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A game is played with a single fair die. A player wins $20 if a 2 turns up, $40 if a 4 turns up, and loses $30 if a 6 turns up. If any other face turns up, there is no winning. Find the expected sum of money the player can win.A)$3B) $4C) $5D) $6.50

Sagot :

The dice has 6 faces numbered 1 to 6.

The probability of landing any number is 1/6.

The expected value is the sum of products of the probability and the winnings/losings.

So the expected value in this case is given by:

[tex]\begin{gathered} E(X)=\frac{1}{6}\times20+\frac{1}{6}\times40+\frac{1}{6}\times(-30)+\frac{1}{6}\times0+\frac{1}{6}\times0+\frac{1}{6}\times0 \\ E(X)=\frac{20+40-30}{6} \\ E(X)=5 \end{gathered}[/tex]

So the player can expect to win $5 per game in the long run.

Option C is correct.