Question

Let $x_i(1\le i\le 5)$ denote the probability of an event $E_i$. Let
$y=5\sum _{i=1}^5{x}_i^2-2\sum _{i=1}^5{x}_i+5$, Then

• A. max $y=20$
• B. min $y=4$
• C. max $y=1$
• D. min $y=0$

Answer 1

Answer: (A), (B)

max $y=20$
min $y=4$

$5x^2-2x=5(x^2-2\times x\times \frac{1}{5}+\frac{1}{25}-\frac{1}{25})=5{(x-\frac{1}{5})}^2-\frac{1}{5}$
This is maximum when $x=1$ to give maximum value of $3$ and minimum when $x=\frac{1}{5}$ to give minimum value of $-\frac{1}{5}$.
There are five such $x$ terms. Hence, the overall maximum value would have been $5\times 3+5=20$ and the overall minimum value would have been $5\ast (-\frac{1}{5})+5=4$