If you toss a fair coin three times, what is the probability of getting at least one head?

To answer this question, we can first find the probability of getting no heads at all, and then subtract it from 1 to find the probability of getting at least one head.

There are two possible outcomes for each coin toss: a head (H) or a tail (T). Since there are three coin tosses, there are a total of 2³ = 8 possible outcomes:

HHH, HHT, HTH, THH, HTT, THT, TTH, TTT

Now let’s find the probability of getting no heads (i.e., getting all tails). There is only one such outcome: TTT. Since the coin is fair, each toss has a probability of $\frac{1}{2}$ of resulting in a head and a probability of $\frac{1}{2}$ of resulting in a tail.

The probability of getting all tails (TTT) is the product of the probabilities of each individual coin toss:

$P(TTT) = P(T) \cdot P(T) \cdot P(T) = \left(\frac{1}{2}\right)^3 = \frac{1}{8}$

Now, we want to find the probability of getting at least one head. This is simply the complement of getting no heads, which means we subtract the probability of getting no heads (TTT) from 1:

$P(\text{at least one head}) = 1 - P(TTT) = 1 - \frac{1}{8} = \frac{7}{8}$

So, the probability of getting at least one head when tossing a fair coin three times is $\frac{7}{8}$.