Property 3. Law of Compound or Joint Probability

Example 1

Law of total probability and the reverse (Bayes) question — Property 3. Law of Compound or Joint Probability

Plant 1 supplies 40% 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

$P(1) = 0.40, P(2) = 0.60$
$P(L|1) = 0.080$
$P(L|2) = 0.060$

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
$P(L) = 0.080(0.40) + 0.060(0.60) = 0.06800$
Formula
$P(1|L) = P(L|1)P(1)/P(L)$
Substituting
$P(1|L) = 0.03200/0.06800 = 0.4706$

Answer:

P(L) = 0.0680; P(plant 1 | low break) = 0.471

Why the other options are there

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

Reference: FE Reference Handbook — Probability and Statistics → Property 3. Law of Compound or Joint Probability

Example 2

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

Plant 1 supplies 60% of the concrete cylinders and has a 7.0% 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

$P(1) = 0.60, P(2) = 0.40$
$P(L|1) = 0.070$
$P(L|2) = 0.060$

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
$P(L) = 0.070(0.60) + 0.060(0.40) = 0.06600$
Formula
$P(1|L) = P(L|1)P(1)/P(L)$
Substituting
$P(1|L) = 0.04200/0.06600 = 0.6364$

Answer:

P(L) = 0.0660; P(plant 1 | low break) = 0.636

Why the other options are there

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

Reference: FE Reference Handbook — Probability and Statistics → Property 3. Law of Compound or Joint Probability

Example 3

Law of total probability and the reverse (Bayes) question — Property 3. Law of Compound or Joint Probability (3)

Plant 1 supplies 50% 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

$P(1) = 0.50, P(2) = 0.50$
$P(L|1) = 0.050$
$P(L|2) = 0.060$

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
$P(L) = 0.050(0.50) + 0.060(0.50) = 0.05500$
Formula
$P(1|L) = P(L|1)P(1)/P(L)$
Substituting
$P(1|L) = 0.02500/0.05500 = 0.4545$

Answer:

P(L) = 0.0550; P(plant 1 | low break) = 0.455

Why the other options are there

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

Reference: FE Reference Handbook — Probability and Statistics → Property 3. Law of Compound or Joint Probability

Example 4

Law of total probability and the reverse (Bayes) question — Property 3. Law of Compound or Joint Probability (4)

Plant 1 supplies 60% of the concrete cylinders and has a 7.0% 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

$P(1) = 0.60, P(2) = 0.40$
$P(L|1) = 0.070$
$P(L|2) = 0.100$

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
$P(L) = 0.070(0.60) + 0.100(0.40) = 0.08200$
Formula
$P(1|L) = P(L|1)P(1)/P(L)$
Substituting
$P(1|L) = 0.04200/0.08200 = 0.5122$

Answer:

P(L) = 0.0820; P(plant 1 | low break) = 0.512

Why the other options are there

0.1700 (rates added)
0.60 (prior reported as posterior)

Reference: FE Reference Handbook — Probability and Statistics → Property 3. Law of Compound or Joint Probability

Example 5

Law of total probability and the reverse (Bayes) question — Property 3. Law of Compound or Joint Probability (5)

Plant 1 supplies 30% of the concrete cylinders and has a 2.0% 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

$P(1) = 0.30, P(2) = 0.70$
$P(L|1) = 0.020$
$P(L|2) = 0.060$

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
$P(L) = 0.020(0.30) + 0.060(0.70) = 0.04800$
Formula
$P(1|L) = P(L|1)P(1)/P(L)$
Substituting
$P(1|L) = 0.00600/0.04800 = 0.1250$

Answer:

P(L) = 0.0480; P(plant 1 | low break) = 0.125

Why the other options are there

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

Reference: FE Reference Handbook — Probability and Statistics → Property 3. Law of Compound or Joint Probability

Example 6

Law of total probability and the reverse (Bayes) question — Property 3. Law of Compound or Joint 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 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

$P(1) = 0.30, P(2) = 0.70$
$P(L|1) = 0.080$
$P(L|2) = 0.050$

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
$P(L) = 0.080(0.30) + 0.050(0.70) = 0.05900$
Formula
$P(1|L) = P(L|1)P(1)/P(L)$
Substituting
$P(1|L) = 0.02400/0.05900 = 0.4068$

Answer:

P(L) = 0.0590; P(plant 1 | low break) = 0.407

Why the other options are there

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

Reference: FE Reference Handbook — Probability and Statistics → Property 3. Law of Compound or Joint Probability

Example 7

Law of total probability and the reverse (Bayes) question — Property 3. Law of Compound or Joint Probability (7)

Plant 1 supplies 35% of the concrete cylinders and has a 4.0% 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

$P(1) = 0.35, P(2) = 0.65$
$P(L|1) = 0.040$
$P(L|2) = 0.090$

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
$P(L) = 0.040(0.35) + 0.090(0.65) = 0.07250$
Formula
$P(1|L) = P(L|1)P(1)/P(L)$
Substituting
$P(1|L) = 0.01400/0.07250 = 0.1931$

Answer:

P(L) = 0.0725; P(plant 1 | low break) = 0.193

Why the other options are there

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

Reference: FE Reference Handbook — Probability and Statistics → Property 3. Law of Compound or Joint Probability

Example 8

Law of total probability and the reverse (Bayes) question — Property 3. Law of Compound or Joint Probability (8)

Plant 1 supplies 40% of the concrete cylinders and has a 4.0% 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

$P(1) = 0.40, P(2) = 0.60$
$P(L|1) = 0.040$
$P(L|2) = 0.080$

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
$P(L) = 0.040(0.40) + 0.080(0.60) = 0.06400$
Formula
$P(1|L) = P(L|1)P(1)/P(L)$
Substituting
$P(1|L) = 0.01600/0.06400 = 0.2500$

Answer:

P(L) = 0.0640; P(plant 1 | low break) = 0.250

Why the other options are there

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

Reference: FE Reference Handbook — Probability and Statistics → Property 3. Law of Compound or Joint Probability

Example 9

Law of total probability and the reverse (Bayes) question — Property 3. Law of Compound or Joint Probability (9)

Plant 1 supplies 50% 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.

Given

$P(1) = 0.50, P(2) = 0.50$
$P(L|1) = 0.080$
$P(L|2) = 0.100$

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
$P(L) = 0.080(0.50) + 0.100(0.50) = 0.09000$
Formula
$P(1|L) = P(L|1)P(1)/P(L)$
Substituting
$P(1|L) = 0.04000/0.09000 = 0.4444$

Answer:

P(L) = 0.0900; P(plant 1 | low break) = 0.444

Why the other options are there

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

Reference: FE Reference Handbook — Probability and Statistics → Property 3. Law of Compound or Joint Probability

Example 10

Law of total probability and the reverse (Bayes) question — Property 3. Law of Compound or Joint Probability (10)

Plant 1 supplies 50% of the concrete cylinders and has a 4.0% 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.

Given

$P(1) = 0.50, P(2) = 0.50$
$P(L|1) = 0.040$
$P(L|2) = 0.110$

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
$P(L) = 0.040(0.50) + 0.110(0.50) = 0.07500$
Formula
$P(1|L) = P(L|1)P(1)/P(L)$
Substituting
$P(1|L) = 0.02000/0.07500 = 0.2667$

Answer:

P(L) = 0.0750; P(plant 1 | low break) = 0.267

Why the other options are there

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

Reference: FE Reference Handbook — Probability and Statistics → Property 3. Law of Compound or Joint Probability

Property 3. Law of Compound or Joint Probability

Core formulas for this FE topic

Worked exam-style examples

Property 3. Law of Compound or Joint Probability

Core formulas for this FE topic

P(A,B)=P(A)P(B∣A)=P(B)P(A∣B)P(A, B) = P(A)P(B | A) = P(B)P(A | B)P(A,B)=P(A)P(B∣A)=P(B)P(A∣B)

P(B∣A)=theprobabilitythatBoccursgiventhefactthatAhasoccurredP(B | A) = the probability that B occurs given the fact that A has occurredP(B∣A)=theprobabilitythatBoccursgiventhefactthatAhasoccurred

P(A∣B)=theprobabilitythatAoccursgiventhefactthatBhasoccurredP(A | B) = the probability that A occurs given the fact that B has occurredP(A∣B)=theprobabilitythatAoccursgiventhefactthatBhasoccurred

IfeitherP(A)orP(B)iszero,thenP(A,B)=0.If either P(A) or P(B) is zero, then P(A, B) = 0.IfeitherP(A)orP(B)iszero,thenP(A,B)=0.

Worked exam-style examples