Conditional Probability and the Rules of Probability: Understanding Independent Events (CCSS.S-CP.2)
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Common Core High School Statistics And Probability › Conditional Probability and the Rules of Probability: Understanding Independent Events (CCSS.S-CP.2)
In a club, 25% play chess (event $A$), 40% are in math club (event $B$), and 10% do both. Are $A$ and $B$ independent?
No
Not enough information
Yes
Only if the sample is random
Explanation
Compute $P(A)P(B)=0.25 \times 0.40=0.10$ and compare to $P(A \cap B)=0.10$. Since they are equal, $A$ and $B$ are independent.
At a school, $P(A)=0.70$ for playing a sport, and $P(A \cap B)=0.21$ for playing a sport and volunteering (event $B$). If $A$ and $B$ are independent, what is $P(B)$?
0.49
0.21
0.09
0.3
Explanation
Independence gives $P(A \cap B)=P(A)P(B)$. So $P(B)=\dfrac{P(A \cap B)}{P(A)}=\dfrac{0.21}{0.70}=0.30$.
On a team, 35% arrive early to practice (event $A$), 25% do strength training (event $B$), and 10% do both. What is $P(A \mid B)$?
0.4
0.1
0.35
0.75
Explanation
Use $P(A \mid B)=\dfrac{P(A \cap B)}{P(B)}=\dfrac{0.10}{0.25}=0.40$.
At a school, 40% of students play a sport ($A$), 30% are in the band ($B$), and 12% do both. Are $A$ and $B$ independent?
Yes, because $0.12 = 0.40 \times 0.30$
Yes, because $0.12 = 0.40 + 0.30$
No, because $0.12 < 0.40$
No, because $0.12 > 0.30$
Explanation
They are independent if $P(A\cap B) = P(A)P(B)$. Here $0.12 = 0.40 \times 0.30 = 0.12$, so $A$ and $B$ are independent.
In a tech use survey, 60% of students text daily ($A$), 50% stream videos daily ($B$), and 28% do both. Are $A$ and $B$ independent?
Yes, because $0.28 = 0.60 \times 0.50$
No, because $0.28 \ne 0.60 \times 0.50$
Yes, because $0.28 < 0.60$
No, because $0.28 = 0.60 + 0.50$
Explanation
Check $P(A\cap B)$ versus $P(A)P(B)$. Since $0.28 \ne 0.60 \times 0.50 = 0.30$, the events are not independent.
In a city, 70% use public transit weekly ($A$) and 40% bike at least monthly ($B$). If $A$ and $B$ are independent, what is $P(A\cap B)$?
30%
10%
28%
70%
Explanation
If independent, $P(A\cap B) = P(A)P(B) = 0.70 \times 0.40 = 0.28$ (28%).
In a game, 55% of players pick red ($A$) and 40% win a prize ($B$). It is observed that $P(A\mid B) = 0.55$. Are $A$ and $B$ independent?
Yes, because $P(A\cap B) = 0.55$
No, because $P(A\cap B) = P(A)+P(B)$
No, because $P(A\mid B)$ should equal $P(B)$
Yes, because $P(A\mid B) = P(A)$
Explanation
When $P(B)>0$, $A$ and $B$ are independent if and only if $P(A\mid B)=P(A)$. Here $0.55=0.55$, so they are independent.
At a club, 35% attend a workshop ($A$) and 20% join a committee ($B$). If $A$ and $B$ are independent, what is $P(A\mid B)$?
35%
20%
7%
55%
Explanation
If $A$ and $B$ are independent, then $P(A\mid B) = P(A) = 35%$.