Question

A sample space consists of integers $1,2,3,....100.$ The probability of choosing an integer $k$ is proportional to $\ln k$. The conditional probability of choosing the integer ${}^{\prime }{2}^{\prime }$, given that an even integer is chosen is

• A. $\frac{\ln 2}{50\ln 2+\ln 50!}$
• B. $\frac{\ln 2}{\ln 2+\ln 50!}$
• C. $\frac{\ln 2}{50\ln 2+\ln 50}$
• D. $\frac{\ln 2}{\ln 2+\ln 50}$

Answer 1

The correct option is A $\frac{\ln 2}{50\ln 2+\ln 50!}$

$A=$ integer $2$ is chosen
$B=$ Even number is chosen
$P\left(1\right)=\lambda \ln 1,P\left(2\right)=\lambda \ln 2,....P\left(n\right)=\lambda \ln \left(n\right)$
$P\left(\frac{A}{B}\right)=\frac{P(A\cap B)}{P\left(B\right)}=\frac{\lambda \ln 2}{\lambda \ln \left(\mathrm{2.4.6....100}\right)}$
$=\frac{\ln 2}{50\ln 2+\ln 50!}$