Question

If the probability of choosing an integer $k$ out of $2m$ integers $1,2,3,...,2m$ is inversely proportional to $k^4(1\le k\le m).$ If $x_1$ is the probability that chosen number is odd and $x_2$ is the probability that chosen number is even, then

• A. $x_1>\frac{1}{2}\text{and}{x}_2<\frac{1}{2}$
• B. $x_1>\frac{2}{3}\text{and}{x}_2<\frac{1}{3}$
• C. $x_1<\frac{1}{2}\text{and}{x}_2>\frac{1}{2}$
• D. $x_1<\frac{1}{3}\text{and}{x}_2>\frac{2}{3}$

Answer 1

The correct option is A $x_1>\frac{1}{2}\text{and}{x}_2<\frac{1}{2}$

Let the probability of choosing one integer $k$ be $P\left(k\right)=\frac{C}{k^4},$ where $C$ is proportionality constant. Then,
$\sum _{k=1}^{2m}\frac{C}{k^4}=1$
$\Rightarrow C\sum _{k=1}^{2m}\frac{1}{k^4}=1$

Since, $x_1$ be the probability that chosen number is odd number.
$\therefore x_1=\sum _{k=1}^{m}P(2k-1)=C\sum _{k=1}^m\frac{1}{(2k-1{)}^4}$

Also,
$x_2=1-x_1=\sum _{k=1}^{m}P\left(2k\right)=C\sum _{k=1}^m\frac{1}{(2k{)}^4}$

Now,
$C\sum _{k=1}^m\frac{1}{(2k{)}^4}<C\sum _{k=1}^m\frac{1}{(2k-1{)}^4}$
$\Rightarrow 1-x_1<x_1$
$\Rightarrow x_1>\frac{1}{2}$
$\Rightarrow x_2<\frac{1}{2}$