In a certain lottery,10,000 tickets are sold | vedclass

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Question

(A) Total number of tickets sold $= 10,000$. Number of prize-winning tickets $= 10$. Number of tickets that do not win a prize $= 10,000 - 10 = 9,990$

Vedclass · Accepted Answer

$\frac{^{9990}C_2}{^{10000}C_2}$

Vedclass · Answer

(A) Total number of tickets sold $= 10,000$. Number of prize-winning tickets $= 10$. Number of tickets that do not win a prize $= 10,000 - 10 = 9,990$. If you buy $2$ tickets,the total number of ways to choose $2$ tickets from $10,000$ is $^{10000}C_2$. The number of ways to choose $2$ tickets from the $9,990$ non-winning tickets is $^{9990}C_2$. Therefore,the probability of not getting a prize is $P = \frac{^{9990}C_2}{^{10000}C_2}$.

In a certain lottery,$10,000$ tickets are sold and $10$ equal prizes are awarded. What is the probability of not getting a prize if you buy $2$ tickets?

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