Question

heather has a deck of 10 cards numbered 1 through 10. she is playing a game of chance. this game is this: heather chooses one card from the deck at random. she wins an amount of money equal to the value of the card if an odd numbered card is drawn. she loses $4.20 if an even numbered card is drawn. (a) find the expected value of playing the game. dollars (b) what can heather expect in the long run, after playing the game many times? (she replaces the card in the deck each time.) heather can expect to gain money. she can expect to win dollars per draw. heather can expect to lose money. she can expect to lose dollars per draw. heather can expect to break even (neither gain nor lose money).

Explanation:

Step1: Calculate probabilities

There are 5 odd - numbered cards and 5 even - numbered cards out of 10 cards. So the probability of drawing an odd - numbered card $P(O)=\frac{5}{10}=0.5$, and the probability of drawing an even - numbered card $P(E)=\frac{5}{10}=0.5$.

Step2: Determine winning and losing amounts

If an odd - numbered card is drawn, the winning amounts are 1, 3, 5, 7, 9. The expected winning amount when an odd - numbered card is drawn is $\frac{1 + 3+5+7+9}{5}=5$. If an even - numbered card is drawn, the losing amount is $4.20$.

Step3: Calculate the expected value

The formula for the expected value $E(X)$ of a discrete random variable is $E(X)=\sum_{i}x_ip_i$. Here, $E(X)=(5\times0.5)+(- 4.20\times0.5)$.
$E(X)=2.5-2.1$.
$E(X) = 0.4$.

Answer:

(a) $0.4$
(b) Heather can expect to gain money. She can expect to win $0.4$ dollars per draw.