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- Homework Statement
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- Relevant Equations
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summary: you toss six coins. if you have 3 heads, you don't get any money. if you have 4 or more heads, you get the number of heads amount of dollars. if you have 2 or less heads, you toss six coins again and get the number of heads amount of dollars.

the number of ways to get 0 head ##\frac{6!}{(6)!1!}=1##

the number of ways to get 1 head ##\frac{6!}{(6-1)!1!}=6##

the number of ways to get 2 heads ##\frac{6!}{(6-2)!2!}=15##

the number of ways to get 3 heads ##\frac{6!}{(6-3)!3!}=20##

the number of ways to get 4 heads ##\frac{6!}{(6-4)!4!}=15##

the number of ways to get 5 heads ##\frac{6!}{(6-2)!2!}=6##

the number of ways to get 6 heads ##\frac{6!}{(6-2)!2!}=1##

$$P(X<3)=\frac{22}{2^6}$$

$$P(X=3)=\frac{20}{2^6}$$

$$P(X>3)=\frac{22}{2^6}$$

$$E(X<3)=E(X)=3$$

$$E(X=3)=0$$

$$E(X>3)=\frac{15}{21}\cdot 4+\frac{6}{21}\cdot 5+\frac{1}{21}\cdot 6=\frac{32}{7}$$

the expected value of the game is

$$P(X<3)E(X<3)+P(X=3)E(X=3)+P(X>3)E(X>3)$$

$$=\frac{22}{2^6}\cdot 3 +\frac{20}{2^6}\cdot 0 + \frac{22}{2^6} \frac{32}{7}=\frac{583}{224}\approx 2.6$$