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In a certain lottery 10,000 tickets are sold and ten equal prizes are awarded. What is the probability of not getting a prize if you buy (a) one ticket (b) two tickets (c) 10 tickets.
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Total number of tickets $=10000$
Number of prizes $=10$
$\therefore \quad$ Number of tickets which do not carry a prize $=10000-10=9990$
(i) Number of ways when 1 ticket without prize is
received $={ }^{9990} C_1=9990$
Total number of tickets $=10,000$
$\therefore$ Probability of not getting prize with one ticket $=\frac{9990}{10000}=\frac{999}{1000}$
(ii) Number of ways when 2 tickets without prize are received $={ }^{9990} C_2$
2 tickets out of 10000 may be taken out in ${ }^{10000} C_2$ ways.
$\therefore$ Probability of not getting a prize with two tickets $=\frac{{ }^{9990} C_2}{10000 C_2}$
(iii) 10 tickets out of 9990 may be taken in ${ }^{9990} C_{10}$ ways.
10 tickets out of 10000 may be taken in ${ }^{10000} C_{10}$ ways.
$\therefore$ Probability of not getting a prize with 10 tickets
$=\frac{{ }^{9990} C_{10}}{{ }^{10000} C_{10}}$
Number of prizes $=10$
$\therefore \quad$ Number of tickets which do not carry a prize $=10000-10=9990$
(i) Number of ways when 1 ticket without prize is
received $={ }^{9990} C_1=9990$
Total number of tickets $=10,000$
$\therefore$ Probability of not getting prize with one ticket $=\frac{9990}{10000}=\frac{999}{1000}$
(ii) Number of ways when 2 tickets without prize are received $={ }^{9990} C_2$
2 tickets out of 10000 may be taken out in ${ }^{10000} C_2$ ways.
$\therefore$ Probability of not getting a prize with two tickets $=\frac{{ }^{9990} C_2}{10000 C_2}$
(iii) 10 tickets out of 9990 may be taken in ${ }^{9990} C_{10}$ ways.
10 tickets out of 10000 may be taken in ${ }^{10000} C_{10}$ ways.
$\therefore$ Probability of not getting a prize with 10 tickets
$=\frac{{ }^{9990} C_{10}}{{ }^{10000} C_{10}}$
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