Question

Let $X_n=\{1,2,3,....,n\}$ and let a subset $A$ of $X_n$ be chosen so that every pair of elements of $A$ differ by at least $3.$ (For example, if $n=5,A$ can be $\varphi ,\left\{2\right\}\text{or}\{1,5\}$ among others). When $n=10,$ let the probability that $1\in A$ be $p$ and let the probability that $2\in A$ be $q.$ Then

• A. $p>q\text{and}p-q=\frac{1}{6}$
• B. $p<q\text{and}q-p=\frac{1}{6}$
• C. $p>q\text{and}p-q=\frac{1}{10}$
• D. $p<q\text{and}q-p=\frac{1}{10}$

Answer 1

The correct option is C $p>q\text{and}p-q=\frac{1}{10}$

When $n=10$
Let $A_r$ be number of ways of selecting $r$ numbers.
Number of selection of $A$ is
$=n\left(A_0\right)+n\left(A_1\right)+n\left(A_2\right)+n\left(A_3\right)+n\left(A_4\right)$
$=1+10+(7+6+5+....+1)+\left\{\right(4+3+2+1)+(3+2+1)+(2+1)+1\}+1$
$=11+\frac{7\cdot 8}{2}+10+6+3+1+1=60$
$N\left(p\right)=n\left(\text{no. of ways 1 is selected}\right)=1+7+4+3+2+1+1=19$
$N\left(q\right)=n\left(\text{no. of ways 2 is selected}\right)=1+6+3+2+1=13$
Therefore,
$p=\frac{19}{60}$ and $q=\frac{13}{60}$
$\therefore p>q$
and $p-q=\frac{1}{10}$