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Chapter 9: Problem 512

Over a period of some years, a car manufacturing firm finds that \(18 \%\) of their cars develop body squeaks within the guarantee period. In a randomly selected shipment, 20 cars reach the end of the guarantee period and none develop squeaks. What is the probability of this?

Short Answer

Expert verified
The probability of having none of the 20 cars develop body squeaks within the guarantee period is approximately 0.00002667 or 0.00267%.

Step by step solution

01

Identify the given values

We are given the following values: - The probability of success (developing body squeaks): p = 18% - The sample size (number of cars): n = 20 - The number of successes we want to calculate the probability for: k = 0
02

Convert the percentage to a decimal value

Convert the given percentage (18%) to a decimal value by dividing by 100: p = 18/100 = 0.18
03

Calculate the probability of failure

Since there are only two possible outcomes, the probability of failure (not developing body squeaks) can be calculated as: q = 1 - p q = 1 - 0.18 = 0.82
04

Apply the binomial probability formula

The binomial probability formula is given by: \[P(X = k) = \binom{n}{k} \cdot p^k \cdot q^{(n-k)}\] where - \(P(X = k)\) is the probability of k successes, - \(n\) is the sample size, - \(k\) is the number of successes, - \(p\) is the probability of success, and - \(q\) is the probability of failure (1-p). Plugging in the given values, we get: \[P(X = 0) = \binom{20}{0} \cdot 0.18^0 \cdot 0.82^{(20-0)}\]
05

Calculate the binomial coefficient

Calculate the binomial coefficient, \(\binom{20}{0}\), using the formula: \[\binom{n}{k} = \frac{n!}{k!(n-k)!}\] where - \(n\) is the sample size, and - \(k\) is the number of successes. For our example, this means: \[\binom{20}{0} = \frac{20!}{0!(20-0)!} = \frac{20!}{0!20!} = 1\]
06

Evaluate the probability

Now, plug the binomial coefficient and given values back into the binomial probability formula: \[P(X = 0) = 1 \cdot 0.18^0 \cdot 0.82^{20}\] \[P(X = 0) = 1 \cdot 1 \cdot 0.82^{20}\] \[P(X = 0) = 0.82^{20}\] Using a calculator, we find: \[P(X = 0) \approx 0.00002667\] Therefore, the probability of having none of the 20 cars develop body squeaks within the guarantee period is approximately 0.00002667 or 0.00267%.

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