- #1

vcsharp2003

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- Homework Statement:
- The question is part (b) as shown in screenshot.

- Relevant Equations:
- ##p(A) + p(not ~ A) = 1##

Let W be the event that a prize is received. Then ##p(W) + p(not ~ W) = 1##. We need to find ##p(not W)## and so let's try to find ##p(W)## and then we can subtract it from 1 to get ##p(not ~ W)##.

The experiment is buying 2 tickets. So, $$p(W) = \frac { {}^{10}C_2} { {}^{10000}C_2}$$

Thus $$p(not ~ W) = 1 - p(W) $$ $$p(not ~ W) = 1 - \frac { {}^{10}C_2} { {}^{10000}C_2}$$

The logic used is that we select 2 tickets from 10 winning tickets i.e. ##{}^{10}C_2##, which is the number of ways that 2 tickets can win. Also the total number of ways of buying 2 tickets from 10,000 tickets is ##{}^{10000}C_2##.

But the answer in the back is given as ##p(not ~ W) = \frac { {}^{9990}C_2} { {}^{10000}C_2}##

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