### Home > INT2 > Chapter 9 > Lesson 9.3.4 > Problem9-120

9-120.

Mr. Kyi has placed three red, seven blue, and two yellow beads in a hat. If a person selects a red bead, they win $3$. If that person selects a blue bead, they lose $1$. If the person selects a yellow bead, they win $10$. What is the expected value for one draw? Is this game fair? If not, then who has the advantage in this game? Explain.

Review the math notes box in lesson 3.2.1 to see how to compute expected value.

$(\frac{3}{12})(3) \ + \ (\frac{7}{12})(-1) \ + \ (\frac{2}{12})(10) \ = \ \frac{11}{10} \ \approx \ 1.83.$

The game is not fair because the expected value is not zero. The player will come out
ahead over time, so students may want to play the game!