Gọi $X_i$ là số sản phẩm loại $1$ trong $2$ sản phẩm lấy ra $(i = \overline{0;2})$
$\Rightarrow P_i$ là xác suất tương ứng
Ta có:
$+)\quad X_0:$
$P_0 = \dfrac{C_4^1}{C_{10}^1}\cdot \dfrac{C_4^1}{C_{10}^1}= \dfrac{4}{15}$
$+)\quad X_1:$
$P_1 = C_2^1\cdot \dfrac{C_6^1}{C_{10}^1}\cdot \dfrac{C_4^1}{C_{10}^1}=\dfrac{12}{25}$
$+)\quad X_2:$
$P_2= \dfrac{C_6^1}{C_{10}^1}\cdot \dfrac{C_6^1}{C_{10}^1}= \dfrac{9}{25}$
Ta được bảng phân phối xác suất:
$\begin{array}{|c|c|c|}\hline X&0&1&2\\\hline P&\dfrac{4}{25}&\dfrac{12}{25}&\dfrac{9}{25}\\\hline\end{array}$
b) $E(X)= X_0.P_0 + X_1.P_1 + X_2.P_2$
$\to E(X)= 0\cdot \dfrac{4}{25} + 1\cdot \dfrac{12}{25} + 2\cdot\dfrac{9}{25}$
$\to E(X) = \dfrac{6}{5}$
c) $Var(X)= E(X^2) - [E(X)]^2$
$\to Var(X)= X_0^2.P_0 + X_1^2.P_1 + X_2^2.P_2 - [E(X)]^2$
$\to Var(X)= 0^2\cdot \dfrac{4}{25} + 1^2\cdot \dfrac{12}{25} + 2^2\cdot\dfrac{9}{25}- \left(\dfrac65\right)^2$
$\to Var(X)= \dfrac{12}{25}$
d) Ta có:
$P_{\max}= P_1 = \dfrac{12}{25}$
$\Rightarrow Mod(X) = X_1 =1$
$Med(X)= 2$
