 times, each toss has 2 possible outcomes: heads (H) or tails (T). Therefore, the total number of possible outcomes for 3 tosses is: $2^3 = 8$ The possible outcomes are: $\{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$

Step 2: Count favorable outcomes

We are interested in the outcomes with exactly 2 heads. The favorable outcomes are: $\{HHT, HTH, THH\}$ So, there are 3 favorable outcomes.

Step 3: Calculate the probability

The probability of obtaining exactly 2 heads is the ratio of favorable outcomes to total outcomes: $P(\text{exactly 2 heads}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{3}{8}$

Final Answer:

The probability of obtaining exactly two heads when an unbiased coin is tossed 3 times is: $\frac{3}{8}$

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Would you like further clarification or more details on probability questions?

Here are some related questions to explore:

1. What is the probability of obtaining no heads in 3 tosses?
2. How do you calculate the probability of getting exactly 1 head?
3. What is the probability of obtaining 3 heads in 3 tosses?
4. How can you generalize this problem for $n$ tosses?
5. What is the formula for the probability of obtaining exactly $k$ heads in $n$ tosses?

Tip: Remember, for binomial probability, the formula is: $P(k \text{ heads}) = \binom{n}{k} \left(\frac{1}{2}\right)^n$ Where $\binom{n}{k}$ is the binomial coefficient!