The probability of a man hitting a target is 1/2. How many times must he fire so that the probability of hitting the target at least once is more than 90%?

The probability of hitting the target is $\frac{1}{2}$ and the probability of missing is $\frac{1}{2}$.

The probability of hitting at least once equals 1 minus the probability of missing all shots:

$P(\text{hitting at least once}) = 1 - P(\text{missing all shots})$

If the man fires $n$ times:

$P(\text{missing all } n \text{ shots}) = \left(\frac{1}{2}\right)^n$

Therefore:

$P(\text{hitting at least once}) = 1 - \left(\frac{1}{2}\right)^n$

The probability of hitting at least once must be more than 90%:

$1 - \left(\frac{1}{2}\right)^n > 0.90$

$\left(\frac{1}{2}\right)^n < 0.10$

For $n = 3$:

$\left(\frac{1}{2}\right)^3 = \frac{1}{8} = 0.125$

$P(\text{hitting at least once}) = 1 - 0.125 = 0.875 = 87.5\%$

This is less than 90%.

For $n = 4$:

$\left(\frac{1}{2}\right)^4 = \frac{1}{16} = 0.0625$

$P(\text{hitting at least once}) = 1 - 0.0625 = 0.9375 = 93.75\%$

This is more than 90%.

The man must fire 4 times to ensure the probability of hitting the target at least once is more than 90%.