Question

Roll a fair six-sided die. You win $10 if you roll a number that is at least 5.You will lose $3 if you roll the number 3. If you roll any other number, youwill neither win nor lose anything. Find the expected value (winnings) of thisprobability experiment to the nearest cent.O $7.00O $2.83O $6.17O $1.17

Answer 1

GIVEN:

You roll a fair six-sided die.

You win $10 if you roll at least 4

You lose $3 if you roll the numnber 3.

If you roll any other number, you will neither win nor lose.

Required;

Find the expected value of this probability experiment.

Step-by-step solution;

We shall take the expected winning or loss from each condition given;

`\begin{gathered} Probability\text{ }of\text{ }at\text{ }least\text{ }5: \\ \\ P[rolling\text{ }a\text{ }5]=(1)/(6) \\ \\ P[rolling\text{ }a\text{ }6]=(1)/(6) \\ \\ P[rolling\text{ }a\text{ }5\text{ }or\text{ }6]=(1)/(6)+(1)/(6) \\ \\ =(1)/(3) \end{gathered}`

For this outcome, we now have;

`\begin{gathered} Expected\text{ }value=P[at\text{ }least\text{ }5]*10 \\ \\ EV=(1)/(3)*10 \\ \\ EV=3.33 \end{gathered}`

The probability of rolling a 3 is;

`P[rolling\text{ }a\text{ }3]=(1)/(6)`

Therefore, the expected value is now;

`\begin{gathered} EV=P[rolling\text{ }a\text{ }3]*(-3) \\ \\ EV=(1)/(6)*(-3) \\ \\ EV=-0.5 \end{gathered}`

If you roll any other number you will neither win nor lose.

Therefore the expected value for rolling any other number will be zero (that is, the probability multiplied by $0).

Therefore, the expected value (winnings) of this probability experiment would be;

`\begin{gathered} Expected\text{ }value=3.33+(-0.50) \\ \\ Expected\text{ }value=2.83 \end{gathered}`

ANSWER:

The expected value of playing this game with the conditions given is $2.83