Car windshields are inspected at the manufacturing plant before installation. Ea

Question

Car windshields are inspected at the manufacturing plant before installation. Each windshield is deemed usable or defective. Based on historical information, the manufacturer knows that the windshields have a probability of being defective of 8%. What is the probability that 3 windshields will be found defective in 20 tested? What is the probability that 2 defective windshields will be found in 4 or fewer tests? What is the probability it will take more than 3 tests before the first defective windshield is found?

Explanation / Answer

a)

Note that the probability of x successes out of n trials is          
          
P(n, x) = nCx p^x (1 - p)^(n - x)          
          
where          
          
n = number of trials =    20      
p = the probability of a success =    0.08      
x = the number of successes =    3      
          
Thus, the probability is          
          
P (    3   ) =    0.141438577 [answer]

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b)

This should mean 2 defective windshield in 2, 3, or 4 tests.

Note that the probability of x successes out of n trials is          
          
P(n, x) = nCx p^x (1 - p)^(n - x)          
          
where          
          
n = number of trials =    2      
p = the probability of a success =    0.08      
x = the number of successes =    2      
          
Thus, the probability is          
          
P ( 2 out of 2) =    0.0064

Now, if n = 3,

P(2 out of 3) = 0.017664

If n = 4,

P(2 out of 4) = 0.03250176

Thus, the probability of 4 or fewer tests is

P(4 or fewer) = 0.0064 + 0.017664 + 0.03250176 = 0.05656576 [answer]

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c)

Here,

P(more than 3) = 1 - P(1) - P(2) - P(3)

and using the geometric series to get the probabilities.

Thus,

P(more than 3) = 1 - 0.08 - (1-0.08)*0.08 - (1-0.08)^2 * 0.08

= 0.778688 [ANSWER]

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