Conditional Probability and the Rules of Probability: Finding Conditional Probability in Models (CCSS.S-CP.6)
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Common Core High School Statistics And Probability › Conditional Probability and the Rules of Probability: Finding Conditional Probability in Models (CCSS.S-CP.6)
A survey of 40 students recorded whether they play a sport (Yes/No) and whether they have a part-time job (Yes/No): 16 play a sport and have a job; 6 play a sport and do not have a job; 8 do not play a sport but have a job; 10 do not play a sport and do not have a job.
What is $P(\text{Have a job}\mid \text{Play a sport})$?
$\frac{16}{40}$
$\frac{16}{22}$
$\frac{22}{40}$
$\frac{6}{22}$
Explanation
Among those who play a sport (22 students), 16 have a job. So $P(\text{job}\mid \text{sport})=\frac{16}{22}=\frac{8}{11}\approx 0.73$. Interpretation: About 73% of sport-playing students have a job.
A librarian classifies 30 books as Fiction/Nonfiction and Hardcover/Paperback: 9 fiction hardcover, 6 fiction paperback, 5 nonfiction hardcover, 10 nonfiction paperback.
What is $P(\text{Fiction}\mid \text{Hardcover})$?
$\frac{9}{30}$
$\frac{14}{30}$
$\frac{9}{19}$
$\frac{9}{14}$
Explanation
Among hardcover books (14 total), 9 are fiction. So $P(\text{Fiction}\mid \text{Hardcover})=\frac{9}{14}\approx 0.64$. Interpretation: About 64% of hardcover books are fiction.
At a science fair with 48 participants, results by gender (Girl/Boy) and prize (Won/Did not win) are: 12 girls won, 10 girls did not; 8 boys won, 18 boys did not.
What is $P(\text{Won}\mid \text{Girl})$?
$\frac{12}{22}$
$\frac{12}{48}$
$\frac{22}{48}$
$\frac{10}{22}$
Explanation
Among girls (22 total), 12 won a prize. So $P(\text{Won}\mid \text{Girl})=\frac{12}{22}=\frac{6}{11}\approx 0.55$. Interpretation: About 55% of girls won a prize.
In a housing survey of 36 households, categories were Dwelling (Apartment/House) and Dog ownership (Dog/No Dog): 5 apartment households have a dog, 11 apartments have no dog, 13 houses have a dog, 7 houses have no dog.
What is $P(\text{Dog}\mid \text{Apartment})$?
$\frac{5}{36}$
$\frac{13}{36}$
$\frac{5}{16}$
$\frac{16}{36}$
Explanation
Among apartments (16 total), 5 have a dog. So $P(\text{Dog}\mid \text{Apartment})=\frac{5}{16}=0.3125$. Interpretation: About 31% of apartment households have a dog.
In a taste test of 50 students, preference (Chocolate/Vanilla) and wearing glasses (Glasses/No glasses) were recorded: 12 with glasses prefer chocolate, 8 with glasses prefer vanilla, 15 without glasses prefer chocolate, 15 without glasses prefer vanilla.
What is $P(\text{Chocolate}\mid \text{Glasses})$?
$\frac{12}{50}$
$\frac{20}{50}$
$\frac{12}{20}$
$\frac{15}{20}$
Explanation
Among students with glasses (20 total), 12 prefer chocolate. So $P(\text{Chocolate}\mid \text{Glasses})=\frac{12}{20}=\frac{3}{5}=0.6$. Interpretation: 60% of students with glasses prefer chocolate.
A school surveyed 40 students about sports and subject preference. Of the 18 who play a sport, 11 prefer math and 7 prefer English. Of the 22 who do not play a sport, 9 prefer math and 13 prefer English. Let event A be "prefers math," and event B be "plays a sport."
What is $P(A\mid B)$?
$11/40$
$11/18$
$20/40$
$9/22$
Explanation
$P(A\mid B)$ is the fraction of B's outcomes that also belong to A. Among the 18 who play a sport (B), 11 prefer math (A), so $P(A\mid B)=11/18$.
A library checked out 36 books. There were 16 hardcover and 20 paperback. Among the hardcover books, 9 were fiction and 7 were nonfiction. Among the paperback books, 6 were fiction and 14 were nonfiction. Let event A be "hardcover," and event B be "nonfiction."
What is $P(A\mid B)$?
$9/16$
$7/36$
$15/36$
$7/21$
Explanation
Condition on B (nonfiction): there are 21 nonfiction books in total, and 7 of them are hardcover. Thus $P(A\mid B)=7/21$.
In a class of 30 students, 12 are in band and 18 are not in band. Among those in band, 3 are left-handed and 9 are right-handed. Among those not in band, 5 are left-handed and 13 are right-handed. Let event A be "left-handed," and event B be "not in band."
What is $P(A\mid B)$?
$5/18$
$3/12$
$8/30$
$5/30$
Explanation
Condition on B (not in band): there are 18 not in band, and among them 5 are left-handed. So $P(A\mid B)=5/18$.
A store sold 24 items in a day: 10 snacks and 14 drinks. Of the snacks, 6 were cold and 4 were hot. Of the drinks, 12 were cold and 2 were hot. Let event A be "cold item," and event B be "drink."
What is $P(A\mid B)$?
$12/24$
$6/10$
$6/7$
$14/24$
Explanation
Condition on B (drink): there are 14 drinks, and 12 of them are cold. Thus $P(A\mid B)=12/14=6/7$.
A city survey recorded 48 commuters: 20 take the bus and 28 drive. Among bus riders, 15 arrive on time and 5 are late. Among drivers, 18 arrive on time and 10 are late. Let event A be "arrives on time," and event B be "drives."
What is $P(A\mid B)$?
$15/20$
$9/14$
$33/48$
$18/48$
Explanation
Condition on B (drives): there are 28 drivers, and among them 18 arrive on time. So $P(A\mid B)=18/28=9/14$.