Math Problem Statement

Question number 4.

A recording company obtains the blank CDs used to produce its labels from three compact disk manufacturers: I, II, and III. The quality control department of the company has determined that 7% of the compact disks produced by manufacturer I are defective, 5% of those produced by manufacturer II are defective, and 4% of those produced by manufacturer III are defective. Manufacturers I, II, and III supply 37%, 45%, and 18%, respectively, of the compact disks used by the company. What is the probability that a randomly selected label produced by the company will contain a defective compact disk? 0.0225

0.0259

0.0556

0.1600

0.0072

None of the above.

Question number 5.

There are three colored cookie jars. One jar is blue, another green and the last one pink. The blue jar contains 8 chocolate chip and 12 sugar cookies. The green jar contains 5 chocolate chip, 10 sugar and 7 peanut butter cookies. The pink jar contains 14 chocolate chip, 13 sugar and 15 peanut butter cookies. One of the three cookie jars is chosen at random. The probabilities that the blue jar, green jar, or pink jar will be chosen are 1/2, 1/4, and 1/4 respectively. A cookie is then chosen at random from the chosen jar. What is the probability that the pink jar was chosen, if it is known that the cookie was a sugar cookie? 0.3095

0.1576

0.2269

0.0774

0.1032

None of the above.

Solution

Solution to Question 4:

The problem asks for the probability that a randomly selected label contains a defective compact disk.

  1. Determine the contribution from each manufacturer:

    • Manufacturer I: 0.370.37 (supply probability) × 0.070.07 (defective rate) = 0.02590.0259
    • Manufacturer II: 0.450.45 × 0.050.05 = 0.02250.0225
    • Manufacturer III: 0.180.18 × 0.040.04 = 0.00720.0072
  2. Add these contributions together: P(Defective)=0.0259+0.0225+0.0072=0.0556P(\text{Defective}) = 0.0259 + 0.0225 + 0.0072 = 0.0556

Thus, the probability that a randomly selected label contains a defective compact disk is 0.05560.0556.
The correct answer is: 0.0556.


Solution to Question 5:

We need to calculate the probability that the pink jar was chosen given that the cookie selected is a sugar cookie.

  1. Step 1: Calculate total probabilities for sugar cookies from each jar:

    • Blue jar: P(Blue)=12P(\text{Blue}) = \frac{1}{2}, P(Sugar | Blue)=1220P(\text{Sugar | Blue}) = \frac{12}{20}, so P(Sugar and Blue)=12×1220=620=0.3P(\text{Sugar and Blue}) = \frac{1}{2} \times \frac{12}{20} = \frac{6}{20} = 0.3.
    • Green jar: P(Green)=14P(\text{Green}) = \frac{1}{4}, P(Sugar | Green)=1022P(\text{Sugar | Green}) = \frac{10}{22}, so P(Sugar and Green)=14×1022=1088=0.1136P(\text{Sugar and Green}) = \frac{1}{4} \times \frac{10}{22} = \frac{10}{88} = 0.1136.
    • Pink jar: P(Pink)=14P(\text{Pink}) = \frac{1}{4}, P(Sugar | Pink)=1342P(\text{Sugar | Pink}) = \frac{13}{42}, so P(Sugar and Pink)=14×1342=13168=0.0774P(\text{Sugar and Pink}) = \frac{1}{4} \times \frac{13}{42} = \frac{13}{168} = 0.0774.
  2. Step 2: Total probability of selecting a sugar cookie: P(Sugar)=0.3+0.1136+0.0774=0.491P(\text{Sugar}) = 0.3 + 0.1136 + 0.0774 = 0.491

  3. Step 3: Use Bayes’ theorem to find P(Pink | Sugar)P(\text{Pink | Sugar}): P(Pink | Sugar)=P(Sugar and Pink)P(Sugar)=0.07740.4910.1576P(\text{Pink | Sugar}) = \frac{P(\text{Sugar and Pink})}{P(\text{Sugar})} = \frac{0.0774}{0.491} \approx 0.1576

Thus, the probability that the pink jar was chosen given that the cookie is a sugar cookie is 0.15760.1576.
The correct answer is: 0.1576.


Follow-Up Questions:

  1. Can you explain the difference between conditional and unconditional probabilities?
  2. How would the probabilities change in Question 5 if the pink jar had an additional sugar cookie?
  3. What is Bayes' theorem and how is it applied to real-world situations?
  4. If Manufacturer III’s defect rate changes, how would that affect the result in Question 4?
  5. Can you calculate the expected number of defective CDs if 1,000 CDs are ordered?

Tip: For problems involving probability trees, visualizing the branches with their probabilities helps clarify the relationships.

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Math Problem Analysis

Mathematical Concepts

Probability
Weighted Averages
Bayes' Theorem

Formulas

P(Defective) = P(Manufacturer I) * P(Defective | Manufacturer I) + P(Manufacturer II) * P(Defective | Manufacturer II) + P(Manufacturer III) * P(Defective | Manufacturer III)

Theorems

Law of Total Probability

Suitable Grade Level

Grades 10-12