Question

Consider the following game: You reach into a jar of money, and select a single bill at random to keep. There are 9 five-dollar bills, 5 ten-dollar bills, and 3 twenty-dollar bills in the jar. What should the cost of this game be in order for the game to be fair

Answer 1

Answer:

$E(x)=\ 9.118$

Step-by-step explanation:

From the question we are told that:

Available bills

 $\ 5=N0 9\\\\\ 10=N0 5$

 $\ 20=N0 3$

Therefore

Total Bills

 $n=5+9+3$

 $n=17$

Probability of selecting each bill

 $For\ 5$

 $P(\ 5)=\frac{9}{17}$

 $For\ 10$

 $P(\ 10)=\frac{5}{17}$

 $For\ 20$

 $P(\ 20)=\frac{3}{17}$

Generally the equation for Expected winning is mathematically given by

 $E(x)=\sum(X)*P(X)$

 $E(x)=5*\frac{9}{17}+10*\frac{5}{17}+20*\frac{3}{17}$

 $E(x)=\ 9.118$