7. Suppose that in a lot of 20 manufactured items there are 2 having major defec

Question

7. Suppose that in a lot of 20 manufactured items there are 2 having major defects, 5 having minor (but not major) defects, and 13 that are without defects. As a check on the lot, a sample of 4 is drawn at random without replacement. Let X denote the number of items with major defects and Y with minor defects. a) Compute the joint probability mass function fix,y). (Use a two-way table to display the distribution.) Compute the marginal distributions for X and for Y. Compute the conditional distribution of Y given that you have drawn both items with major defects. b) c)

Explanation / Answer

total manufactured items = 20

Major defects = 2 = X

Minor Defects = 5 = Y

No defectts = 13

HEre we selected 4 Items.

Here are the combinations of x and y are

(0,0), (0,1), (0,2), (0,3), (0,4) ,(1,0), (1,1), (1,2), (1,3), (2,0), (2,1), (2,2)

Here

f(x,y) = 2Cx5Cy13C(4- x-y)/ 20C4 [ this is calculated as x is from 2 items, y is from 5 items and rest are from 13 no defects items.

(b) Now we will calculate the marginal distribution of X and Y .

so That means

f(x) = 0.6316 ; X = 0

= 0.3368; X = 1

= 0.0316 ; X = 2

Now,

Marginal distriubtion of y

f(y) = 0.2817 ; y = 0

= 0.4696 ; y = 1

= 0.2167; y = 2

= 0.03096 ; y = 3

= 0.001032 ; y = 4

(c) Now here X = 2

so conditional distribution y here is

f(Y l X = 2) = 0.016099/ 0.031579 = 0.5098 ; Y = 0

= 0.0134/ 0.031579 = 0.4248 ; Y = 1

= 0.002064/0.031579 = 0.0654 ; Y = 2

X/Y 0 1 2 3 4 0 0.147575 0.29515 0.160991 0.026832 0.001032 1 0.11806 0.160991 0.053664 0.004128 0 2 0.016099 0.013416 0.002064 0 0

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