Math, asked by gadibathinanagalaksh, 4 hours ago

a sample of 4items is selected at random from a box containing 12 items of which 5are defective. find the expected no.of defective items​

Answers

Answered by niranjanack042
2

Step-by-step explanation:

selected from among 12 = 12/(81)(4!) = 495. Those possible with 0 defective items = [5!/(5) (017/(31)(41) = 35. Those possible with 1 defective item = [5!/(41) (11[7V(49(31) = 175. Those possible with 2 defective items = [5/(31) (2)][7/(51)(21) = 210. Those possible with 3 defective items = [5/(2) (3)|[7/(61)(11) = 70. Those possible with 4 defective items = [5!/(11) (41)|[7U(79(01) = 5. Thus, there are 460 possible combinations of 4 items, selected as described, which include at least 1 and up to 4 defective items. Just 35 possible combinations of 4 items include none that are defective.

Answered by amitnrw
5

Given: A sample of 4 items is selected of random from a box containing

12 items of which 5 are defective

To Find:  the expected no. E of the defectives

Solution:

Total = 12

Defectives = 5   ,  Non defectives = 7

0 defectives =  ⁷C₄.⁵C₀/¹²C₄  =    35/495

1 defectives =  ⁷C₃.⁵C₁/¹²C₄  =    175/495

2 defectives =  ⁷C₂.⁵C₂/¹²C₄  =   210/495

3 defectives =  ⁷C₁.⁵C₃/¹²C₄  =    70/495

4 defectives =  ⁷C₀.⁵C₄/¹²C₄  =    5/495

expected no. E of the defective​s

= 0 *  35/495  + 1 *   175/495 + 2 *  210/495 + 3 * 70/495 + 4 * 5/495

= ( 175 + 420 + 210 + 20)/495

= 825/495

= 1.667

expected no. E of the defective​s  is 1.667

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