Question

A game is played using one die. If the die is rolled and shows a 2, the player wins $7. If the die shows any number other than 2, the player wins nothing. If there is a charge of $1 to play the game, what is the game's expected value? Round to the nearest cent. A game is played using one die. If the die is rolled and shows a 2, the player wins $7. If the die shows any number other than 2, the player wins nothing. If there is a charge of $1 to play the game, what is the game's expected value? Round to the nearest cent. @Mathematics

Answer 1

14 i guess

Answer 2

is that right ?

Answer 3

E(x) = sum of all x and their probabilties

Answer 4

Expectation Value =\[\sum_{i=1}^{n}x _{i}p _{i}\], where x represent the random variables given, p represent the associated probability with each variable.

Answer 5

i got 0.17

Answer 6

how ?

Answer 7

is it 1/7

Answer 8

????

Answer 9

the probability of getting a 2 is 1/6 x (6) - the probabilty of not getting a 2 5/6 x (-1) for the dollar lost by playing the game

Answer 10

QUESTION 7 OF 10

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 $4 if the spinner stops on red and $2 if it stops on green. The player loses $1 if it stops on blue and $5 if it stops on yellow.

What is the expected value rounded to the nearest cent?

Answer 11

So, p(rolling a )=1/6, p(rolling any other number is 5/6. Therefore E(x) =(7))(1/6)+(0)(5/6)=7/6. With a dollar a play, are you feeling lucky?

Answer 12

its not a 7 because you dont gain 7 dollars if you win you only gain 6 because you paid a dollar to play.

Answer 13

Sorry didn't get to finish. Since it cost a dollar to play, and the expected value is 7/6. which is $1.17, rounded to the nearest cent. So you expect to net around $.17, which I see is the answer you also got.

Answer 14

:) perfect