Question

Let numbers $1,2,\mathrm{3...4}n$ be pasted on $4n$ blocks. The probability of drawing a number is proportional to $r$, then the probability of drawing an even number in one draw is $(n\in N)$

• A. $\frac{n+1}{2n+1}$
• B. $\frac{2n+1}{4n+1}$
• C. $\frac{n+2}{n+3}$
• D. $\frac{2n+3}{4n+1}$

Answer 1

Answer: (B)

$\frac{2n+1}{4n+1}$

If $P\left(r\right)$ be the probability that a number $r$ is drawn in one draw, it is given that $P\left(r\right)=kr;$
where $k$ is a constant.
$\therefore P\left(1\right)+P\left(2\right)+P\left(3\right)+\cdots +p\left(4n\right)=k\cdot 1+k\cdot 2+...+k\cdot \left(4n\right)=k[1+2+3+\cdots +4n]$
$1=\frac{4n(4n+1)k}{2}$
$\therefore \sum _{r=0}^{n}P\left(r\right)=1$ or
$P\left(S\right)=1$ $\Rightarrow K=\frac{1}{2n(4n+1)}$
Now ${}^{\prime }{A}^{\prime }$ event of drawing $2,4,6,...4n$
$P\left(A\right)=P(2$ or $4$ or $6....$ or $4n)$
$\therefore P\left(A\right)=P\left(2\right)+P\left(4\right)+\cdots +P\left(4n\right)$
$=2k(1+2+...+2n)$ $P\left(A\right)=k\left(2n\right)(2n+I)$

$P\left(A\right)=\left(2n\right)(2n+I)$

$\therefore P\left(A\right)=\frac{2n+1}{4n+1}$