Conditional Probability and the Rules of Probability: Applying the Addition Rule for Probability (CCSS.S-CP.7)
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Common Core High School Statistics And Probability › Conditional Probability and the Rules of Probability: Applying the Addition Rule for Probability (CCSS.S-CP.7)
In a survey of 80 students, 34 play soccer (event A), 29 play basketball (event B), and 12 play both. What is $P(A \cup B)$?
0.7875
0.3625
0.6375
0.51
Explanation
Use the Addition Rule: $P(A \cup B)=\frac{34+29-12}{80}=\frac{51}{80}=0.6375$. We add the category counts and subtract the overlap to avoid double-counting those in both A and B.
Out of 120 concert-goers, 70 like rock (event A), 55 like jazz (event B), and 25 like both. What is $P(A \cap B)$?
0.8333
0.2674
0.25
0.2083
Explanation
$P(A \cap B)$ is the probability a person is in both categories, so $\frac{25}{120}=0.2083$. The Addition Rule would give $P(A \cup B)$, not the intersection.
A library has 200 members. Ninety borrow fiction (event A), 70 borrow non-fiction (event B), and 30 borrow both. What is $P(\text{not } A)$?
0.55
0.45
0.2
0.35
Explanation
$P(\text{not } A)=1-P(A)$. Since 90 members are in A (including the overlap), $P(A)=\frac{90}{200}=0.45$, so $P(\text{not } A)=1-0.45=0.55$.
Among 150 campers, 80 can swim (event A), 60 can hike (event B), and 25 can do both. What is $P(A \cup B)$?
0.9333
0.7667
0.6333
0.2333
Explanation
Apply the Addition Rule: $P(A \cup B)=\frac{80+60-25}{150}=\frac{115}{150}\approx 0.7667$. Subtract the overlap to avoid double-counting those who can do both.
In an electives fair with 90 students, 50 choose art (event A), 30 choose music (event B), and 12 choose both. What is $P(A \cup B)$?
0.8889
0.2444
0.6222
0.7556
Explanation
Use the Addition Rule: $P(A \cup B)=\frac{50+30-12}{90}=\frac{68}{90}\approx 0.7556$. We subtract the overlap so students counted in both A and B are not double-counted.
In a survey of 200 people, 120 like tea (event A), 90 like coffee (event B), and 50 like both. What is $P(A\cup B)$?
0.85
0.2
0.8
1.05
Explanation
Use the Addition Rule: $P(A\cup B)=\frac{120+90-50}{200}=\frac{160}{200}=0.80$. We subtract the overlap because those people were counted in both groups. This means 80% like tea or coffee (or both).
At a library with 150 visitors, 70 borrow fiction (event A), 55 borrow non-fiction (event B), and 25 borrow both. What is $P(A\cap B)$?
25/150
100/150
125/150
30/150
Explanation
$P(A\cap B)$ counts those in both categories: $\frac{25}{150}$. This means 1 out of 6 visitors borrowed both fiction and non-fiction.
At a camp with 180 participants, 65 attend swimming (event A), 80 attend archery (event B), and 20 attend both. What is $P(\text{not }A)$?
65/180
115/180
55/180
114/180
Explanation
The complement of A is everyone not in A: $P(\text{not }A)=\frac{180-65}{180}=\frac{115}{180}$. This means 115 out of 180 campers do not attend swimming.
In a class of 90 students, 48 take Spanish (event A), 32 take French (event B), and 12 take both. What is $P(A\cup B)$?
80/90
70/90
22/90
68/90
Explanation
Use the Addition Rule: $P(A\cup B)=\frac{48+32-12}{90}=\frac{68}{90}$. We subtract the overlap so students counted in both aren't double-counted. This means 68 of 90 students take at least one of the two languages.
In a town of 240 residents, 150 own a bicycle (event A), 180 own a car (event B), and 110 own both. What is $P(A\cup B)$?
210/240
220/240
330/240
20/240
Explanation
Apply the Addition Rule: $P(A\cup B)=\frac{150+180-110}{240}=\frac{220}{240}$. We subtract the overlap to avoid double-counting. This means 220 out of 240 residents own a bicycle or a car (or both).