7. A survey was conducted involving 420 families. It was found that $95\%$ of the families own a TV, and $80\%$ of the families own a computer. A family is randomly selected. If it is known that each surveyed family has at least one of the mentioned devices, what is the probability that the selected family owns a computer but not a TV? Express the probability as a percentage.

No calculators or reference materials are allowed.
First method:

| $\frac{95}{100}$ | 420 = 399, which means 399 families own a TV. | $1 \text{ POINT}$ |
| :---: | :---: | :---: |
| $\frac{80}{100}$ | $420=336$, which means 336 families own a computer. | 1 POINT |
| $399-$ | $-336=735$ | 1 POINT |
| 735 | $420=315$ | 1 POINT |
| $315$ | families own both a TV and a computer. |  |
| 336 | $315=21$, which means 21 families own only a computer. | 1 POINT |
| The number | of all possible outcomes is 420. | $1 \text{ POINT}$ |
| The number | of favorable outcomes is 21, | $1 \text{ POINT}$ |
| so the | probability of a favorable outcome is $p=\frac{21}{420}=\frac{1}{20}=0.05=5\%$. | 2 POINTS |
| The probability | that a family owns a computer but does not own a TV is $5\%$. | $\begin{aligned} 1 \text{ POINT} \\ \text{POINTS} \end{aligned}$ |