# There are 3 lottery tickets one can choose from: <mtable columnalign="right right" rowspacing="

There are 3 lottery tickets one can choose from:
$\begin{array}{|rrr|}\hline Alternatives& Chances& Outcome\\ A& 1/1000& 1000\\ & 999/1000& 0\\ B& 1/100& 100\\ & 99/100& 0\\ C& 1/2000& 1000\\ & 1/200& 100\\ & 1989/2000& 0\\ \hline\end{array}$
And the formula for computing the probabilities of final outcomes when a coin is flipped between lotteries yields
${P}_{1000}=\frac{1}{2}\frac{2}{2000}=\frac{1}{2000}$
${P}_{100}=\frac{1}{2}\frac{20}{1000}=\frac{10}{1000}$
${P}_{00}=1-{P}_{1000}-{P}_{100}$
$=\frac{1989}{2000}$
So there are 3 different probabilities here. The first one, ${P}_{1000}$ is the probability of the outcome being $1000$ dollars, which is (the probability of getting lottery A from the coin flip) x (the probability of the outcome being $1000$ dollars). That makes sense from the table above. But the second one, ${P}_{100}$, the probability of the outcome being $100$ dollars should be $\frac{20}{2000}$ or $\frac{1}{100}$. Why then is it $\frac{20}{1000}$???
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rynosluv101swv2s
You have some typos, and this might be clearer
${P}_{\mathrm{}1000}=\frac{1}{2}\frac{1}{1000}=\frac{1}{2000}$
${P}_{\mathrm{}100}=\frac{1}{2}\frac{1}{100}=\frac{1}{200}=\frac{10}{2000}$
${P}_{00}=1-{P}_{\mathrm{}1000}-{P}_{\mathrm{}100}\phantom{\rule{0ex}{0ex}}=\frac{2000}{2000}-\frac{1}{2000}-\frac{10}{2000}\phantom{\rule{0ex}{0ex}}=\frac{1989}{2000}$
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