$\begin{array}{1 1}(A)\;0.65\\(B)\;0.75\\(C)\;0.85\\(D)\;0.95\end{array} $

- 10,000 tickets
- 1 ticket is bought
- We define X=The expected value of money for each prize1 ,2 and 3
- X=\((3000\,;2000\;,500)\)
- \(p_i\)=probability of getting '2'prize
- i=1,2,3
- Expected proze money
- =\(\Large\;\sum_{i=1}^{3} p_i x_i\)

P\(x=x_1\)=\(\Large\frac{1}{10,000}\)

P\(x=x_2\)=\(\Large\frac{1}{10,000}\)

P\(x=x_3\)=\(\Large\frac{3}{10,000}\)

\(E(X)=\Large\;3000\times\;\frac{1}{10,000}\;+2000\times\;\frac{1}{10,000}\;+500\times\;\frac{1}{10,000}\)

=\(\Large\frac{65}{100}\)

=\(0.65\)

Ask Question

Tag:MathPhyChemBioOther

Take Test

...