Question

1. in a factory that produces a certain type of item there are three machines, we refer to them as machine a, machine b, and machine c. 65% of the total production of the factory comes from machine a, 25% comes from machine b and 10% from machine c. some of the items produced at the factory are defective, and it is known that 5% of the items produced by machine a are defective, 3% of the items produced by machine b are defective and 1% of the items produced machine c are defective.

(a) (6) summarize this information in a tree diagram.
(b) (3) an item produced by the factory is inspected and it is found to be defective. what is the probability that it was made by machine a, what is the probability that it was made by machine b, and what is the probability that it was made by machine c?

1. (exam 1 bonus) there are two independent events a and b in a sample space s such that p(a) = 0.8 and p(b) = 0.3.

(a) (4) draw a venn diagram and label the non - overlapping regions with the correct probabilities.
(b) (2) determine the following:
i. p((a∩b)∪(a∩b))
ii. p(a|b)

1. (exam 1 bonus) a and b are events in a sample space s such that,

• p(a) = 0.4
• p(a∩b) = 0.42
• p(a∩b) = 0.25

(a) (4) draw a venn diagram and label the non - overlapping regions with the correct probabilities.
(b) (2) determine the following:
i. p(a|b)
ii. p(b|a)

Answer:

3.

(a)

1. First - level branches:

• The first - level branches of the tree diagram represent the three machines. Let \(P(A)=0.65\), \(P(B) = 0.25\), and \(P(C)=0.10\), where \(A\), \(B\), and \(C\) are the events that an item is produced by machine \(A\), machine \(B\), and machine \(C\) respectively.
• From each of these branches, there are two second - level branches representing defective and non - defective items.
• For machine \(A\), the probability of a defective item \(P(D|A)=0.05\) and the probability of a non - defective item \(P(\overline{D}|A)=1 - 0.05=0.95\).
• For machine \(B\), \(P(D|B)=0.03\) and \(P(\overline{D}|B)=1 - 0.03 = 0.97\).
• For machine \(C\), \(P(D|C)=0.01\) and \(P(\overline{D}|C)=1 - 0.01=0.99\).

(b)

1. Use Bayes' theorem:

• First, find the probability of a defective item \(P(D)\) using the law of total probability.
• By the law of total probability \(P(D)=P(D|A)P(A)+P(D|B)P(B)+P(D|C)P(C)\).
• Substitute the values: \(P(D)=(0.05\times0.65)+(0.03\times0.25)+(0.01\times0.10)\).
• Calculate each term: \(0.05\times0.65 = 0.0325\), \(0.03\times0.25=0.0075\), \(0.01\times0.10 = 0.001\).
• Then \(P(D)=0.0325 + 0.0075+0.001=0.041\).
• Now, use Bayes' theorem to find \(P(A|D)\), \(P(B|D)\), and \(P(C|D)\).
• By Bayes' theorem \(P(A|D)=\frac{P(D|A)P(A)}{P(D)}\). Substitute the values: \(P(A|D)=\frac{0.05\times0.65}{0.041}=\frac{0.0325}{0.041}\approx0.7927\).
• \(P(B|D)=\frac{P(D|B)P(B)}{P(D)}=\frac{0.03\times0.25}{0.041}=\frac{0.0075}{0.041}\approx0.1829\).
• \(P(C|D)=\frac{P(D|C)P(C)}{P(D)}=\frac{0.01\times0.10}{0.041}=\frac{0.001}{0.041}\approx0.0244\).

4.

(a)

1. Given probabilities:

• Given \(P(A) = 0.8\) and \(P(B')=0.3\), then \(P(B)=1 - P(B')=0.7\).
• Since \(A\) and \(B\) are independent, \(P(A\cap B)=P(A)P(B)=0.8\times0.7 = 0.56\), \(P(A\cap B')=P(A)P(B')=0.8\times0.3 = 0.24\), \(P(A'\cap B)=P(A')P(B)=(1 - 0.8)\times0.7=0.14\), and \(P(A'\cap B')=P(A')P(B')=(1 - 0.8)\times0.3 = 0.06\).
• In the Venn - diagram, label the region of \(A\cap B\) with \(0.56\), \(A\cap B'\) with \(0.24\), \(A'\cap B\) with \(0.14\), and \(A'\cap B'\) with \(0.06\).

(b)

1. i. Calculate \(P((A\cap B')\cup(A'\cap B))\):

• Since \((A\cap B')\) and \((A'\cap B)\) are mutually - exclusive, \(P((A\cap B')\cup(A'\cap B))=P(A\cap B')+P(A'\cap B)\).
• Substitute the values: \(P(A\cap B') = 0.24\) and \(P(A'\cap B)=0.14\), so \(P((A\cap B')\cup(A'\cap B))=0.24 + 0.14=0.38\).

1. ii. Calculate \(P(A'|B')\):

• Since \(A\) and \(B\) are independent, \(A'\) and \(B'\) are also independent. So \(P(A'|B')=P(A')=1 - P(A)=0.2\).

5.

(a)

1. Given probabilities and relationships:

• Given \(P(A') = 0.4\), so \(P(A)=1 - 0.4 = 0.6\). Given \(P(A\cap B')=0.42\) and \(P(A'\cap B')=0.25\).
• We know that \(P(B')=P(A\cap B')+P(A'\cap B')=0.42 + 0.25=0.67\), then \(P(B)=1 - 0.67 = 0.33\).
• \(P(A\cap B)=P(A)-P(A\cap B')=0.6 - 0.42 = 0.18\), \(P(A'\cap B)=P(A')-P(A'\cap B')=0.4 - 0.25 = 0.15\).
• In the Venn - diagram, label the region of \(A\cap B\) with \(0.18\), \(A\cap B'\) with \(0.42\), \(A'\cap B\) with \(0.15\), and \(A'\cap B'\) with \(0.25\).

(b)

1. i. Calculate \(P(A|B')\):

• By the formula for conditional probability \(P(A|B')=\frac{P(A\cap B')}{P(B')}\).
• Since \(P(A\cap B') = 0.42\) and \(P(B')=0.67\), \(P(A|B')=\frac{0.42}{0.67}\approx0.6269\).

1. ii. Calculate \(P(B|A)\):

• By the formula for conditional probability \(P(B|A)=\frac{P(A\cap B)}{P(A)}\).
• Since \(P(A\cap B)=0.18\) and \(P(A)=0.6\), \(P(B|A)=\frac{0.18}{0.6}=0.3\).