$P\left(H\right)=\frac{2}{3},P\left(T\right)=\frac{1}{3}$

$2^{\mathrm{n}\mathrm{d}}$ tail and $4^{\mathrm{t}\mathrm{h}}$ tail occur at $4^{\mathrm{t}\mathrm{h}}$ and ${10}^{\mathrm{t}\mathrm{h}}$ tosses respectively implies that there is one tail in first three tosses, $2^{\mathrm{n}\mathrm{d}}$ tail at $4^{\mathrm{t}\mathrm{h}}$ toss, one tail at next five tosses and $4^{\mathrm{t}\mathrm{h}}$ tail at ${10}^{\mathrm{t}\mathrm{h}}$ toss.

So, the required probability $={}_{ }{}^{3}C{}_1\left(\frac{1}{3}\right){\left(\frac{2}{3}\right)}^2\times \frac{1}{3}\times {}_{ }{}^{5}C{}_1{\left(\frac{1}{3}\right)}^1{\left(\frac{2}{3}\right)}^4\times \frac{1}{3}$

$=\frac{320}{3^9}$