Probability Test Review Notes
Test Review
Name: Elizabeth Rudd
Unit 5 - Probability
Problem 1: Class Ranks at University X
- Students at University X are categorized into one of four class ranks:
- Freshman
- Sophomore
- Junior
- Senior
- Given Statistics:
- Freshmen: 35%
- Sophomores: 30%
- Total percentage of Freshmen and Sophomores = 35% + 30% = 65%
- To find the probability that a student selected at random is either a junior or a senior, we recognize that:
- Probability that a student is a junior or senior = 100% - (Probability of Freshmen + Probability of Sophomores)
- Thus:
P(Junior ext{ or } Senior) = 1 - 0.65 = 0.35
- Options for the probability that a student is either a junior or senior:
- (A) 30%
- (B) 35%
- (C) 65%
- (D) 70%
- Correct Answer: 35% (B)
Problem 2: Conditional Probability
- Given Events:
- P(A) = 0.5
- P(B) = 0.6
- P(A and B) = 0.1
- We need to find the conditional probability of A given B, denoted as P(A|B).
- The formula for conditional probability is:
P(A|B) = \frac{P(A \cap B)}{P(B)} - Applying the values:
P(A|B) = \frac{0.1}{0.6} \approx 0.1667 - Options for the conditional probability of A given B:
- (A) 0.5
- (B) 0.3
- (C) 0.2
- (D) 0.167
- Correct Answer: 0.167 (D)
Problem 3: Probability of Winning a Lottery
- In the Math-o-licious Lottery:
- Probability of winning with one ticket = 0.11
- If a student buys one ticket each month for five months, we want to find the probability of winning at least one prize.
- To solve this, first calculate the probability of not winning with one ticket:
P(Not ext{ win}) = 1 - P(Win) = 1 - 0.11 = 0.89 - The probability of not winning in five months is:
P(Not ext{ win in 5 months}) = (0.89)^5 \approx 0.55 - Therefore, the probability of winning at least one prize is:
P(Win ext{ at least one}) = 1 - P(Not ext{ win in 5 months}) = 1 - 0.55 = 0.45 - Options for the probability of winning at least one prize:
- (A) 0.55
- (B) 0.50
- (C) 0.44
- (D) 0.45
- Correct Answer: 0.45 (D)
Problem 4: Independent Events
- Given Events A and B:
- Since A and B are independent, the probability of both events A and B occurring, denoted as P(A and B), is given by:
P(A \cap B) = P(A) \times P(B) - Applying the values:
P(A \cap B) = 0.2 \times 0.4 = 0.08 - Options for the probability of both A and B:
- (A) 0.08
- (B) 0.12
- (C) 0.60
- (D) 0.80
- Correct Answer: 0.08 (A)