Conditional Probability and the Rules of Probability: Conditional Probability and Independence (CCSS.S-CP.3)
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Common Core High School Statistics And Probability › Conditional Probability and the Rules of Probability: Conditional Probability and Independence (CCSS.S-CP.3)
In a school of 200 students, 80 take Art, 50 take Music, and 30 take both. What fraction of Music students also take Art?
30/80
30/50
50/200
80/200
Explanation
Use the definition $P(\text{Art} \mid \text{Music}) = \frac{P(\text{Art} \cap \text{Music})}{P(\text{Music})} = \frac{30}{50}$.
A survey records how many students are in the Art club (event $A$), in the Band (event $B$), and in both. Which formula represents $P(A \mid B)$?
$P(A \mid B) = \frac{P(A)}{P(B)}$
$P(A \mid B) = \frac{P(A \cup B)}{P(B)}$
$P(A \mid B) = \frac{P(A \cap B)}{P(A)}$
$P(A \mid B) = \frac{P(A \cap B)}{P(B)}$
Explanation
By definition, $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$.
In a group of 150 students, 90 have a pet (event $A$), 60 play an instrument (event $B$), and 36 do both. Are $A$ and $B$ independent?
Yes; $P(A \mid B) = P(A)$ and $P(B \mid A) = P(B)$
No; because $\frac{36}{150} \neq \frac{90}{150}$
No; because $\frac{36}{60} \neq \frac{90}{150}$
Yes; because $P(A \cap B) = P(A) + P(B)$
Explanation
Compute $P(A)=90/150=0.6$, $P(B)=60/150=0.4$, $P(A \mid B)=36/60=0.6$, and $P(B \mid A)=36/90=0.4$. Since $P(A \mid B)=P(A)$ and $P(B \mid A)=P(B)$, the events are independent.
Out of 220 students, 140 take Biology, 100 take Chemistry, and 60 take both. What fraction of Chemistry students also take Biology?
60/140
100/220
60/100
140/220
Explanation
We want $P(\text{Biology} \mid \text{Chemistry}) = \frac{P(\text{Bio} \cap \text{Chem})}{P(\text{Chem})} = \frac{60}{100}$.
Which equality must be true if events $A$ and $B$ are independent?
$P(A \mid B) = P(A)$
$P(A \mid B) = P(B)$
$P(A \cap B) = P(A) + P(B)$
$P(B \mid A) = \frac{P(B)}{P(A)}$
Explanation
Independence means conditioning on one event does not change the probability of the other: $P(A \mid B) = P(A)$ (equivalently $P(B \mid A) = P(B)$).
A school has 200 students. Event $A$: likes math (80 students). Event $B$: in chess club (50 students). There are 20 students who like math and are in chess club. Are $A$ and $B$ independent?
Not independent because $P(A \mid B)=20/200$
Not independent because $P(A \mid B)=P(B)$
Independent because $P(A \mid B)=P(B)$
Independent because $P(A \mid B)=P(A)$
Explanation
Compute $P(A)=80/200=0.4$ and $P(A \mid B)=20/50=0.4$. Since $P(A \mid B)=P(A)$, the events are independent.