Question

A game costs $3 every time it is played. There is a 1 in 10 chance of winning $25, a 1 in 5 chance of winning $5, and a 7 in 10 chance of winning nothing. What is the expected value of the game?

Question 9 options:

$0.50


$5.00


-$0.50


$0.00

Answer 1

To find the expected value of the game, we will calculate the expected winnings and then subtract the cost of playing the game.

First, let's determine the expected winnings from playing the game:

1. The probability of winning $25 is \( \frac{1}{10} \) and the winnings from this event are $25.
2. The probability of winning $5 is \( \frac{1}{5} \) (which is equivalent to \( \frac{2}{10} \) when using a common denominator) and the winnings from this event are $5.
3. The probability of winning nothing is \( \frac{7}{10} \) and the winnings from this event are $0.

Now we can calculate the expected winnings (EV) using the formula:
\[
\text{EV} = (\text{Probability of winning $25}) \times (\text{Winnings from $25}) + (\text{Probability of winning $5}) \times (\text{Winnings from $5}) + (\text{Probability of winning $0}) \times (\text{Winnings from $0})
\]

Plugging in the values:
\[
\text{EV} = \left( \frac{1}{10} \times 25 \right) + \left( \frac{1}{5} \times 5 \right) + \left( \frac{7}{10} \times 0 \right)
\]

Calculating each term:
\[
\text{EV} = \left( \frac{1}{10} \times 25 \right) + \left( \frac{2}{10} \times 5 \right) + 0
\]
\[
= \frac{25}{10} + \frac{10}{10} + 0
\]
\[
= 2.5 + 1
\]
\[
= 3.5
\]

So the total expected winnings is $3.50.

Now, we need to account for the cost of playing the game, which is $3:
\[
\text{Net Expected Value} = \text{Expected Winnings} - \text{Cost}
\]
\[
= 3.5 - 3
\]
\[
= 0.5
\]

Thus, the expected value of the game is $0.50.

Therefore, the correct option is:
**$0.50**