Question

In a certain lottery 10,000 tickets are sold and ten equal prizes are awared. What is the probability of not getting a prize if you buy (a) one ticket (b) two tickets (c) 10 tickets.

Answer 1

Total number of tickets = 10,000
Number of prize bearing tickets = 10
$\text{(a) Let A be the event that one ticket is prize bearing ticket.}$
$\therefore n\left(A\right){=}^{10}{C}_1\therefore P\left(A\right)=\frac{{}^{10}{C}_1}{{}^{10000}{C}_1}=\frac{10}{10000}=\frac{1}{1000}\text{Now}P\left(\overline{A}\right)=1-\frac{1}{1000}=\frac{999}{1000}\left(b\right)\text{Let B be the event that two tickets are prize bearing tickets.}\therefore n\left(B\right){=}^{10}{C}_2\therefore P\left(B\right)=\frac{{}^{10}{c}_2}{{}^{10000}{C}_2}=\frac{10!}{2!8!}\times \frac{2!9998!}{10000!}=\frac{10\times 9}{2}\times \frac{2}{10000\times 9999}\text{Now}P\left(\overline{B}\right)=1-\frac{1}{1111000}=\frac{1110999}{1111000}\text{(C) Let C be the event that ten tickets are not prize bearing ticket.}\therefore \text{Number of non prize bearing tickets}=10000-10=9990\therefore P\left(C\right)=\frac{{}^{9990}{C}_{10}}{{}^{10000}{C}_{10}}$