设随机变量 $X$ 的分布列如下:

| $\mathrm{X}$ | 0 | 1 | 2 |
| :--- | :--- | :--- | :--- |
| $\mathrm{P}$ | $2 p+\frac{1}{3}$ | $\frac{1}{3}-p$ | $\frac{1}{3}-p$ |

若 $p \geq \frac{1}{5}$, 则 $E(X)_{\text {的最大值是 }}, D(X)_{\text {的最大值是 }}$
(1)由题意可得 $p \geq rac{1}{5}$, 解得 $rac{1}{5} \leq p \leq rac{1}{3}$.因为 $E(X)=0 	imes\left(2 p+rac{1}{3}
ight)+1 	imes\left(rac{1}{3}-p
ight)+2 	imes\left(rac{1}{3}-p
ight)=1-3 p \leqslant rac{2}{5}$,所以 $E(X)_{	ext {的最大值是 }} rac{2}{5}$,(2)因为 $D(x)=[0-(1-3 p)]^{2} 	imes\left(2 p+rac{1}{3}
ight)+[1-(1-3 p)]^{2} 	imes\left(rac{1}{3}-p
ight)+[2-(1-3 p)]^{2} 	imes\left(rac{1}{3}-p
ight)$$=-9 p^{2}+p+rac{2}{3}$,因为 $rac{1}{5} \leq p \leq rac{1}{3}$, 所以 $D(x) \leq rac{38}{75}$,所以 $D(x)$ 的最大值是 $rac{38}{75}$.