Answer:
[tex]3.125\,\%[/tex].
Explanation:
After every half-life, the quantity of the sample will become [tex]1/2[/tex] of the quantity at the beginning of the half-life. If the half-life for Fermium-253 is 3 days, it can be deduced that:
[tex]\begin{array}{|c|c|c|}\cline{1-3}\text{Number of days} & \text{Number of half-lives} & \text{\%\text{ of Sample that's still Fermium-253}} \\ \cline{1-3} 0 & 0 & 100\% \\ \cline{1-3}3 & 3/3 = 1 & (1/2) \times 100\% = 50\%\\ \cline{1-3} 6 & 6/3 = 2 & (1/2) \times ((1/2) \times 100\%)= 25\% \\ \cline{1-3} \vdots & \vdots & \vdots \\\cline{1-3}\end{array}[/tex].
Let [tex]n \ge 0[/tex]. In general, after [tex]n[/tex] half-lives (for the fermium-253 in this question, that would be [tex]3\, n[/tex] days,) the percentage of the sample that's still the initial substance would be:
[tex]\displaystyle 100\% \times \left(\frac{1}{2}\right)^{n}[/tex].
There are [tex]15 / 3 = 5[/tex] half lives in 15 days. (In other words, [tex]n = 5[/tex].) Therefore, the percentage of this sample that's still fermium-253 would be:
[tex]\begin{aligned}&100\% \times \left(\frac{1}{2}\right)^5 \\ &= 100\%\times \frac{1}{32} = 100\% \times 0.03125 \\&= 3.125\%\end{aligned}[/tex].