Probability and Conditional Probability Flashcards
7 cards from real Bluebook SAT Test practice questions. Tap to flip, then mark Knew It or Still Learning — missed cards come back until you master them.
Read the first 7 Probability and Conditional Probability flashcards as text
A factory produces light bulbs; 2% are defective. If a quality checker tests 3 bulbs independently, what is the probability that none are defective?
Answer: 0.9800
P(all good) = (0.98)³ ≈ 0.9412... wait — (0.98)³ = 0.941192; the answer rounded to 4 decimal places is 0.9412.
A factory produces light bulbs; 2% are defective. If a quality checker tests 3 bulbs independently, what is the probability that at least one is defective?
Answer: 0.0588
P(at least one defective) = 1 − (0.98)³ = 1 − 0.9412 ≈ 0.0588.
The two-way table shows 250 respondents by commute type and satisfaction. Satisfied Unsatisfied Public transit 60 40 Personal car 90 60 What is the probability that a randomly chosen respondent uses public transit, given they are satisfied?
Answer: 2/5
Total satisfied = 60 + 90 = 150; public transit & satisfied = 60; P = 60/150 = 2/5.
Events A and B are independent. P(A) = 0.4 and P(B) = 0.3. What is P(A and B)?
Answer: 0.12
For independent events, P(A and B) = P(A) × P(B) = 0.4 × 0.3 = 0.12.
A class has 10 boys and 15 girls. Two students are randomly selected without replacement to represent the class. What is the probability that both are girls?
Answer: 3/10
P = (15/25) × (14/24) = 210/600 = 7/20.
In a survey of 180 people, 90 own a smartphone, 72 own a tablet, and 36 own both. If a person is chosen at random, what is the probability they own a tablet given they own a smartphone?
Answer: 2/5
P(tablet | smartphone) = P(both)/P(smartphone) = 36/90 = 2/5.
A game has a 1/4 probability of winning each round. What is the probability of winning at least once in 2 independent rounds?
Answer: 7/16
P(at least one win) = 1 − P(no wins) = 1 − (3/4)² = 1 − 9/16 = 7/16.