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

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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?

Mathematics

2 answers:

cluponka [151] · Answer 1 · 2022-07-29T12:21:44+03:00

Answer:

Thus, the expected value of points for Derek and Mia are $\dfrac{-1}{9}$ and $\dfrac{1}{9}$ respectively.

Step-by-step 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,

$1.\ P(GG)=P(G)\times P(G)\\\\P(GG)=\dfrac{2}{3}\times \dfrac{2}{3}\\\\P(GG)=\dfrac{4}{9}$

$2.\ P(GY)=P(G)\times P(Y)\\\\P(GY)=\dfrac{2}{3}\times \dfrac{1}{3}\\\\P(GY)=\dfrac{2}{9}$

$3.\ P(YG)=P(Y)\times P(G)\\\\P(YG)=\dfrac{1}{3}\times \dfrac{2}{3}\\\\P(YG)=\dfrac{2}{9}$

$4.\ P(YY)=P(Y)\times P(Y)\\\\P(YY)=\dfrac{1}{3}\times \dfrac{1}{3}\\\\P(YY)=\dfrac{1}{9}$

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.

<h3>Also, the expected value of a random variable X is $E(X)=\sum_{i=1}^{n} x_i\times P(x_i)$ .</h3>

Then, the expected value of points for Derek is,

$E(D)= (-1)\times \dfrac{4}{9}+1\times \dfrac{2}{9}+1\times \dfrac{2}{9}+(-1)\times \dfrac{1}{9}\\\\E(D)= \dfrac{-5}{9}+\dfrac{4}{9}\\\\E(D)=\dfrac{-1}{9}$

And the expected value of points for Mia is,

$E(M)= 1\times \dfrac{4}{9}+(-1)\times \dfrac{2}{9}+(-1)\times \dfrac{2}{9}+1\times \dfrac{1}{9}\\\\E(M)= \dfrac{5}{9}-\dfrac{4}{9}\\\\E(M)=\dfrac{1}{9}$ .