$3 p ( x =0)= p ( x =1)$
3. ${ }^{33} C _{0}( q )^{33}={ }^{33} C _{1} pq ^{32}$
$p =\frac{1}{12}, q =\frac{11}{12}, \frac{ q }{ p }=11$
$\frac{ p ( x =15)}{ p ( x =18)}-\frac{ p ( x =16)}{ p ( x =17)}$
$\frac{{ }^{33} C_{15} p^{15} q^{18}}{{ }^{33} C_{18} p^{18} q^{15}}-\frac{{ }^{33} C_{16} p^{16} q^{17}}{{ }_{17} P ^{17} q^{16}}=\left(\frac{q}{p}\right)^{3}-\left(\frac{q}{p}\right)$
$=(11)^{3}-11$
$=1320$
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Statement $-1 :$ The probability that the chosen numbers when arranged in some order will form an $A.P.$ is $\frac{1}{{85}}$ .
Statement $-2 :$ If the four chosen numbers form an $A.P.$, then the set of all possible values of common difference is $\left( { \pm 1, \pm 2, \pm 3, \pm 4, \pm 5} \right)$ છે.