Randomly choose two numbers between 0 and 1. What is the probability that their mean is no greater than 2/3?

Guest Aug 6, 2020

#1**+1 **

**Randomly choose two numbers between 0 and 1. What is the probability that their mean is no greater than 2/3?**

\(\text{Let number one $= x$ } \\ \text{Let number two $= y$ } \\ \text{Let the mean $= \dfrac{x+y}{2}$ } \)

Their mean is no greater than \(\dfrac{2}{3}\):

\(\dfrac{x+y}{2} \le \dfrac{2}{3}\)

\(\begin{array}{|rcll|} \hline \dfrac{x+y}{2} &=& \dfrac{2}{3} \\ x+y &=& \dfrac{4}{3} \quad | \quad y=1 \\ x+1 &=& \dfrac{4}{3} \\ x &=& \dfrac{4}{3}-1 \\ \mathbf{x} &=& \mathbf{\dfrac{1}{3}} \\ \hline \end{array} \begin{array}{|rcll|} \hline \dfrac{x+y}{2} &=& \dfrac{2}{3} \\ x+y &=& \dfrac{4}{3} \quad | \quad x=1 \\ 1+y &=& \dfrac{4}{3} \\ y &=& \dfrac{4}{3}-1 \\ \mathbf{y} &=& \mathbf{\dfrac{1}{3}} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \text{The probability} &=& \dfrac{A_{\color{red}red}}{A} \\\\ && A_{\color{red}red} = \dfrac{\left(1+\frac13\right)}{2}*\dfrac23+\dfrac13*1 \\\\ && \mathbf{A_{\color{red}red} = \dfrac{7}{9}} \\\\ && A = 1*1 \\\\ && \mathbf{A = 1} \\\\ \text{The probability} &=& \dfrac{\frac{7}{9}}{1} \\\\ \mathbf{\text{The probability}} &=& \mathbf{\dfrac{7}{9}} \\ \hline \end{array}\)

heureka Aug 10, 2020