Conditional Probability and the Rules of Probability: Using Two-Way Tables for Independence and Conditional Probability (CCSS.S-CP.4)
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Common Core High School Statistics And Probability › Conditional Probability and the Rules of Probability: Using Two-Way Tables for Independence and Conditional Probability (CCSS.S-CP.4)
In a school survey, 40% of students prefer science (event $S$) and 30% are in tenth grade (event $T$). If $S$ and $T$ are independent, what is $P(S \cap T)$?
0.12
0.7
0.1
0.34
Explanation
By independence, $P(S \cap T)=P(S)P(T)=0.40 \times 0.30=0.12$.
Which formula correctly expresses $P(A \cap B)$ for any two events $A$ and $B$?
$P(A)+P(B)$
$P(A) \cdot P(B\mid A)$
$\frac{P(A\mid B)}{P(B)}$
$P(A\mid B)-P(B)$
Explanation
The multiplication rule gives $P(A \cap B)=P(A)P(B\mid A)$ (equivalently, $P(B)P(A\mid B)$).
Which statement about independence is true?
If $P(A\mid B)=P(B\mid A)$, then $A$ and $B$ are independent.
If $A$ and $B$ are independent, then $P(A\mid B)=1$.
If $A$ and $B$ are independent, then $P(A \cap B)=P(A)P(B)$.
If $P(A \cap B)=0$, then $A$ and $B$ must be independent.
Explanation
By definition of independence, $P(A \cap B)=P(A)P(B)$ and equivalently $P(A\mid B)=P(A)$ when $P(B)>0$.
At a high school, 55% of students play a sport (event $S$). Among those who play a sport, 40% also participate in band (event $B$). What is $P(S \cap B)$?
0.95
0.15
0.33
0.22
Explanation
Use the multiplication rule with a conditional probability: $P(S \cap B)=P(S)P(B\mid S)=0.55 \times 0.40=0.22$.
In a club, 60% of members are eleventh graders (event $E$). Among eleventh graders, 30% prefer math (event $M$). What is $P(M \cap E)$?
0.18
0.15
0.75
0.3
Explanation
Apply the multiplication rule with a conditional probability: $P(M \cap E)=P(E)P(M\mid E)=0.60 \times 0.30=0.18$.