A glass manufacturer finds that 1 in every 1000 glass items produced is warped. Find the probability that​ (a) the first warped glass item is the 12th item​ produced, (b) the first warped item is the​ first, second, or third item​ produced, and​ (c) none of the first 10 glass items produced are defective.

Answer :

Answer:

(a) The probability that the first warped glass item is the 12th item produced is [tex]0.0009890[/tex]

              (b) The probability that the first warped item is the first, second, or third item produced is [tex]0.002997001[/tex].

             (c) The probability that none of the first 10 glass items produced are defective is [tex]0.9900488503187[/tex].

Step-by-step explanation:

From the question,

Glass manufacture found that 1 in every 1000 glass items produced is warped.

So,               P(1 in every 1000) = [tex]\frac{1}{1000}[/tex] = [tex]0.001[/tex]

Now,

(a) Find the probability that the first warped glass item is the 12th item produced.

Here, There are occurence of data in a successive order so we use the Geometric Distribution.

Formula of Geometric Distribution = [tex](1-P)^{(x-1)}\times P[/tex]

                                                                                 (Where P is the probability)

Now,

         P(12) = [tex](1-0.001)^{(12-1)} \times 0.001[/tex]

         P(12) = 0.0009890

(b) Find the probability that the first warped item is the first, second, or third item produced.

P(1) = [tex](1-0.001)^{0} \times 0.001[/tex] = [tex]0.001[/tex]

P(2) = [tex](1-0.001)^{1} \times 0.001 = 0.000999[/tex]

P(3) = [tex](1-0.001)^{2} \times 0.001 = 0.000998001[/tex]

Use the addition rule;

[tex]P(X\leq 3) = P(1) +P(2)+P(3)[/tex]

               = [tex]0.001+0.000999+0.000998001[/tex]

              = [tex]0.002997001[/tex]

(c) Find the probability that none of the first 10 glass items produced are defective.

P(1) = [tex](1-0.001)^{0}\times 0.001 = 0.001[/tex]

P(2) = [tex](1-0.001)^{1}\times 0.001 = 0.000999[/tex]

using the same formula,

P(3) = [tex]0.000998001[/tex]

P(4) = [tex]0.00099700299[/tex]

P(5) = [tex]0.0009960059960[/tex]

P(6) = [tex]0.0009950099900[/tex]

P(7) = [tex]0.0009930209650[/tex]

P(8) = [tex]0.0009920279440[/tex]

P(9) = [tex]0.0009910359161[/tex]

P(10) = [tex]0.0009900448802[/tex]

Using addition rule

[tex]P(X \leq10 )= P(1)+P(2)+P(3)+P(4)+P(5)+P(6)+P(7)+P(8)+P(9)+P(10)[/tex]

                  = [tex]0.0099511496813[/tex]

Using the complement rule

P(X > 10) = [tex]1- P(X\leq 10)[/tex]

              = [tex]1-0.0099511496813[/tex]

              = [tex]0.9900488503187[/tex]

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