1.A coin is weighted so that the probability of heads is ¼. In a game of chance, a coin is tossed five times in succession, the player winning $12, $19.50, $16.75, $8, $21.50 and $25 for five, four, three, two, one and no heads respectively. What is the expected value of this game for the player?

Solution:

E(x)=12*(1/32)+19.5*(5/32)+16.75*(10/32)+8*(10/32)+21.5*(5/32)+25*(1/32)= 15.30

So the expected value of this game for the player is $15.30