A bag contains 3 gold marbles, 7 siver marbles, and 28 bronze marbles, Someone offers to play this game: You randomly
select one marble from the bag. If it is gold, you win . If it is silver, you win . If it is bronze, you lose .
What is your Expocted Value if you play this game?
Round your answer to the nearest cent.

Ask by Hammond Daniel. in the United States

Mar 29,2025

Question

A bag contains 3 gold marbles, 7 siver marbles, and 28 bronze marbles, Someone offers to play this game: You randomly
select one marble from the bag. If it is gold, you win . If it is silver, you win . If it is bronze, you lose .
What is your Expocted Value if you play this game?
Round your answer to the nearest cent.

Ask by Hammond Daniel. in the United States

Mar 29,2025

UpStudy AI · Accepted Answer

Let total number of marbles be

$$
3 + 7 + 28 = 38.
$$

The probability of drawing each marble is:

- Gold: $$\frac{3}{38}$$
- Silver: $$\frac{7}{38}$$
- Bronze: $$\frac{28}{38}$$

The expected value (EV) of the game is calculated by summing the products of each outcome's value and its probability:

$$
\text{EV} = \left(\frac{3}{38} \times 4\right) + \left(\frac{7}{38} \times 2\right) + \left(\frac{28}{38} \times (-1)\right).
$$

Calculating each term:

$$
\frac{3}{38} \times 4 = \frac{12}{38},
$$
$$
\frac{7}{38} \times 2 = \frac{14}{38},
$$
$$
\frac{28}{38} \times (-1) = -\frac{28}{38}.
$$

Summing these:

$$
\text{EV} = \frac{12}{38} + \frac{14}{38} - \frac{28}{38} = \frac{12 + 14 - 28}{38} = \frac{-2}{38} = -\frac{1}{19}.
$$

In decimal form,

$$
-\frac{1}{19} \approx -0.0526.
$$

Rounded to the nearest cent, the expected value is $$-\$0.05$$.
The expected value of the game is approximately $$-\$0.05$$.