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Chapter 2: Q25SE (page 91)

Suppose that 30 percent of the bottles produced in a certain plant are defective. If a bottle is defective, the probability is 0.9 that an inspector will notice it and remove it from the filling line. If a bottle is not defective, the probability is 0.2 that the inspector will think that it is defective and remove it from the filling line.

a. If a bottle is removed from the filling line, what is the probability that it is defective?

b. If a customer buys a bottle that has not been removed from the filling line, what is the probability that it is defective?

Short Answer

Expert verified

a.The probability that the bottle is defective given that it has been removed from the filling line is\(P\left( {D\left| R \right.} \right) = 0.6585\)

b. The probability that defective bottle is not removed is \(P\left( {D\left| {NR} \right.} \right) = 0.0508\)

Step by step solution

01

Given information

Probability of defective bottles\(P\left( D \right) = 0.3\)\(\)

The probability that the defective bottle will be removed is\(P\left( {R\left| D \right.} \right) = 0.9\)

The probability that the not defective bottle will be removed is \(P\left( {R\left| {ND} \right.} \right) = 0.2\)

02

Computing the probability for part (a)

a.

It is given that the probability of defective bottles\(P\left( D \right) = 0.3\)

So, the probability of not defective bottle is\(P\left( {ND} \right) = 0.7\)

The probability that the bottle is defective given that it has been removed from the filling line is

\(P\left( {D\left| R \right.} \right) = \frac{{P\left( {R\left| D \right.} \right)P\left( D \right)}}{{P\left( {R\left| D \right.} \right)P\left( D \right) + P\left( {R\left| {ND} \right.} \right)P\left( {ND} \right)}}\)

\(\begin{array}{c}P\left( {D\left| R \right.} \right) = \frac{{0.9 \times 0.3}}{{0.9 \times 0.3 + 0.2 \times 0.7}}\\ = \frac{{0.27}}{{0.41}}\\ = 0.6585\end{array}\)

The probability that the bottle is defective given that it has been removed from the filling line is 0.6585

03

Computing the probability for (b)

b.

From the given information we can find the following probabilities

Probability of bottle not removed when it is defective

\(\begin{array}{c}P\left( {NR\left| D \right.} \right) = 1 - P\left( {R\left| D \right.} \right)\\ = 0.1\end{array}\)

Probability of bottle not removed when it is not defective

\(\begin{array}{c}P\left( {NR\left| {ND} \right.} \right) = 1 - P\left( {R\left| {ND} \right.} \right)\\ = 0.8\end{array}\)

The aim is to find the probability that defective bottle is not removed

\(P\left( {D\left| {NR} \right.} \right) = \frac{{P\left( {NR\left| D \right.} \right)P\left( D \right)}}{{P\left( {NR\left| D \right.} \right)P\left( D \right) + P\left( {NR\left| {ND} \right.} \right)P\left( {ND} \right)}}\)

\(P\left( {D\left| {NR} \right.} \right) = \frac{{0.03}}{\begin{array}{l}0.03 + 0.56\\ = \frac{{0.03}}{{0.59}}\\ = 0.0508\end{array}}\)

The probability that defective bottle is not removed is 0.0508

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