Question

There are 3 red balls, 5 green balls, and 1 yellow ball in a hat. If you draw the yellow ball, you gain $10. If you draw any of the red balls, you gain $4. If you draw a green ball, you lose $3. If you were to play this game 25 times, replacing the drawn ball each time, how much money can you expect to gain or lose?

Round to the nearest cent.
Do not round until the final answer.
Enter an expected loss as a negative number.

Answer 1

$19.44

Explanation:

Solution:-

- The hat contains red, green and yellow balls.

- The colored balls are distributed such that there are:

Color Number of balls

Red Balls ( R ) 3

Green Balls ( G ) 5

Yellow Balls ( Y ) 1

Total 9

- The amount that we either win or loose when a ball is drawn from the hat is also distributed according to color. Positive number are shown as gain and negative numbers are given as loss.

Color Win/Loss ( $ )

Red Balls ( R ) 4

Green Balls ( G ) -3

Yellow Balls ( Y ) 10

- We are to play the game for n = 25 times.

- The probability of selecting any ball from the hat remains constant under the condition of ( replacement ). Meaning that any ball drawn is placed back into the hat.

- The probability of selecting any colored ball is independent from selecting other colored ball. Moreover, the probability of selecting the same colored ball is also independent across every successive trial/draw.

- The probabilities for each colored ball are determined from the basic definition ( Favorable outcomes / Total outcomes ).

Color Probability ( p = number of balls / Total )

Red Balls ( R ) 3 / 9

Green Balls ( G ) 5 / 9

Yellow Balls ( Y ) 1 / 9

- We will define 3 random variables as follows:

X: The number of red balls drawn from hat

Y: The number of Green balls drawn from hat

Z: The number of Yellow balls drawn from the hat

Note: Each random variable follows "Binomial distribution" under the conditions applied above ( Replacement and independent events ).

- The expected number of colored balls drawn in n = 25 trials/draws.

Color Expected number = n*p

Red Balls ( R ) E ( X ) = 25*(3/9) = 8.33

Green Balls ( G ) E ( Y ) = 25*(5/9) = 13.889

Yellow Balls ( Y ) E ( Z ) = 25*(1/9) = 2.7778

- We will define a random variable W: The amount of gain/loss from the game.

- The expected value of a random variable ( W ) i.e the expected gain/loss from playing the game n = 25 times is defined as:

E ( W ) = Sum ( expected number*win/loss )

Color Expected number = n*p Win/Loss ( $ ) Expected Gain

Red Balls ( R ) E ( X ) = 25*(3/9) = 8.33 4 8.33*4 = $300/9

Green Balls ( G ) E ( Y ) = 25*(5/9) = 13.888 -3 13.8*-3 = -$375/9

Yellow Balls ( Y ) E ( Z ) = 25*(1/9) = 2.7778 10 10*2.8 = $250/9

Where,

E ( W ) = Sum of last column

= 300/9 - 375/9 + 250/9

= 175 / 9 = $19.44444

Answer: There is an expected gain of $19.44 from playing the game 25 times.