Problem#3 Consider a batch of 30 devices out of which 10 devices are defective (

Question

Problem#3

Consider a batch of 30 devices out of which 10 devices are defective (D). The batch is inspected by choosing randomly a sample of 5 devices (the 5 items are taken at the same time).

1- How many different samples (of five devices) are possible?

2-How many sample of five devices contain exactly one defective device?

3-How many sample of five devices contain at least one defective device?

4-What is the probability to select a sample containing only one defective device?

5-What is the probability to select a sample that has no defective device?

Explanation / Answer

Basically: Non Defective= 20, Defective= 10, Total=30

1) 30C5= (30! / 5!25!)= 30x29x28x27x26/5x4x3x2x1= 1,42,506 ways

2) Choose 4 out of 20 and 1 out 10 = 20C4 x 10C1= (20!/4!16!) x (10!/9!1!) = 4845x10= 48,450 ways

3) At least 1 Defective has the following possibilities:

a) 4 non-defective& 1 Defective= 48,450 (as done above)

b) 3 non-defective& 2 Defective = 20C3 x 10C2= (20!/17!3!) x (10!/8!2!)= 1140x45= 51,300

c) 2 non-defective& 3 Defective = 20C2 x 10C3= (20!/18!2!) x (10!/7!3!)= 190x120= 22,800

d) 1 non-defective& 4 Defective = 20C1 x 10C4= (20!/19!1!) x (10!/6!4!)= 20x210= 4,200

e) 0 non-defective& 5 Defective = 20C0 x 10C5= (20!/20!0!) x (10!/5!5!)= 1x252= 252

Therefore, total no of ways= 48,450+51,300+22,800+4,200+252 = 1,27,002 ways

4) Probabilty = no of favourable options/Total options = 20C1 x 10C4/30C5= 48450/142506= 24250/71253

5) Probability = 20C5/30C5= 15504/142506 = 2584/23571

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