$\begin{array}{1 1}(A)\;\large\frac{9990C_{10}}{10000C_{10}}\\(B)\;\large\frac{9990C_{3}}{10000C_{3}}\\(C)\;\large\frac{9990C_{5}}{10000C_{5}}\\(D)\;\text{None of these}\end{array} $

- $nC_r=\large\frac{n!}{r!(n-r)!}$
- 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 =$1000C_{10}$

Number of ways in which 10 tickets are without prize =$9990C_{10}$

Step 2:

Required probability =$\large\frac{9990C_{10}}{10000C_{10}}$

Hence (A) is the correct answer.

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