$\begin{array}{1 1}(A)\;\large\frac{999}{1000}\\(B)\;\large\frac{909}{1000}\\(C)\;\large\frac{979}{1000}\\(D)\;\large\frac{989}{1000}\end{array} $

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- $nC_r=\large\frac{n!}{r!(n-r)!}$
- Required probability=$\large\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}=\frac{n(E)}{n(S)}$

Step 1:

Given total tickets =10,000

Out of which 10 tickets have prizes 9990 are blank

Total number of outcomes =1 ticket from 10,000 can be selected $10000C_1$ ways

Number of ways in which 1 ticket is without prize =$9990C_1$

Step 2:

$\therefore$ Required probability =$\large\frac{n(E)}{n(S)}$

$\Rightarrow \large\frac{9990C_1}{10000C_1}$

$\Rightarrow \large\frac{9990}{10000}$

$\Rightarrow \large\frac{999}{1000}$

Hence (A) is the correct answer.

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