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
From the integers 1–20, one is chosen at random. What is the probability it is either odd or a multiple of 6?
Answer: 13/20
Odd: 1,3,5,7,9,11,13,15,17,19 (10); Multiples of 6: 6,12,18 (3); Overlap (odd multiples of 6): none; Union = 13; P = 13/20.
A student guesses on 5 true/false questions. What is the probability of answering all 5 correctly?
Answer: 1/32
P = (1/2)^5 = 1/32.
The two-way table shows 480 shoppers by age and purchase type. Online In-Store Under 40 180 60 40 and over 80 160 Given a shopper buys in-store, what is the probability they are 40 or over?
Answer: 8/11
In-store total = 60 + 160 = 220; P(40+ | in-store) = 160/220 = 8/11.
Events C and D satisfy P(C) = 0.3, P(D | C) = 0.6. What is P(C and D)?
Answer: 0.18
P(C and D) = P(C) × P(D | C) = 0.3 × 0.6 = 0.18.
In a deck of 40 cards numbered 1–40, one card is drawn at random. What is the probability the number is a multiple of 7?
Answer: 1/8
Multiples of 7 in 1–40: 7,14,21,28,35 — 5 numbers; P = 5/40 = 1/8.
A bag contains 8 tiles: 2 red, 3 blue, and 3 green. Two tiles are drawn at random without replacement. What is the probability the first is red and the second is green?
Answer: 3/28
P = (2/8)(3/7) = 6/56 = 3/28.
Of 200 commuters, 90 take the bus and 80 take the train, with 20 taking both. What is the probability a randomly chosen commuter takes neither?
Answer: 1/4
Bus or train = 90 + 80 − 20 = 150; neither = 200 − 150 = 50; P = 50/200 = 1/4.
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