Property 2. Law of Total Probability

Example 1

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

Plant 1 supplies 50% 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.

Given

$P(1) = 0.50, P(2) = 0.50$
$P(L|1) = 0.030$
$P(L|2) = 0.050$

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
$P(L) = 0.030(0.50) + 0.050(0.50) = 0.04000$
Formula
$P(1|L) = P(L|1)P(1)/P(L)$
Substituting
$P(1|L) = 0.01500/0.04000 = 0.3750$

Answer:

P(L) = 0.0400; P(plant 1 | low break) = 0.375

Why the other options are there

0.0800 (rates added)
0.50 (prior reported as posterior)

Reference: FE Reference Handbook — Probability and Statistics → Property 2. Law of Total Probability

Example 2

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

Plant 1 supplies 60% of the concrete cylinders and has a 5.0% 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

$P(1) = 0.60, P(2) = 0.40$
$P(L|1) = 0.050$
$P(L|2) = 0.060$

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
$P(L) = 0.050(0.60) + 0.060(0.40) = 0.05400$
Formula
$P(1|L) = P(L|1)P(1)/P(L)$
Substituting
$P(1|L) = 0.03000/0.05400 = 0.5556$

Answer:

P(L) = 0.0540; P(plant 1 | low break) = 0.556

Why the other options are there

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

Reference: FE Reference Handbook — Probability and Statistics → Property 2. Law of Total Probability

Example 3

Law of total probability and the reverse (Bayes) question — Property 2. Law of Total Probability (3)

Plant 1 supplies 60% of the concrete cylinders and has a 6.0% 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

$P(1) = 0.60, P(2) = 0.40$
$P(L|1) = 0.060$
$P(L|2) = 0.050$

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
$P(L) = 0.060(0.60) + 0.050(0.40) = 0.05600$
Formula
$P(1|L) = P(L|1)P(1)/P(L)$
Substituting
$P(1|L) = 0.03600/0.05600 = 0.6429$

Answer:

P(L) = 0.0560; P(plant 1 | low break) = 0.643

Why the other options are there

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

Reference: FE Reference Handbook — Probability and Statistics → Property 2. Law of Total Probability

Example 4

Law of total probability and the reverse (Bayes) question — Property 2. Law of Total Probability (4)

Plant 1 supplies 50% 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

$P(1) = 0.50, P(2) = 0.50$
$P(L|1) = 0.070$
$P(L|2) = 0.070$

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
$P(L) = 0.070(0.50) + 0.070(0.50) = 0.07000$
Formula
$P(1|L) = P(L|1)P(1)/P(L)$
Substituting
$P(1|L) = 0.03500/0.07000 = 0.5000$

Answer:

P(L) = 0.0700; P(plant 1 | low break) = 0.500

Why the other options are there

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

Reference: FE Reference Handbook — Probability and Statistics → Property 2. Law of Total Probability

Example 5

Law of total probability and the reverse (Bayes) question — Property 2. Law of Total Probability (5)

Plant 1 supplies 45% of the concrete cylinders and has a 6.0% low-break rate; plant 2 supplies the rest with a 12.0% low-break rate. Find the overall probability of a low break, and the probability that a low-breaking cylinder came from plant 1.

Given

$P(1) = 0.45, P(2) = 0.55$
$P(L|1) = 0.060$
$P(L|2) = 0.120$

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
$P(L) = 0.060(0.45) + 0.120(0.55) = 0.09300$
Formula
$P(1|L) = P(L|1)P(1)/P(L)$
Substituting
$P(1|L) = 0.02700/0.09300 = 0.2903$

Answer:

P(L) = 0.0930; P(plant 1 | low break) = 0.290

Why the other options are there

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

Reference: FE Reference Handbook — Probability and Statistics → Property 2. Law of Total Probability

Example 6

Law of total probability and the reverse (Bayes) question — Property 2. Law of Total Probability (6)

Plant 1 supplies 35% of the concrete cylinders and has a 6.0% 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.

Given

$P(1) = 0.35, P(2) = 0.65$
$P(L|1) = 0.060$
$P(L|2) = 0.090$

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
$P(L) = 0.060(0.35) + 0.090(0.65) = 0.07950$
Formula
$P(1|L) = P(L|1)P(1)/P(L)$
Substituting
$P(1|L) = 0.02100/0.07950 = 0.2642$

Answer:

P(L) = 0.0795; P(plant 1 | low break) = 0.264

Why the other options are there

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

Reference: FE Reference Handbook — Probability and Statistics → Property 2. Law of Total Probability

Example 7

Law of total probability and the reverse (Bayes) question — Property 2. Law of Total Probability (7)

Plant 1 supplies 60% of the concrete cylinders and has a 5.0% 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

$P(1) = 0.60, P(2) = 0.40$
$P(L|1) = 0.050$
$P(L|2) = 0.060$

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
$P(L) = 0.050(0.60) + 0.060(0.40) = 0.05400$
Formula
$P(1|L) = P(L|1)P(1)/P(L)$
Substituting
$P(1|L) = 0.03000/0.05400 = 0.5556$

Answer:

P(L) = 0.0540; P(plant 1 | low break) = 0.556

Why the other options are there

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

Reference: FE Reference Handbook — Probability and Statistics → Property 2. Law of Total Probability

Example 8

Law of total probability and the reverse (Bayes) question — Property 2. Law of Total Probability (8)

Plant 1 supplies 60% of the concrete cylinders and has a 5.0% 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

$P(1) = 0.60, P(2) = 0.40$
$P(L|1) = 0.050$
$P(L|2) = 0.080$

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
$P(L) = 0.050(0.60) + 0.080(0.40) = 0.06200$
Formula
$P(1|L) = P(L|1)P(1)/P(L)$
Substituting
$P(1|L) = 0.03000/0.06200 = 0.4839$

Answer:

P(L) = 0.0620; P(plant 1 | low break) = 0.484

Why the other options are there

0.1300 (rates added)
0.60 (prior reported as posterior)

Reference: FE Reference Handbook — Probability and Statistics → Property 2. Law of Total Probability

Example 9

Law of total probability and the reverse (Bayes) question — Property 2. Law of Total Probability (9)

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 2. Law of Total Probability

Example 10

Law of total probability and the reverse (Bayes) question — Property 2. Law of Total Probability (10)

Plant 1 supplies 55% of the concrete cylinders and has a 3.0% 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

$P(1) = 0.55, P(2) = 0.45$
$P(L|1) = 0.030$
$P(L|2) = 0.060$

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
$P(L) = 0.030(0.55) + 0.060(0.45) = 0.04350$
Formula
$P(1|L) = P(L|1)P(1)/P(L)$
Substituting
$P(1|L) = 0.01650/0.04350 = 0.3793$

Answer:

P(L) = 0.0435; P(plant 1 | low break) = 0.379

Why the other options are there

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

Reference: FE Reference Handbook — Probability and Statistics → Property 2. Law of Total Probability

Property 2. Law of Total Probability

Handbook notes for this section

Core formulas for this FE topic

Worked exam-style examples

Property 2. Law of Total Probability

Handbook notes for this section

Core formulas for this FE topic

P(A+B)=P(A)+P(B)−P(A,B)P(A + B) = P(A) + P(B) - P(A, B)P(A+B)=P(A)+P(B)−P(A,B)

P(A+B)=theprobabilitythateitherAorBoccuraloneorthatbothoccurtogetherP(A + B) = the probability that either A or B occur alone or that both occur togetherP(A+B)=theprobabilitythateitherAorBoccuraloneorthatbothoccurtogether

P(A)=theprobabilitythatAoccursP(A) = the probability that A occursP(A)=theprobabilitythatAoccurs

P(B)=theprobabilitythatBoccursP(B) = the probability that B occursP(B)=theprobabilitythatBoccurs

P(A,B)=theprobabilitythatbothAandBoccursimultaneouslyP(A, B) = the probability that both A and B occur simultaneouslyP(A,B)=theprobabilitythatbothAandBoccursimultaneously

Worked exam-style examples