Suppose that 2 beams are randomly chosen without replacement from the following

Question

Suppose that 2 beams are randomly chosen without replacement from the following group of 15 beams: 10 new 3 recycled (but ok) 2 defective a) How many possible outcomes exist? Next, let X represent the number of new beams that are selected. Let Y represent the number of recycled beams that are selected. b) Define the sample space (i.e., possible outcomes) for X and Y c) Find pxy(x, y). Make a table akin to Table 1 in Problem 1. Hint: You need to find xC0,0),Pxy (1,0),Pxr(2,0), pxr(1,1), pxy(0,1), and Pxy (0,2). d) Find E[X].

Explanation / Answer

(a) Here there are2 beams are selcted out of 3 different type of beams.

If say x,y,z are the type of beams where x is new, y is recycled and z is defective.

so total type of combinations are = 2+3 -1 C3-1 = 4C2 = 6

which are

(2,0,0), (0,2,0), (0,0,2), (1,1,0), (1,0,1), (0,1,1)

so total of 6 possible outcoms exists.

(b) Here sample space for x and y are (2,0), (0,2) , (0,0) ,(1,1) , (1,0) & (0,1)

(c) p (0,0) = 10C03C0 * 2C2/15C2 = 1/105

p(1,0) =  10C13C0 * 2C1/15C2 = 20/105 = 4/21

p(2,0) = 10C23C0 * 2C0/15C2 = 45/105 = 3/7

p(1,1) =  10C13C1 * 2C0/15C2 = 30/105 = 2/7

p(0,1) = 10C03C1 * 2C1/15C2  = 6/105 = 2/35

p(0,2) = 10C03C2 * 2C0/15C2 = 3/105 = 1/35

(d) Here p(x) = (6 + 3 + 1)/105 = 2/21 ; x = 0

= 50/105 = 10/21 ; x = 1

= 3/7 ; x = 2

E[X] = 2/21 * 0 + 10/21 * 1 + 3/7 * 2 = 10/21 + 6/7 = 28/21 = 4/3

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