Mahmoud tosses three coins. What is the probability that he will get at least one head?
Let's first find the probability that Mahmoud will get no heads. Ok, so the probability of getting no heads is the probability of getting tails three times in a row. And since the probability of getting tails is $\frac{1}{2}$, then the probability of getting tails three times in a row is $\frac{1}{2}*\frac{1}{2}*\frac{1}{2} = \frac{1}{8}$. So the probability of getting at least one head is $1 - \frac{1}{8} = \frac{7}{8}$ because the event of getting at least one head and the event of getting no heads are complementary.

