Question

As a project for math class, two students devised a game in which 3 black marbles and 2 red marbles are put into a bag. First the players must decide who is playing black marbles and who is playing red marbles. Then each player takes a turn at drawing a marble, noting the color, replacing the marble in the bag, and then drawing a second marble and noting the color before returning it to the bag. The point scheme for the game is detailed in the table below.

Point Values for Marble Game

Black Marble Points Red Marble Points

Both black: +2 points Both red: +4 points
Different colors: –1 point Different colors: –1 point

If Seth is challenged to a game by a classmate, which statement below is correct in all aspects in helping him make the correct choice?


A) Since E(black) = 0.24 and E(red) = 0.16, Seth should choose to play black marbles.

B) E(red) will be twice that of E(black), so Seth should choose to play red marbles.

C) Since E(red) = E(black), it is a fair game, so it doesn’t matter which color Seth chooses.

D) Both options will lose points because there are two ways to lose points and only one way to gain points. He should choose neither color.

Answer 1

The correct answer is A. 

Answer 2

Answer:

Option: A is the correct answer.

A) Since E(black) = 0.24 and E(red) = 0.16, Seth should choose to play black marbles.

Step-by-step explanation:

We are asked to find the aspects in helping him make the correct choice is.

Now for this we need to find the expectation of black and red and see whose expectation is more.

Now to calculate the expectation of red we take the condition that at least one red is to be obtained.

Hence, Expectation(E) of black is calculated as:

$E(black)=\dfrac{3}{5}\times \dfrac{3}{5}\times 2-\dfrac{2}{5}\times \dfrac{3}{5}-\dfrac{2}{5}\times \dfrac{3}{5}\\\\\\E(black)=\dfrac{18}{25}-\dfrac{12}{25}\\\\\\E(black)=\dfrac{6}{25}\\\\E(black)=0.24$

Now, Expectation of red is calculated as:

$E(red)=\dfrac{2}{5}\times \dfrac{2}{5}\times 4-\dfrac{2}{5}\times \dfrac{3}{5}-\dfrac{2}{5}\times \dfrac{3}{5}\\\\E(red)=\dfrac{16}{25}-\dfrac{12}{25}\\\\\\E(red)=\dfrac{4}{25}\\\\\\E(red)=0.16$

Hence, expectation of black is more than that of red.

Hence, option: A is correct option.