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**Soroban** Hello, mathsnerd101!

There are 48 books: .$\displaystyle \begin{Bmatrix}\text{15 novels} \\ \text{33 others} \end{Bmatrix}$

There are: .$\displaystyle _{48}C_2 \:=\:1128$ possible outcomes.

The opposite of "at least one novel" is "*no* novels" (all Others).

No novels: .$\displaystyle _{33}C_2 \:=\:528$ ways.

Hence: .$\displaystyle 1128 - 528 \:=\:600$ ways to get at least one novel.

Therefore: .$\displaystyle P(\text{at least 1 novel}) \;=\;\frac{600}{1128} \;=\;\frac{25}{47}$

There are: .$\displaystyle 6^2 \:=\:36$ possible outcomes.

There are 6 ways to get a "7": .$\displaystyle P(7) \:=\:\tfrac{6}{36} \:=\:\tfrac{1}{6}$

. . She will win $50 one-sixth of the time: .$\displaystyle P(+\$50) \:=\:\tfrac{1}{6}$

She pays $10 for *every game*: .$\displaystyle P(\text{-}\$10) \:=\:\tfrac{36}{36} \:=\:1$

Her expected value is: .$\displaystyle E \;=\;(50)\left(\tfrac{1}{6}\right) + (\text{-}10)(1) \:=\:-\frac{5}{3}$

She can expect to lose an average of $1.67 per game.

He pays $10 for *every game*: .$\displaystyle P(-\$10) \:=\:1$

GAME 1 - Half the outcomes are even: .$\displaystyle P(+\$20) \:=\:\tfrac{1}{2}$

Therefore: .$\displaystyle E \;=\;(20)\left(\tfrac{1}{2}\right) + (-10)(1) \;=\;0$

The game is fair . . . Jack can expect to break even.

GAME 2 - There are 5 ways to get a "6" and 5 ways to get an "8"

. . Hence: .$\displaystyle P(\text{6 or 8}) \:=\:P(+\$30) \:=\:\tfrac{10}{36} \:=\:\tfrac{5}{18}$

Therefore: .$\displaystyle E \;=\;\left(\tfrac{5}{18}\right)(+30) + (1)(-10) \;=\;-\frac{5}{3}$

Jack can expect to lose an average of $1.67 per game.

GAME 3 - There are 15 outcomes with a sum less than or equal to 5.

. . Hence: .$\displaystyle P(\leq 5) \:=\:P(+\$40) \:=\:\tfrac{15}{36} \:=\:\tfrac{5}{12}$

Therefore: .$\displaystyle E \;=\;(+40)\left(\frac{5}{12}\right) + (1)(-10) \;=\;\frac{20}{3}$

Jack can expect to win an average of $6.67 per game.

*It's a no-brainer . . .*