Question

1. A box contains 4 red and 6 green balls. Two balls are drawn. You receive $10 if both are red and you pay $1 otherwise. What are your expected winnings?

Answer 1

Answer:

Your Winnings are $ -1

Step-by-step explanation:

Hello, great question. These types are questions are the beginning steps for learning more advanced Algebraic Equations.

Since we are not told what two balls were actually drawn, then the expected winnings would be of the greatest probability occurring. To figure this out we need to calculate the probability of getting two red balls consecutively and the probability of getting two different colors consecutively. Since there are a total of 10 balls the probabilities of getting one ball would be the following,

Red: $\frac{4}{10} = 40%$

Green: $\frac{6}{10} = 60%$

Now we need to know the probability of us getting 2 reds together, two greens, or one of each.

2 Reds: $\frac{4}{10} *\frac{4}{10} = \frac{16}{100} = 16%$

2 Greens:  $\frac{6}{10} *\frac{6}{10} = \frac{36}{100} = 36%$

1 Each:  $\frac{4}{10} *\frac{6}{10} = \frac{24}{100} = 24%$

So we can see that the highest probability is getting 2 Greens in  a row. Meaning you would have to pay $1

I hope this answered your question. If you have any more questions feel free to ask away at Brainly.