Lesson 4-4 – Conditional Probability & Independent Events

0.0(0)
studied byStudied by 0 people
learnLearn
examPractice Test
spaced repetitionSpaced Repetition
heart puzzleMatch
flashcardsFlashcards
Card Sorting

1/29

flashcard set

Earn XP

Description and Tags

These Q&A flashcards review definitions, rules, and illustrative examples for independent events, mutually exclusive events, conditional probability, and compound probability calculations from Lesson 4-4.

Study Analytics
Name
Mastery
Learn
Test
Matching
Spaced

No study sessions yet.

30 Terms

1
New cards

What is the definition of independent events in probability?

Two events are independent when the occurrence of one does not change the probability of the other occurring.

2
New cards

How does replacing an item (“with replacement”) affect independence?

With replacement keeps probabilities the same for each trial, so the events are independent.

3
New cards

How does NOT replacing an item (“without replacement”) affect independence?

Without replacement alters the remaining probabilities, so the events become dependent.

4
New cards

Are the events “flip a head on a coin” and “roll a 5 on a die” independent? Why?

Yes; the chance of rolling a 5 (1⁄6) is the same whether or not a head is flipped.

5
New cards

Why are “robbing a bank” and “going to prison” NOT independent?

Because robbing a bank greatly increases the probability of going to prison compared with not robbing a bank.

6
New cards

Can events be both independent and mutually exclusive?

No. If two events are mutually exclusive they cannot occur together, which forces dependence.

7
New cards

State the AND (intersection) rule for independent events.

P(A ∩ B) = P(A) · P(B).

8
New cards

State the AND (intersection) rule for dependent events.

P(A ∩ B) = P(A) · P(B | A).

9
New cards

State the OR (union) rule for any two events A and B.

P(A ∪ B) = P(A) + P(B) – P(A ∩ B).

10
New cards

When is the OR rule P(A ∪ B) = P(A)+P(B) used without subtraction?

When A and B are mutually exclusive; then P(A ∩ B)=0.

11
New cards

Write the formula for conditional probability.

P(B | A) = P(A ∩ B) / P(A), provided P(A) ≠ 0.

12
New cards

How can you prove two events A and B are independent using conditional probability?

Show that P(B | A) = P(B) (or equivalently P(A | B) = P(A)).

13
New cards

Define mutually exclusive (disjoint) events.

Events that cannot occur together in a single trial; their intersection is empty.

14
New cards

Give an example of mutually exclusive events with a die.

Rolling a 3 and rolling a 4 on the same single die roll.

15
New cards

Why do we subtract P(A ∩ B) in the OR rule?

Because the outcomes in the overlap are counted twice—once in P(A) and once in P(B)—so one copy must be removed.

16
New cards

In a 2×2 contingency table, how do you find P(grade | female)?

Divide the number of girls who choose grades by the total number of girls.

17
New cards

If P(girl)=0.484 and P(girl ∩ grades)=0.400, are gender and choice independent?

No, because P(grades | girl)=0.827 ≠ P(grades)=0.548 (for the whole sample).

18
New cards

Sock drawer: What rule is usually applied when the wording says “at least one”?

Use the complement rule: P(at least one) = 1 – P(none).

19
New cards

What is the probability of drawing two blue socks (4 blue, 5 gray, 3 red, no replacement)?

(4/12) · (3/11) = 12/132 ≈ 0.091 (9.09%).

20
New cards

What does P(stop at least once during 5 days) equal if P(red)=0.61 each day?

1 – (0.39)^5 ≈ 0.991 (99.1%).

21
New cards

Superbowl fans: If 65% root Rams, 46% Bengals, 29% both, what % root for at least one?

P(R ∪ B) = 0.65 + 0.46 – 0.29 = 0.82 or 82%.

22
New cards

Using the same data, what % root for neither team?

1 – 0.82 = 0.18 or 18%.

23
New cards

DUI tests: 78% breath, 36% blood, 22% both. What % receive some test?

0.78 + 0.36 – 0.22 = 0.92 or 92%.

24
New cards

DUI tests: What % receive only a blood test?

0.36 – 0.22 = 0.14 or 14%.

25
New cards

If two outcomes share the same sample space but have at least one common element, are they mutually exclusive?

No, because at least one outcome makes them overlap.

26
New cards

Die roll: Why are “multiple of 3” and “even number” not mutually exclusive?

Because 6 is both even and a multiple of 3, so the events overlap.

27
New cards

Traffic lights: Why can Monday’s and Tuesday’s light colors be treated as independent?

Each day’s light color is assumed to have the same probabilities and is not affected by previous days.

28
New cards

Give the joint probability of two independent red lights on consecutive days if P(red)=0.61.

0.61 × 0.61 = 0.3721 (37.21%).

29
New cards

Explain why independence involves two separate actions while mutual exclusivity involves one action.

Independence addresses whether one event in a sequence affects another; mutual exclusivity addresses whether two events can happen simultaneously in a single trial.

30
New cards

When testing independence, what conclusion do you draw if P(A ∩ B) = P(A)·P(B)?

The events A and B are independent because the joint probability equals the product of individual probabilities.