There are 5 defective items in a consignment of 18 items In

There are 5 defective items in a consignment of 18 items. In a quality check, the inspector picks a lot of 2 items at random from the consignment. If there is any defective item in the selected lot, the inspector picks another lot of 2 items at random from the remaining consignment of 16 items. If the total number of defective items picked by the inspector exceeds 2, then the consignment is rejected. Find the

(i) expected number of items picked;

(ii) probability distribution of X, the total number of defective item picked;

(iii) probability that the consignment is rejected;

(iv) expected number of defective items picked.

Solution

X = number of defective items picked

P (X = 0) = ¹³ C ₂ / ¹⁸ C ₂ = 26 / 51

P (X = 1) = ¹³ C ₁ ⁵ C ₁/ ¹⁸ C ₂ * ¹² C ₂ / ¹⁶ C ₂ = 143 / 612

P (X = 2) =⁵ C ₂/ ¹⁸ C ₂ * ¹³ C ₂ / ¹⁶ C ₂ + ¹³ C ₁ ⁵ C ₁/ ¹⁸ C ₂ * ¹² C ₁ ⁴ C ₁/ ¹⁶ C ₂ = 65 / 306

P(X=3) = ⁵ C ₂/ ¹⁸ C ₂ * ¹³ C ₁ ³ C ₁/ ¹⁶ C ₂ + ¹³ C ₁ ⁵ C ₁/ ¹⁸ C ₂ * ⁴ C ₂ / ¹⁶ C ₂ = 13 / 306

P(X = 4) = ⁵ C ₂/ ¹⁸ C ₂ * ³ C ₂/ ¹⁶ C ₂ = 1/ 612

This gives the probability distribution of X and answers part (ii)

(iii) P ( Consignment is rejected ) = P (X >2) = P(X = 3) + P (X = 4)

= 13/ 306 + 1 / 612

= 27 / 612

= 3 / 68

(iv) Expected number of defective items picked =

= 0 ( 26/51) + 1 (143/612) + 2 (65/306) + 3 (13/306) + 4 (1/612)

= 0.792 items

Part (i) is to be solved separately,

2 items will be picked if there are no defective items i.e P( X=0) and 4 items will be picked in all other cases.

Thus.

E (number of items picked) = 2 (26/51) + 4 ( 1 - 26/51)

= 52/51 + 4(25/51)

= 152 / 51

= 2.98 items will be picked on average.

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