Two players a and b play a marble game. each player has both a red and blue marble. they present one marble to each other. if both present red, a wins $3. if both present blue, a wins $1. if the colors of the two marbles do not match, b wins $2. is it better to be a, or b, or does it matter?

Step-by-step explanation:

A has one red and blue marble and B has one red and blue marble.

Hence selecting one marble is equally likely with prob = 0.5

Since A and B are independent the joint event would be product of probabilities.

Let A be the amount A wins.

If each selects one, the sample space would be

             (R,R)  (R,B)  (B,R) (B,B)

Prob     0.25  0.25  0.25  0.25

A              3       -2     -2       1

E(A)      0.75   -0.5    -0.5   0.25   =    0

The game is a fair game with equal expected values for A and B.

It does not matter whether to be A or B

(x3 + 6x2 + 2x - 6) • (x - 1)

step-by-step explanation:

polynomial long division :

3.2     polynomial long division

dividing :   x4+5x3-4x2-8x+6

                              ("dividend")

by         :     x-1     ("divisor")

dividend     x4   +   5x3   -   4x2   -   8x   +   6

- divisor   * x3     x4   -   x3            

remainder         6x3   -   4x2   -   8x   +   6

- divisor   * 6x2         6x3   -   6x2        

remainder             2x2   -   8x   +   6

- divisor   * 2x1             2x2   -   2x    

remainder               -   6x   +   6

- divisor   * -6x0               -   6x   +   6

remainder                     0

quotient :   x3+6x2+2x-6   remainder:   0

144.50

step-by-step explanation:

21.5 x 13.4=288.10 divide by 2