Question

Derek and Mia place two green marbles and one yellow marble in a bag. Somebody picks a marble out of the bag without looking and records its color (G for green and Y for yellow). They replace the marble and then pick another marble. If the two marbles picked have the same color, Derek loses 1 point and Mia gains 1 point. If they are different colors, Mia loses 1 point and Derek gains 1 point. What is the expected value of the points for Derek and Mia?

Answer 1

Answer: P(GG)= 4/9

P(GY)= 2/9

P(YG)= 2/9

P(YY)= 1/9

Derek, E(X) = -1/9

Mia, E(X) = 1/9

Step-by-step explanation: just did it on edge

Answer 2

Thus, the expected value of points for Derek and Mia are
`(-1)/(9)` and
`(1)/(9)` respectively.

Explanation:

Number of green marbles = 2 and Number of Yellow marbles = 1

Then, total number of marbles = 2+1 = 3

A person selects two marbles one after another after replacing them.

So, the probabilities of selecting different combinations of colors are,

`1.\ P(GG)=P(G)* P(G)\\\\P(GG)=(2)/(3)* (2)/(3)\\\\P(GG)=(4)/(9)`

`2.\ P(GY)=P(G)* P(Y)\\\\P(GY)=(2)/(3)* (1)/(3)\\\\P(GY)=(2)/(9)`

`3.\ P(YG)=P(Y)* P(G)\\\\P(YG)=(1)/(3)* (2)/(3)\\\\P(YG)=(2)/(9)`

`4.\ P(YY)=P(Y)* P(Y)\\\\P(YY)=(1)/(3)* (1)/(3)\\\\P(YY)=(1)/(9)`

Now, we have that,

If two marbles are of same color, then Mia gains 1 point and Derek loses 1 point.

If two marbles are of different color, then Derek gains 1 point and Mia loses 1 point.

Also, the expected value of a random variable X is
`E(X)=\sum_(i=1)^(n) x_i* P(x_i)`.

Then, the expected value of points for Derek is,

`E(D)= (-1)* (4)/(9)+1* (2)/(9)+1* (2)/(9)+(-1)* (1)/(9)\\\\E(D)= (-5)/(9)+(4)/(9)\\\\E(D)=(-1)/(9)`

And the expected value of points for Mia is,

`E(M)= 1* (4)/(9)+(-1)* (2)/(9)+(-1)* (2)/(9)+1* (1)/(9)\\\\E(M)= (5)/(9)-(4)/(9)\\\\E(M)=(1)/(9)`.

Thus, the expected value of points for Derek and Mia are
`(-1)/(9)` and
`(1)/(9)` respectively.