Property 1. General Character of Probability

Example 1

Law of total probability and the reverse (Bayes) question — Property 1. General Character of Probability

Plant 1 supplies 50% of the concrete cylinders and has a 6.0% low-break rate; plant 2 supplies the rest with a 8.0% low-break rate. Find the overall probability of a low break, and the probability that a low-breaking cylinder came from plant 1.

Given

$P(1) = 0.50, P(2) = 0.50$
$P(L|1) = 0.060$
$P(L|2) = 0.080$

Find

P(L) by the law of total probability, then P(1|L)

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

Formula
$P(L) = P(L|1)P(1) + P(L|2)P(2)$
Substituting
$P(L) = 0.060(0.50) + 0.080(0.50) = 0.07000$
Formula
$P(1|L) = P(L|1)P(1)/P(L)$
Substituting
$P(1|L) = 0.03000/0.07000 = 0.4286$

Answer:

P(L) = 0.0700; P(plant 1 | low break) = 0.429

Why the other options are there

0.1400 (rates added)
0.50 (prior reported as posterior)

Reference: FE Reference Handbook — Probability and Statistics → Property 1. General Character of Probability

Example 2

Law of total probability and the reverse (Bayes) question — Property 1. General Character of Probability (2)

Plant 1 supplies 50% of the concrete cylinders and has a 4.0% low-break rate; plant 2 supplies the rest with a 3.0% low-break rate. Find the overall probability of a low break, and the probability that a low-breaking cylinder came from plant 1.

Given

$P(1) = 0.50, P(2) = 0.50$
$P(L|1) = 0.040$
$P(L|2) = 0.030$

Find

P(L) by the law of total probability, then P(1|L)

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

Formula
$P(L) = P(L|1)P(1) + P(L|2)P(2)$
Substituting
$P(L) = 0.040(0.50) + 0.030(0.50) = 0.03500$
Formula
$P(1|L) = P(L|1)P(1)/P(L)$
Substituting
$P(1|L) = 0.02000/0.03500 = 0.5714$

Answer:

P(L) = 0.0350; P(plant 1 | low break) = 0.571

Why the other options are there

0.0700 (rates added)
0.50 (prior reported as posterior)

Reference: FE Reference Handbook — Probability and Statistics → Property 1. General Character of Probability

Example 3

Law of total probability and the reverse (Bayes) question — Property 1. General Character of Probability (3)

Plant 1 supplies 35% of the concrete cylinders and has a 5.0% low-break rate; plant 2 supplies the rest with a 4.0% low-break rate. Find the overall probability of a low break, and the probability that a low-breaking cylinder came from plant 1.

Given

$P(1) = 0.35, P(2) = 0.65$
$P(L|1) = 0.050$
$P(L|2) = 0.040$

Find

P(L) by the law of total probability, then P(1|L)

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

Formula
$P(L) = P(L|1)P(1) + P(L|2)P(2)$
Substituting
$P(L) = 0.050(0.35) + 0.040(0.65) = 0.04350$
Formula
$P(1|L) = P(L|1)P(1)/P(L)$
Substituting
$P(1|L) = 0.01750/0.04350 = 0.4023$

Answer:

P(L) = 0.0435; P(plant 1 | low break) = 0.402

Why the other options are there

0.0900 (rates added)
0.35 (prior reported as posterior)

Reference: FE Reference Handbook — Probability and Statistics → Property 1. General Character of Probability

Example 4

Law of total probability and the reverse (Bayes) question — Property 1. General Character of Probability (4)

Plant 1 supplies 50% of the concrete cylinders and has a 5.0% 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.

Given

$P(1) = 0.50, P(2) = 0.50$
$P(L|1) = 0.050$
$P(L|2) = 0.050$

Find

P(L) by the law of total probability, then P(1|L)

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

Formula
$P(L) = P(L|1)P(1) + P(L|2)P(2)$
Substituting
$P(L) = 0.050(0.50) + 0.050(0.50) = 0.05000$
Formula
$P(1|L) = P(L|1)P(1)/P(L)$
Substituting
$P(1|L) = 0.02500/0.05000 = 0.5000$

Answer:

P(L) = 0.0500; P(plant 1 | low break) = 0.500

Why the other options are there

0.1000 (rates added)
0.50 (prior reported as posterior)

Reference: FE Reference Handbook — Probability and Statistics → Property 1. General Character of Probability

Example 5

Law of total probability and the reverse (Bayes) question — Property 1. General Character of Probability (5)

Plant 1 supplies 50% of the concrete cylinders and has a 5.0% low-break rate; plant 2 supplies the rest with a 4.0% low-break rate. Find the overall probability of a low break, and the probability that a low-breaking cylinder came from plant 1.

Given

$P(1) = 0.50, P(2) = 0.50$
$P(L|1) = 0.050$
$P(L|2) = 0.040$

Find

P(L) by the law of total probability, then P(1|L)

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

Formula
$P(L) = P(L|1)P(1) + P(L|2)P(2)$
Substituting
$P(L) = 0.050(0.50) + 0.040(0.50) = 0.04500$
Formula
$P(1|L) = P(L|1)P(1)/P(L)$
Substituting
$P(1|L) = 0.02500/0.04500 = 0.5556$

Answer:

P(L) = 0.0450; P(plant 1 | low break) = 0.556

Why the other options are there

0.0900 (rates added)
0.50 (prior reported as posterior)

Reference: FE Reference Handbook — Probability and Statistics → Property 1. General Character of Probability

Example 6

Law of total probability and the reverse (Bayes) question — Property 1. General Character of Probability (6)

Plant 1 supplies 30% 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.

Given

$P(1) = 0.30, P(2) = 0.70$
$P(L|1) = 0.080$
$P(L|2) = 0.060$

Find

P(L) by the law of total probability, then P(1|L)

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

Formula
$P(L) = P(L|1)P(1) + P(L|2)P(2)$
Substituting
$P(L) = 0.080(0.30) + 0.060(0.70) = 0.06600$
Formula
$P(1|L) = P(L|1)P(1)/P(L)$
Substituting
$P(1|L) = 0.02400/0.06600 = 0.3636$

Answer:

P(L) = 0.0660; P(plant 1 | low break) = 0.364

Why the other options are there

0.1400 (rates added)
0.30 (prior reported as posterior)

Reference: FE Reference Handbook — Probability and Statistics → Property 1. General Character of Probability

Example 7

Law of total probability and the reverse (Bayes) question — Property 1. General Character of Probability (7)

Plant 1 supplies 60% of the concrete cylinders and has a 7.0% low-break rate; plant 2 supplies the rest with a 7.0% low-break rate. Find the overall probability of a low break, and the probability that a low-breaking cylinder came from plant 1.

Given

$P(1) = 0.60, P(2) = 0.40$
$P(L|1) = 0.070$
$P(L|2) = 0.070$

Find

P(L) by the law of total probability, then P(1|L)

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

Formula
$P(L) = P(L|1)P(1) + P(L|2)P(2)$
Substituting
$P(L) = 0.070(0.60) + 0.070(0.40) = 0.07000$
Formula
$P(1|L) = P(L|1)P(1)/P(L)$
Substituting
$P(1|L) = 0.04200/0.07000 = 0.6000$

Answer:

P(L) = 0.0700; P(plant 1 | low break) = 0.600

Why the other options are there

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

Reference: FE Reference Handbook — Probability and Statistics → Property 1. General Character of Probability

Example 8

Law of total probability and the reverse (Bayes) question — Property 1. General Character of Probability (8)

Plant 1 supplies 45% of the concrete cylinders and has a 7.0% low-break rate; plant 2 supplies the rest with a 4.0% low-break rate. Find the overall probability of a low break, and the probability that a low-breaking cylinder came from plant 1.

Given

$P(1) = 0.45, P(2) = 0.55$
$P(L|1) = 0.070$
$P(L|2) = 0.040$

Find

P(L) by the law of total probability, then P(1|L)

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

Formula
$P(L) = P(L|1)P(1) + P(L|2)P(2)$
Substituting
$P(L) = 0.070(0.45) + 0.040(0.55) = 0.05350$
Formula
$P(1|L) = P(L|1)P(1)/P(L)$
Substituting
$P(1|L) = 0.03150/0.05350 = 0.5888$

Answer:

P(L) = 0.0535; P(plant 1 | low break) = 0.589

Why the other options are there

0.1100 (rates added)
0.45 (prior reported as posterior)

Reference: FE Reference Handbook — Probability and Statistics → Property 1. General Character of Probability

Example 9

Law of total probability and the reverse (Bayes) question — Property 1. General Character of Probability (9)

Plant 1 supplies 55% of the concrete cylinders and has a 6.0% 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.

Given

$P(1) = 0.55, P(2) = 0.45$
$P(L|1) = 0.060$
$P(L|2) = 0.060$

Find

P(L) by the law of total probability, then P(1|L)

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

Formula
$P(L) = P(L|1)P(1) + P(L|2)P(2)$
Substituting
$P(L) = 0.060(0.55) + 0.060(0.45) = 0.06000$
Formula
$P(1|L) = P(L|1)P(1)/P(L)$
Substituting
$P(1|L) = 0.03300/0.06000 = 0.5500$

Answer:

P(L) = 0.0600; P(plant 1 | low break) = 0.550

Why the other options are there

0.1200 (rates added)
0.55 (prior reported as posterior)

Reference: FE Reference Handbook — Probability and Statistics → Property 1. General Character of Probability

Example 10

Law of total probability and the reverse (Bayes) question — Property 1. General Character of Probability (10)

Plant 1 supplies 35% of the concrete cylinders and has a 5.0% 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.

Given

$P(1) = 0.35, P(2) = 0.65$
$P(L|1) = 0.050$
$P(L|2) = 0.100$

Find

P(L) by the law of total probability, then P(1|L)

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

Formula
$P(L) = P(L|1)P(1) + P(L|2)P(2)$
Substituting
$P(L) = 0.050(0.35) + 0.100(0.65) = 0.08250$
Formula
$P(1|L) = P(L|1)P(1)/P(L)$
Substituting
$P(1|L) = 0.01750/0.08250 = 0.2121$

Answer:

P(L) = 0.0825; P(plant 1 | low break) = 0.212

Why the other options are there

0.1500 (rates added)
0.35 (prior reported as posterior)

Reference: FE Reference Handbook — Probability and Statistics → Property 1. General Character of Probability

Property 1. General Character of Probability

Core formulas for this FE topic

Worked exam-style examples