$n$ coins are simultaneously flipped.  The probability that at most one of them shows tails is $\frac{3}{16}$.  Find $n$.
I want to find the probability that at most one of the $n$ coins shows tails, in terms of $n$. That means either all of them show heads, which has probability $\left(\frac{1}{2}\right)^n$, or exactly one of them shows tails, which has probability $n\left(\frac{1}{2}\right)^n$. So the total probability is $\left(\frac{1}{2}\right)^n + n\left(\frac{1}{2}\right)^n$. This should equal $\frac{3}{16}$, so I can set up an equation: $\left(\frac{1}{2}\right)^n + n\left(\frac{1}{2}\right)^n = \frac{3}{16}$. I want to solve for $n$, but this equation looks hard to solve algebraically. Maybe I can try some small values of $n$ and see if they work. If $n = 1$, then the probability is $\frac{1}{2} + \frac{1}{2} = 1$, which is too big. If $n = 2$, then the probability is $\frac{1}{4} + 2\cdot\frac{1}{4} = \frac{3}{4}$, which is still too big. If $n = 3$, then the probability is $\frac{1}{8} + 3\cdot\frac{1}{8} = \frac{1}{2}$, which is also too big. If $n = 4$, then the probability is $\frac{1}{16} + 4\cdot\frac{1}{16} = \frac{5}{16}$, which is close but not equal to $\frac{3}{16}$. If $n = 5$, then the probability is $\frac{1}{32} + 5\cdot\frac{1}{32} = \frac{3}{16}$, which is exactly what I want. So $n = 5$ is the solution. 