$\begin{array}{1 1}(A)\;\large\frac{9990C_2}{10000C_2}\\(B)\;\large\frac{9980C_2}{10000C_2}\\(C)\;\large\frac{9970C_2}{10,000C_2}\\(D)\;\large\frac{9000C_2}{10,000C_2}\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 =$10000C_2$

Number of ways in which 2 tickets are without prize =$9990C_2$

Step 2:

$\therefore$ Required probability =$\large\frac{9990C_2}{10000C_2}$

Hence (A) is the correct answer.

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