Question

If $x,y,z\in R$ and x+y+z =5, xy +yz+zx=3, probability for x to be positive only is

• A. $\frac{3}{16}$
• B. $\frac{5}{16}$
• C. $\frac{13}{16}$
• D. $\frac{15}{16}$

Answer 1

The correct option is C $\frac{13}{16}$

Range of x is $[-1,\frac{13}{3}]$
Since probability of x to be $\frac{{\int }_0^{\frac{13}{3}}dx}{{\int }_{-1}^{\frac{13}{3}}dx}=\frac{\frac{13}{3}}{\frac{13}{3}+1}=\frac{13}{16}$ Positive is