20 points 6. There are eight pictures on display in a line. Three of them are po

Question

20 points 6. There are eight pictures on display in a line. Three of them are portraits, four are photos and the remaining are paintings. If pictures of the same kind need to be arranged together, what is the number of ways that we can put these eight pictures in a line? 25 points 7. A random variable X takes values between 0 and 4 with a probability mass function 4 P-cx for x 0,1,2 p(x)-135 for x = 3,4 3x2 a. Find the appropriate value of c. b. Find the expected value of X c. Calculate variance of (2-2X).

Explanation / Answer

6. Since the pictures of the same kind need to be arranged together, let us arrange them first.

The three portraits can be arranged in 3! = 6 ways.

The four photos can be arranged in 4! = 24 ways.

Let us now consider the three portraits to be a BIG portrait. Also the 4 photos form a BIG photo.

We now have 3 pictures - one big portrait, one painting and one big photo.

Number of ways to arrange them = 3! = 6

=> Total number of arrangements = 6 * 24 * 6 = 864.

7.

p(x) = (4/5)x - cx2 for x = 0,1,2

p(x) = 3x2/125 for x = 3,4

a. p(0) = 0

p(1) = (4/5) - c

p(2) = (4/5)*2 - 4c = 8/5 - 4c

p(3) = 3*9/125 = 27/125

p(4) = 3*16/125 = 48/125

Adding p(0) + p(1) + p(2) + p(3) + p(4) = 1

=> 0 + 4/5 - c + 8/5 - 4c + 27/125 + 48/125 = 1

=> - 5c + 12/5 + 75/125 = 1

=> - 5c + 12/5 + 3/5 = 1

=> - 5c + 3 = 1

=> - 5c = -2

=> c = 2/5

b. p(0) = 0

p(1) = (4/5) - 2/5 = 2/5

p(2) = (4/5)*2 - 4*2/5 = 8/5 - 8/5 = 0

p(3) = 27/125

p(4) = = 48/125

E(x) = 0*0 + (2/5)*1 + 0*2 + (27/125)*3 + (48/125)*4

= 2.584

c. If x = 0, 2 - 2x = 2 f(2) = 0

If x = 1, 2 - 2x = 0 f(0) = 0

If x >= 2, 2 - 2x = -2 f(-2) is not defined.

Expected value of 2-2x = 0

Variance of 2-2x = 0

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