Car windshields are inspected at the manufacturing plant before installation. Ea
ID: 3440554 • Letter: C
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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