Question

Carlos has a deck of 10 cards numbered 1 through 10. He’s playing a game of chance

Answer 1

Given a deck of 10 cards numbered 1 through 10, you know that:

- Carlos wins an amount of money equal to the value of the card if an odd-numbered card is drawn.

- He loses $6 if an even-numbered card is drawn.

(a) By definition, the Expected Value is:

$E(x)=\sum_{i=1}^nx_iP(x_i)$

Where:

- An outcome is:

$x_i$

- The probability of the outcome is:

$P(x_i)$

In this case, you can set up this equation in order to find the expected value of playing the game:

$E(x)=(1+3+5+7+9)(\frac{5}{10})-6(\frac{5}{10})$

Evaluating, you get:

$E(x)=(25)(\frac{5}{10})-6(\frac{5}{10})$$E(x)=9.5$

(b) If he replaces the card in the deck each time, you know that the expected value indicates that the more he plays, the more probable is he gets this value:

$E(x)=9.5$

Hence, the answers are:

(a)

$9.5\text{ }dollars$

(b) First option: Carlos can expect to gain money. He can expect to win 9.5 dollars per draw.