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Conditional Probability and Bayes Theorem

Conditional probability answers: "Given that something already happened, how does that change the odds?" $P(A|B)$ reads as "the probability of A given that B already occurred." New information reduces the sample space, which changes the probability. Bayes' Theorem lets you work backwards: if you observe an effect (e.g., a defective item), it tells you the probability of each possible cause (e.g., which machine made it), weighted by how likely each cause was in the first place. Two events are independent if knowing one happened tells you nothing about the other: $P(A|B) = P(A)$.

$$P(A|B)=\frac{P(A\cap B)}{P(B)}$$
$$P(A_i|B)=\frac{P(A_i)\,P(B|A_i)}{\sum_j P(A_j)\,P(B|A_j)} \quad \text{(Bayes)}$$

Drawing Without Replacement

A bag contains 4 red and 6 blue balls. Two balls are drawn without replacement. Find the probability both are red.

$$P=\frac{4}{10}\cdot\frac{3}{9}=\frac{2}{15}$$

Final answer: $2/15$.

Conditional Probability from a Table

In a class, 18 students passed Math, 12 passed Physics, and 8 passed both. If a student passed Physics, find the probability that the student also passed Math.

$$P(M|P)=\frac{P(M\cap P)}{P(P)}=\frac{8}{12}=\frac{2}{3}$$

Final answer: $2/3$.

Bayes: Machine Defects

Machine A makes 60% of bolts with 2% defective rate. Machine B makes 40% with 5% defective rate. If a bolt is defective, find the probability it came from B.

$$P(B|D)=\frac{0.40(0.05)}{0.60(0.02)+0.40(0.05)}=\frac{0.020}{0.032}=0.625$$

Final answer: 62.5%.

Drawing With Replacement

A bag has 3 red and 7 blue balls. Two draws are made with replacement. Find the probability both are red.

With replacement, each draw is independent. The probability of red stays at 3/10 each time.

$$P(\text{both red})=\frac{3}{10}\times\frac{3}{10}=\frac{9}{100}=0.09$$

Final answer: 0.09 or 9%.

Medical Test (Bayes Theorem)

A disease affects 1% of the population. A test is 95% accurate (detects disease when present) with a 3% false positive rate. If a person tests positive, what is the probability they actually have the disease?

Let D = disease, T = positive test. P(D) = 0.01, P(T|D) = 0.95, P(T|D') = 0.03.

$$P(T)=0.01(0.95)+0.99(0.03)=0.0095+0.0297=0.0392$$
$$P(D|T)=\frac{0.01(0.95)}{0.0392}=\frac{0.0095}{0.0392}\approx0.242$$

Final answer: only about 24.2% — most positives are false alarms when the disease is rare.

Checking for Independence

Events A and B have P(A) = 0.4, P(B) = 0.3, and P(A ∩ B) = 0.12. Are A and B independent?

Two events are independent if $P(A \cap B) = P(A) \cdot P(B)$. Check:

$$P(A)\cdot P(B)=0.4\times0.3=0.12=P(A\cap B) \checkmark$$

Final answer: Yes, A and B are independent.

Finding Conditional Probability from Joint

In a survey: P(employed) = 0.65, P(college graduate) = 0.40, P(employed AND college graduate) = 0.30. Find P(college graduate | employed).

$$P(C|E)=\frac{P(C\cap E)}{P(E)}=\frac{0.30}{0.65}\approx0.462$$

Final answer: about 46.2% of employed people in the survey are college graduates.

Bayes: Three Factories

Factory X produces 50% of parts (1% defective), Factory Y produces 30% (2% defective), Factory Z produces 20% (3% defective). A part is found defective. Find the probability it came from Y.

$$P(D)=0.50(0.01)+0.30(0.02)+0.20(0.03)=0.005+0.006+0.006=0.017$$
$$P(Y|D)=\frac{0.30(0.02)}{0.017}=\frac{0.006}{0.017}\approx0.353$$

Final answer: about 35.3%.

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Exam Generator Problems

Additional board-style practice items for this topic.

Question Bank: w18

MSTE - Statistics and Probability / Fundamental Principles of Counting / MSTE May 2019

How many different ways are there to arrange six people in a round table?

  1. 240
  2. 120
  3. 60
  4. 720
Circular permutations of $n$ objects:
$P = (n - 1)! = (6 - 1)! = \boxed{120\text{ ways}}$

Question Bank: w79

MSTE - Statistics and Probability / Probability / MSTE November 2019

An automated sandwich-making machine in a food manufacturer's factory has six major components with individual reliabilities: bread slicer 0.97, butter applicator 0.96, salad filler 0.94, meat filler 0.92, top slice of bread applicator 0.96, and wrapper 0.91. If one of these parts of the production line fails, the system will stop working. Therefore, the reliability of the whole system is:

  1. 0.678
  2. 0.687
  3. 0.704
  4. 0.740
The components act in series, so all six must work. Multiply the reliabilities:
$R = 0.97 \times 0.96 \times 0.94 \times 0.92 \times 0.96 \times 0.91 = 0.7035$
$\boxed{R \approx 0.704}$