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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 \(P\left( {R\left| {ND} \right.} \right) = 0.2\)

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

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