The spinner on a wheel of fortune can land with an equal chance on any one of ten regions. Three regions are red, four are blue, two are yellow, and one is green. A player wins $1 if the spinner stops on red and it it stops on green. The player loses $1 it stops on blue and $2 rit stops on yellow
a. What is the expected value?
(Round to the nearest cent)
b. What does this value mean if the game is played ten times?
O A Over 10 games, the player can expect to lose money
OB. More than 10 games need to be played for the expected value to be a reasonable estimate of expected winnings
OC. Over 10 games, the player can expect to win money
OD. Over 10 games, the player can expect to break even

3 red

4 blue

2 yellow

1 green

There are 10 sections. You get one dollar for landing on red or green. Red is 3/10 of the spinner. Green is 1/10 of the spinner. Altogether this is 4/10. The probability that you will win one dollar is 4/10. This is a less than even chance, meaning less than half.

You lose one dollar if you land on blue. Blue is 4/10 so the probability that you will lose one dollar is 4/10.

Therefore you are just as likely to win one dollar than you are to lose one dollar.

You will lose two dollars if you land on yellow. Yellow is 2/10 so the probability that you will lose two dollars is 2/10. 2/10 is half of 4/10 so you are half as likely to lose 2 dollars than you are to win/lose one dollar.

Let's simplify these.

Win 1 __4/10__2/5

Lose 1__4/10__2/5

Lose 2__2/10__1/5

This means that the probability that you will lose either one or two dollars 6/10 (3/5). Therefore players cna expect to lose money. This is because 6/10 is more than even, in other words more th ma half the chance. 5/10 is half.

So O is the correct answer.

idk

step-by-step explanation: