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:
  • P(A) = 0.2
  • P(B) = 0.4
• 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)