Bài Tập Xác Suất Thống Kê Chương 2

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Bài tập xác suất thống kê chương 2 10 nhận xét Bài 1: Cho biến ngẫu nhiên liên tục $X$ có hàm mật độ xác suất $$f_X(x)= \begin{cases} kx^2 & \mbox{ nếu $0\leq x\leq 3$},\\ 0 & \mbox{ nếu $x$ còn lại}.\\ \end{cases}$$ a) Tìm hằng số $k.$ b) Tìm hàm phân bố xác suất $F_X(x).$ c) Tính $\Bbb P(X>1).$ d) Tính $\Bbb P(0,5\leq X\leq 2|X>1).$ Lời giải bài 1. a) Ta có \begin{equation*}\notag \begin{aligned} \displaystyle\int\limits_{-\infty}^{+\infty}f_X(x)dx&=\displaystyle\int\limits_{-\infty}^{0}f_X(x)dx+\displaystyle\int\limits_{0}^{3}f_X(x)dx+\displaystyle\int\limits_{3}^{+\infty}f_X(x)dx\\ &=\displaystyle\int\limits_{-\infty}^{0}0dx+\displaystyle\int\limits_{0}^{3}kx^2dx+\displaystyle\int\limits_{3}^{+\infty}0dx\\ &=0+\Big(\displaystyle\frac{kx^3}{3}\Big)\Bigg|_0^3+0\\ &=\displaystyle\frac{27k}{3}\\ &=9k. \end{aligned} \end{equation*} Theo tính chất của hàm mật độ xác suất $$\displaystyle\int\limits_{-\infty}^{+\infty}f_X(x)dx=1.$$ Do đó $9k=1$ hay $k=\displaystyle\frac{1}{9}.$ b) Nếu $t< 0$ thì \begin{equation*}\notag \begin{aligned} F_X(t)&=\displaystyle\int\limits_{-\infty}^{t}f_X(x)dx\\ &=\displaystyle\int\limits_{-\infty}^{t}0dx\\ &=0. \end{aligned} \end{equation*} Nếu $0\leq t\leq 3$ \begin{equation*}\notag \begin{aligned} F_X(t)&=\displaystyle\int\limits_{-\infty}^{t}f_X(x)dx\\ &=\displaystyle\int\limits_{-\infty}^{0}f_X(x)dx+\displaystyle\int\limits_{0}^{t}f_X(x)dx\\ &=\displaystyle\int\limits_{-\infty}^{0}0dx+\displaystyle\int\limits_{0}^{t}kx^2dx\\ &=0+\Big(\displaystyle\frac{kx^3}{3}\Big)\Bigg|_0^t\\ &=\displaystyle\frac{kt^3}{3}\\ &=\displaystyle\frac{t^3}{27}. \end{aligned} \end{equation*} Nếu $t>3$ \begin{equation*}\notag \begin{aligned} F_X(t)&=\displaystyle\int\limits_{-\infty}^{t}f_X(x)dx\\ &=\displaystyle\int\limits_{-\infty}^{0}f_X(x)dx+\displaystyle\int\limits_{0}^{3}f_X(x)dx+\displaystyle\int\limits_{3}^{t}f_X(x)dx\\ &=\displaystyle\int\limits_{-\infty}^{0}0dx+\displaystyle\int\limits_{0}^{3}kx^2dx+\displaystyle\int\limits_{3}^{t}0dx\\ &=0+\Big(\displaystyle\frac{kx^3}{3}\Big)\Bigg|_0^3+0\\ &=9k\\ &=1. \end{aligned} \end{equation*} Vậy $$F_X(t)= \begin{cases} 0 & \mbox{ nếu $t< 0$},\\ \displaystyle\frac{t^3}{27} & \mbox{ nếu $0\leq t\leq 3$},\\ 1& \mbox{ nếu $t>3$}.\\ \end{cases}$$ c) \begin{equation*}\notag \begin{aligned} \Bbb P(X>1)&=\displaystyle\int\limits_{1}^{+\infty}f_X(x)dx\\ &=\displaystyle\int\limits_{1}^{3}f_X(x)dx+\displaystyle\int\limits_{3}^{+\infty}f_X(x)dx\\ &=\displaystyle\int\limits_{1}^{3}kx^2dx+\displaystyle\int\limits_{3}^{+\infty}0dx\\ &=\Big(\displaystyle\frac{kx^3}{3}\Big)\Bigg|_1^3+0\\ &=\displaystyle\frac{27k}{3}-\displaystyle\frac{k}{3}\\ &=\displaystyle\frac{26k}{3}\\ &=\displaystyle\frac{26}{3}\times\displaystyle\frac{1}{9}\\ &=\displaystyle\frac{26}{27}. \end{aligned} \end{equation*} d) \begin{equation*}\notag \begin{aligned} \Bbb P(0,5\leq X\leq 2|X > 1)&=\displaystyle\frac{\Bbb P[(0,5\leq X\leq 2)(X > 1)]}{\Bbb P(X > 1)}\\ &=\displaystyle\frac{\Bbb P(1 < X\leq 2)}{\Bbb P(X > 1)}. \end{aligned} \end{equation*} \begin{equation*}\notag \begin{aligned} \Bbb P(1 < X\leq 2)&=\displaystyle\int\limits_{1}^{2}f_X(x)dx\\ &=\displaystyle\int\limits_{1}^{2}kx^2dx\\ &=\Big(\displaystyle\frac{kx^3}{3}\Big)\Bigg|_1^2\\ &=\displaystyle\frac{8k}{3}-\displaystyle\frac{k}{3}\\ &=\displaystyle\frac{7k}{3}\\ &=\displaystyle\frac{7}{3}\times\displaystyle\frac{1}{9}\\ &=\displaystyle\frac{7}{27}. \end{aligned} \end{equation*} Do đó \begin{equation*}\notag \begin{aligned} \Bbb P(0,5\leq X\leq 2|X > 1)&=\displaystyle\frac{\Bbb P(1 < X\leq 2)}{\Bbb P(X > 1)}\\ &=\displaystyle\frac{\displaystyle\frac{7}{27}}{\displaystyle\frac{26}{27}}\\ &=\displaystyle\frac{7}{26}. \end{aligned} \end{equation*} Bài 2: Cho biến ngẫu nhiên liên tục $X$ có hàm mật độ xác suất $$f_X(x)= \begin{cases} kx^2e^{-2x} & \mbox{ nếu $x\geq 0$},\\ 0 & \mbox{ nếu $x80)$, $C=(60 < X < 70).$ Tính các xác suất $\Bbb P(B|A)$, $\Bbb P(B|C).$ Lời giải bài 5. Với mọi $a>0,$ ta có \begin{equation*}\notag \begin{aligned} \Bbb P(X>a)&=\Bbb P(X\geq a)\\ &=\displaystyle\int\limits_{a}^{+\infty}f_X(x)dx\\ &=\displaystyle\int\limits_{a}^{+\infty}\lambda e^{-\lambda x}dx\\ &=\lim\limits_{b\longrightarrow+\infty}\displaystyle\int\limits_{a}^{b}\lambda e^{-\lambda x}dx\\ &=\lim\limits_{b\longrightarrow+\infty}(-e^{-\lambda x})\Big|_a^b\\ &=\lim\limits_{b\longrightarrow+\infty}(e^{-\lambda a}-e^{-\lambda b})\\ &=e^{-\lambda a}. \end{aligned} \end{equation*} a) Theo đề bài \begin{equation*}\notag \begin{aligned} &\;\;\;\;\;\;\;\;\;\Bbb P(X>60)=0,5\\ &\Longleftrightarrow e^{-60\lambda}=0,5\\ &\Longleftrightarrow \lambda=\displaystyle\frac{\ln 2}{60}. \end{aligned} \end{equation*} b) Ta có \begin{equation*}\notag \begin{aligned} \Bbb P(X>70|X\geq 60)&=\displaystyle\frac{\Bbb P[(X>70)(X\geq 60)]}{\Bbb P(X\geq 60)}\\ &=\displaystyle\frac{\Bbb P(X>70)}{\Bbb P(X\geq 60)}\\ &=\displaystyle\frac{e^{-70\lambda}}{e^{-60\lambda}}\\ &=e^{-10\lambda}\\ &=e^{-\frac{\ln 2}{6}}\\ &\approx 0,8909. \end{aligned} \end{equation*} c) Ta có \begin{equation*}\notag \begin{aligned} \Bbb P(B|A)&=\Bbb P(X>80|X>70)\\ &=\displaystyle\frac{\Bbb P[(X>80)(X>70)]}{\Bbb P(X>70)}\\ &=\displaystyle\frac{\Bbb P(X>80)}{\Bbb P(X>70)}\\ &=\displaystyle\frac{e^{-80\lambda}}{e^{-70\lambda}}\\ &=e^{-10\lambda}\\ &=e^{-\frac{\ln 2}{6}}\\ &\approx 0,8909. \end{aligned} \end{equation*} \begin{equation*}\notag \begin{aligned} \Bbb P(B|C)&=\Bbb P(X > 80|60 < X < 70)\\ &=\displaystyle\frac{\Bbb P[(X > 80)(60 < X < 70)]}{\Bbb P(60 < X < 70)}\\ &=\displaystyle\frac{\Bbb P(\emptyset)}{\Bbb P(60 < X < 70)}\\ &=0. \end{aligned} \end{equation*} Bài 6: Cho biến ngẫu nhiên liên tục $X$ với hàm mật độ xác suất $$f_X(x)= \begin{cases} k(1+x)^{-3} & \mbox{ nếu $x\geq 0$},\\ 0 & \mbox{ nếu $x