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The Greens Manufacturing Company makes [tex]$60 \%$[/tex] of a particular type of camera lens, the Parsons Company makes [tex]$15 \%$[/tex] of them, and the Ratten Company makes the remaining [tex]$25 \%$[/tex]. Of all the camera lenses, [tex]$5 \%$[/tex] are made by Greens and are defective, [tex]$10 \%$[/tex] are made by Parsons and are defective, and [tex]$6 \%$[/tex] are made by Ratten and are defective.

\begin{tabular}{|c|c|c|c|}
\hline Lens & Defective & Not Defective & Total \\
\hline Greens & [tex]$5 \%$[/tex] & [tex]$55 \%$[/tex] & [tex]$60 \%$[/tex] \\
\hline Parsons & [tex]$10 \%$[/tex] & [tex]$5 \%$[/tex] & [tex]$15 \%$[/tex] \\
\hline Ratten & [tex]$6 \%$[/tex] & [tex]$19 \%$[/tex] & [tex]$25 \%$[/tex] \\
\hline Total & [tex]$21 \%$[/tex] & [tex]$79 \%$[/tex] & [tex]$100 \%$[/tex] \\
\hline
\end{tabular}

If a camera lens is randomly selected from the general population of all camera lenses, what is the probability that it was made by the Greens Manufacturing Company, given that it is defective?

A. 0.24

B. 0.60

C. 0.71

D. 0.83

Sagot :

GinnyAnsweridn
Let's solve the question step by step.

First, we need to determine the probability that a lens randomly selected is defective. We'll use the provided information about the proportion of lenses each company produces and the probability that a lens from each company is defective.

The probabilities are as follows:
- Probability that a lens is made by Greens: [tex]\( P(\text{Greens}) = 0.60 \)[/tex]
- Probability that a lens is made by Parsons: [tex]\( P(\text{Parsons}) = 0.15 \)[/tex]
- Probability that a lens is made by Ratten: [tex]\( P(\text{Ratten}) = 0.25 \)[/tex]

The probabilities that a lens made by each company is defective are:
- Probability that a lens made by Greens is defective: [tex]\( P(\text{Def} | \text{Greens}) = 0.05 \)[/tex]
- Probability that a lens made by Parsons is defective: [tex]\( P(\text{Def} | \text{Parsons}) = 0.10 \)[/tex]
- Probability that a lens made by Ratten is defective: [tex]\( P(\text{Def} | \text{Ratten}) = 0.06 \)[/tex]

Next, we calculate the total probability that a lens is defective, [tex]\( P(\text{Def}) \)[/tex]. This can be found using the law of total probability:
[tex]\[ P(\text{Def}) = P(\text{Greens}) \cdot P(\text{Def} | \text{Greens}) + P(\text{Parsons}) \cdot P(\text{Def} | \text{Parsons}) + P(\text{Ratten}) \cdot P(\text{Def} | \text{Ratten}) \][/tex]

Substituting in the values:
[tex]\[ P(\text{Def}) = 0.60 \cdot 0.05 + 0.15 \cdot 0.10 + 0.25 \cdot 0.06 \][/tex]
[tex]\[ P(\text{Def}) = 0.03 + 0.015 + 0.015 \][/tex]
[tex]\[ P(\text{Def}) = 0.06 \][/tex]

Now, we need to find the conditional probability that a defective lens was made by Greens, [tex]\( P(\text{Greens} | \text{Def}) \)[/tex]. To do this, we'll use Bayes' theorem:
[tex]\[ P(\text{Greens} | \text{Def}) = \frac{P(\text{Def} | \text{Greens}) \cdot P(\text{Greens})}{P(\text{Def})} \][/tex]

Substituting in the values:
[tex]\[ P(\text{Greens} | \text{Def}) = \frac{0.05 \cdot 0.60}{0.06} \][/tex]
[tex]\[ P(\text{Greens} | \text{Def}) = \frac{0.03}{0.06} \][/tex]
[tex]\[ P(\text{Greens} | \text{Def}) = 0.50 \][/tex]

Now we need to compare this result with the given choices to determine the correct answer. The choices are:
A. 0.24
B. 0.60
C. 0.71
D. 0.83

The closest value to our calculated probability, 0.50, in the given options is:
B. 0.60

Therefore, the correct answer is:
[tex]\( \boxed{2} \)[/tex]