Đề kiểm tra 15 phút Toán 11 chương 2: Tổ hợp xác suất - Đề số 3

ĐK: \(\left\{ \begin{array}{l}0 \le x \le y\\0 \le x \le y + 2\end{array} \right. \Leftrightarrow 0 \le x \le y\,\,\left( {x,y \in N} \right)\)

\(\begin{array}{l}\left\{ \begin{array}{l}C_y^x:C_{y + 2}^x = \dfrac{1}{3}\\C_y^x:A_y^x = \dfrac{1}{{24}}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\dfrac{{y!}}{{x!\left( {y - x} \right)!}}.\dfrac{{x!\left( {y + 2 - x} \right)!}}{{\left( {y + 2} \right)!}} = \dfrac{1}{3}\\\dfrac{{y!}}{{x!\left( {y - x} \right)!}}.\dfrac{{\left( {y - x} \right)!}}{{y!}} = \dfrac{1}{{24}}\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}\dfrac{{\left( {y - x + 1} \right)\left( {y - x + 2} \right)}}{{\left( {y + 1} \right)\left( {y + 2} \right)}} = \dfrac{1}{3}\\\dfrac{1}{{x!}} = \dfrac{1}{{24}}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = 4\,\,\left( {tm} \right)\\\dfrac{{\left( {y - 3} \right)\left( {y - 2} \right)}}{{\left( {y + 1} \right)\left( {y + 2} \right)}} = \dfrac{1}{3}\,\,\left( * \right)\end{array} \right.\\\left( * \right) \Leftrightarrow 3{y^2} - 15y + 18 = {y^2} + 3y + 2\\ \Leftrightarrow 2{y^2} - 18y + 16 = 0 \Leftrightarrow \left[ \begin{array}{l}y = 8\,\,\left( {tm} \right)\\y = 1\,\,\,\left( {ktm} \right)\end{array} \right.\end{array}\)

Vậy hệ phương trình có 1 nghiệm \(\left( {x;y} \right) = \left( {4;8} \right)\)