$\begin{array}{1 1}(A)\;\large\frac{1}{38760}\\(B)\;\large\frac{1}{28760}\\(C)\;\large\frac{1}{18760}\\(D)\;\text{None of these}\end{array} $

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- This problem can be solved by combination method (i.e) $nC_r=\large\frac{n!}{r!(n-r)!}$

Step 1:

Given :

6 numbers out of 20 numbers are chosen

$\therefore$ These numbers can be chosen in $20C_6$ ways and only one combination of the number is correct

Let the probability of winning the prize be E

$\therefore n(E)=1$

Only one prize can be won

Step 2:

$n(S)=20C_6=\large\frac{20!}{6! 14!}$

$\Rightarrow \large\frac{20\times 19\times 18\times 17\times 16\times 15}{6\times 5\times 4\times 3\times 2\times 1}$

$\Rightarrow 38760$

Step 3:

Probability =$\large\frac{n(E)}{n(S)}$

$\Rightarrow \large\frac{1}{38760}$

Hence (A) is the correct answer.

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