Question 1193441: A shipment of 10 items has two defective and eight non defective units. In the inspection of the shipment , a sample of units will be selected and tested. If a defective unit is found , the shipment of 10 units will be rejected.If a sample of five is selected , what is the probability that the shipment will be rejected?
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Answer: 7/9
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Explanation:
We have n = 10 units and we select a sample size of r = 5
Let's find out how many ways we can do this.
Use the nCr combination formula. Order does not matter when selecting the units.
n C r = (n!)/(r!(n-r)!)
10 C 5 = (10!)/(5!*(10-5)!)
10 C 5 = (10!)/(5!*5!)
10 C 5 = (10*9*8*7*6*5!)/(5!*5!)
10 C 5 = (10*9*8*7*6)/(5!)
10 C 5 = (10*9*8*7*6)/(5*4*3*2*1)
10 C 5 = (30240)/(120)
10 C 5 = 252
There are 252 ways to select five items from the pool of ten. There may or may not be defective units in this sample of five.
Let A = 252 so we can use it later.
Now let's only consider the 8 non-defective units
We have n = 8 items to pick from and r = 5 selections
n C r = (n!)/(r!(n-r)!)
8 C 5 = (8!)/(5!*(8-5)!)
8 C 5 = (8!)/(5!*3!)
8 C 5 = (8*7*6*5!)/(5!*3!)
8 C 5 = (8*7*6)/(3!)
8 C 5 = (8*7*6)/(3*2*1)
8 C 5 = (336)/(6)
8 C 5 = 56
There are 56 ways to pick a sample of five items, none of which are defective
Let B = 56
Subtract the values of A and B
A-B = 252-56 = 196
There are 196 ways to have a shipment contain at least one defective unit.
This works because
(number of ways to have no defective units) + (number of ways to have at least one defective) = number total ways
(B) + (number of ways to have at least one defective) = A
number of ways to have at least one defective = A-B
number of ways to have at least one defective = 252-56
number of ways to have at least one defective = 196
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Here's another way to arrive at the figure of 196
Consider the case of picking exactly one defective unit. There are 2C1 = 2 ways to pick the defective unit, since we have 2 to pick from and one slot to fill. I'm going to skip the steps showing the nCr combination formula, which I'll let you do.
We then pick from a pool of the 8 non-defective units to fill the remaining 4 slots
8C4 = 70 ways to pick the non-defective units.
Overall there are 2*70 = 140 ways to pick exactly one defective unit and 4 non-defective units.
Let M = 140 to be used later.
Now consider the case we select exactly 2 defective units
2C2 = 1 way to pick the two defective units
8C3 = 56 ways to pick the three non-defective units
1*56 = 56 ways to pick two defective and three non-defective
Let N = 56
M+N = 140+56 = 196 ways to have a shipment contain at least one defective unit.
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What we then do from here is divide the 196 over 252 which was the number of ways to pick a sample of 5 units (defective or not)
196/252
That fraction reduces to 7/9 when you divide both parts by the GCF 28
This is the probability of getting at least one defective item in a randomly selected sample of five, and thereby making the entire batch rejected.
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