Question

A sample consists of integers $1,2,...,2n$. The probability of choosing the integer $k$ is proportional to $\log k$. Find the probability of choosing the integer $2$ given that an even integer is chosen.

• A. $\frac{\log 2}{n\log 2+\log n!}$
• B. $\frac{\log n!}{\log 2+\log n!}$
• C. $\frac{n\log 2}{n\log 2+\log n!}$
• D. none of these

Answer 1

Answer: (A)

$\frac{\log 2}{n\log 2+\log n!}$

$P\left(1\right)=c\log 1$

$P\left(2\right)=c\log 2$

......................

$P\left(2n\right)=c\log 2n$

Adding

$1=c(\log 2n!)$

So $c=1/(\log 2n!)$

$P\left(Even\right)=c(\log 2+\log 4+..+\log 2n)$

$=c(n\log 2+\log n!)$

So,

$P\left(2/Even\right)=\frac{c\log 2}{c(n\log 2+\log n!)}=\frac{\log 2}{(n\log 2+\log n!)}$