Two independent inspections fail with P(A) = 0.40 and P(B) = 0.25. Find P(A ∪ B) and P(A|B).

Start with the thinking

• Independence gives the intersection by multiplication.
• Independence also means conditioning changes nothing.

Step-by-step solution

1. Intersection

   $$P(A \cap B) = P(A)P(B)$$

2. Substituting

   $$P(A \cap B) = 0.40 \times 0.25 = 0.1000$$

3. Union

   $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

4. Substituting

   $$P(A \cup B) = 0.40 + 0.25 - 0.1000 = 0.5500$$

5. Conditional

   $$P(A|B) = P(A \cap B)/P(B) = 0.1000/0.25 = 0.40$$

Why the other options are there

• 0.65 (intersection not subtracted)
• 0.100 (intersection reported as the union)

Reference: FE Reference Handbook — Probability and Statistics → Bayes' Theorem

Plant 1 supplies 45% of the concrete cylinders and has a 7.0% low-break rate; plant 2 supplies the rest with a 11.0% low-break rate. Find the overall probability of a low break, and the probability that a low-breaking cylinder came from plant 1.

Start with the thinking

• The law of total probability sums the joint probability over every mutually exclusive source.
• The reverse conditional is a joint probability divided by that total.

Step-by-step solution

1. Formula

   $$P(L) = P(L|1)P(1) + P(L|2)P(2)$$

2. Substituting

   $$P(L) = 0.070(0.45) + 0.110(0.55) = 0.09200$$

3. Formula

   $$P(1|L) = P(L|1)P(1)/P(L)$$

4. Substituting

   $$P(1|L) = 0.03150/0.09200 = 0.3424$$

Why the other options are there

• 0.1800 (rates added)
• 0.45 (prior reported as posterior)

Reference: FE Reference Handbook — Probability and Statistics → Bayes' Theorem

Two independent inspections fail with P(A) = 0.30 and P(B) = 0.50. Find P(A ∪ B) and P(A|B).

Start with the thinking

• Independence gives the intersection by multiplication.
• Independence also means conditioning changes nothing.

Step-by-step solution

1. Intersection

   $$P(A \cap B) = P(A)P(B)$$

2. Substituting

   $$P(A \cap B) = 0.30 \times 0.50 = 0.1500$$

3. Union

   $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

4. Substituting

   $$P(A \cup B) = 0.30 + 0.50 - 0.1500 = 0.6500$$

5. Conditional

   $$P(A|B) = P(A \cap B)/P(B) = 0.1500/0.50 = 0.30$$

Why the other options are there

• 0.80 (intersection not subtracted)
• 0.150 (intersection reported as the union)

Reference: FE Reference Handbook — Probability and Statistics → Bayes' Theorem

Plant 1 supplies 55% of the concrete cylinders and has a 8.0% low-break rate; plant 2 supplies the rest with a 10.0% low-break rate. Find the overall probability of a low break, and the probability that a low-breaking cylinder came from plant 1.

Start with the thinking

• The law of total probability sums the joint probability over every mutually exclusive source.
• The reverse conditional is a joint probability divided by that total.

Step-by-step solution

1. Formula

   $$P(L) = P(L|1)P(1) + P(L|2)P(2)$$

2. Substituting

   $$P(L) = 0.080(0.55) + 0.100(0.45) = 0.08900$$

3. Formula

   $$P(1|L) = P(L|1)P(1)/P(L)$$

4. Substituting

   $$P(1|L) = 0.04400/0.08900 = 0.4944$$

Why the other options are there

• 0.1800 (rates added)
• 0.55 (prior reported as posterior)

Reference: FE Reference Handbook — Probability and Statistics → Bayes' Theorem

Two independent inspections fail with P(A) = 0.25 and P(B) = 0.40. Find P(A ∪ B) and P(A|B).

Start with the thinking

• Independence gives the intersection by multiplication.
• Independence also means conditioning changes nothing.

Step-by-step solution

1. Intersection

   $$P(A \cap B) = P(A)P(B)$$

2. Substituting

   $$P(A \cap B) = 0.25 \times 0.40 = 0.1000$$

3. Union

   $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

4. Substituting

   $$P(A \cup B) = 0.25 + 0.40 - 0.1000 = 0.5500$$

5. Conditional

   $$P(A|B) = P(A \cap B)/P(B) = 0.1000/0.40 = 0.25$$

Why the other options are there

• 0.65 (intersection not subtracted)
• 0.100 (intersection reported as the union)

Reference: FE Reference Handbook — Probability and Statistics → Bayes' Theorem

Plant 1 supplies 35% of the concrete cylinders and has a 8.0% low-break rate; plant 2 supplies the rest with a 9.0% low-break rate. Find the overall probability of a low break, and the probability that a low-breaking cylinder came from plant 1.

Start with the thinking

• The law of total probability sums the joint probability over every mutually exclusive source.
• The reverse conditional is a joint probability divided by that total.

Step-by-step solution

1. Formula

   $$P(L) = P(L|1)P(1) + P(L|2)P(2)$$

2. Substituting

   $$P(L) = 0.080(0.35) + 0.090(0.65) = 0.08650$$

3. Formula

   $$P(1|L) = P(L|1)P(1)/P(L)$$

4. Substituting

   $$P(1|L) = 0.02800/0.08650 = 0.3237$$

Why the other options are there

• 0.1700 (rates added)
• 0.35 (prior reported as posterior)

Reference: FE Reference Handbook — Probability and Statistics → Bayes' Theorem

Two independent inspections fail with P(A) = 0.35 and P(B) = 0.20. Find P(A ∪ B) and P(A|B).

Start with the thinking

• Independence gives the intersection by multiplication.
• Independence also means conditioning changes nothing.

Step-by-step solution

1. Intersection

   $$P(A \cap B) = P(A)P(B)$$

2. Substituting

   $$P(A \cap B) = 0.35 \times 0.20 = 0.0700$$

3. Union

   $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

4. Substituting

   $$P(A \cup B) = 0.35 + 0.20 - 0.0700 = 0.4800$$

5. Conditional

   $$P(A|B) = P(A \cap B)/P(B) = 0.0700/0.20 = 0.35$$

Why the other options are there

• 0.55 (intersection not subtracted)
• 0.070 (intersection reported as the union)

Reference: FE Reference Handbook — Probability and Statistics → Bayes' Theorem

Plant 1 supplies 40% of the concrete cylinders and has a 3.0% low-break rate; plant 2 supplies the rest with a 5.0% low-break rate. Find the overall probability of a low break, and the probability that a low-breaking cylinder came from plant 1.

Start with the thinking

• The law of total probability sums the joint probability over every mutually exclusive source.
• The reverse conditional is a joint probability divided by that total.

Step-by-step solution

1. Formula

   $$P(L) = P(L|1)P(1) + P(L|2)P(2)$$

2. Substituting

   $$P(L) = 0.030(0.40) + 0.050(0.60) = 0.04200$$

3. Formula

   $$P(1|L) = P(L|1)P(1)/P(L)$$

4. Substituting

   $$P(1|L) = 0.01200/0.04200 = 0.2857$$

Why the other options are there

• 0.0800 (rates added)
• 0.40 (prior reported as posterior)

Reference: FE Reference Handbook — Probability and Statistics → Bayes' Theorem

Two independent inspections fail with P(A) = 0.40 and P(B) = 0.30. Find P(A ∪ B) and P(A|B).

Start with the thinking

• Independence gives the intersection by multiplication.
• Independence also means conditioning changes nothing.

Step-by-step solution

1. Intersection

   $$P(A \cap B) = P(A)P(B)$$

2. Substituting

   $$P(A \cap B) = 0.40 \times 0.30 = 0.1200$$

3. Union

   $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

4. Substituting

   $$P(A \cup B) = 0.40 + 0.30 - 0.1200 = 0.5800$$

5. Conditional

   $$P(A|B) = P(A \cap B)/P(B) = 0.1200/0.30 = 0.40$$

Why the other options are there

• 0.70 (intersection not subtracted)
• 0.120 (intersection reported as the union)

Reference: FE Reference Handbook — Probability and Statistics → Bayes' Theorem

Plant 1 supplies 60% of the concrete cylinders and has a 8.0% low-break rate; plant 2 supplies the rest with a 6.0% low-break rate. Find the overall probability of a low break, and the probability that a low-breaking cylinder came from plant 1.

Start with the thinking

• The law of total probability sums the joint probability over every mutually exclusive source.
• The reverse conditional is a joint probability divided by that total.

Step-by-step solution

1. Formula

   $$P(L) = P(L|1)P(1) + P(L|2)P(2)$$

2. Substituting

   $$P(L) = 0.080(0.60) + 0.060(0.40) = 0.07200$$

3. Formula

   $$P(1|L) = P(L|1)P(1)/P(L)$$

4. Substituting

   $$P(1|L) = 0.04800/0.07200 = 0.6667$$

Why the other options are there

• 0.1400 (rates added)
• 0.60 (prior reported as posterior)

Reference: FE Reference Handbook — Probability and Statistics → Bayes' Theorem