Given:
Expected value, E(X)= $600
Net Profit = $99400
Cost of playing once = $600
Solution:
Let the probability of winning be P(X) and that of losing [tex]P(\bar{X})[/tex]
[tex]P(\bar{X})[/tex] = 1 - P(X)
Now expected value, E(X) = [tex]Profit\times P(X) + loss\timesP(\bar{X}) [/tex]
E(X) = 99400P(X) + (-400)(1 - P(X))
600 = 99400P(X) -400 +400P(X)
P(X) = 0.01
Therefore, the probability of winning is P(X) = 0.01 or 1%
