Conditional Probability and the Rules of Probability: Explaining Conditional Probability in Everyday Situations (CCSS.S-CP.5)
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Common Core High School Statistics And Probability › Conditional Probability and the Rules of Probability: Explaining Conditional Probability in Everyday Situations (CCSS.S-CP.5)
At a high school, 40% of students play a sport (event $S$), 30% take an art class (event $A$), and 12% do both. Are $S$ and $A$ independent?
Yes, because $P(S\cap A)=0.12$ equals $P(S)\cdot P(A)=0.4\times 0.3$.
No, because you should compare $P(S\cap A)$ to $P(S)+P(A)$.
No, because $0.4\times 0.3=0.10$, which does not match \$0.12$.
Yes, because $P(S)>P(A)$.
Explanation
Independence holds if $P(S\cap A)=P(S)\cdot P(A)$. Here $0.4\times 0.3=0.12$, which matches $P(S\cap A)=0.12$.
In a city, 55% of adults commute by car (event $C$), 25% use the bike-share program (event $B$), and 12% do both. Which calculation checks independence between commuting by car and bike-share use?
Compare $P(C\cap B)$ to $P(C)+P(B)$ by checking \$0.12$ versus $0.55+0.25$.
Compare $P(C\cap B)$ to $P(C)\cdot P(B)$ by checking \$0.12$ versus $0.55\times 0.25$.
Compare $P(C\cap B)$ to $P(C)-P(B)$ by checking \$0.12$ versus $0.55-0.25$.
Decide independence because both $P(C)$ and $P(B)$ are less than \$0.6$.
Explanation
To test independence, check whether $P(C\cap B)=P(C)\cdot P(B)$ (product of marginals).
A health survey reports that 20% of adults smoke (event $S$), 5% have lung cancer (event $L$), and 1% both smoke and have lung cancer. Which statement best describes independence in this context?
They are independent because $P(S\mid L)=P(L)$.
They are not independent because $P(S)+P(L)=0.25$ is not \$0.01$.
They appear independent because $P(S\cap L)=0.01$ equals $P(S)\cdot P(L)=0.20\times 0.05$.
They are independent because more adults are non-smokers than smokers.
Explanation
Independence means $P(S\cap L)=P(S)\cdot P(L)$. Here $0.20\times 0.05=0.01$, matching $P(S\cap L)=0.01$.
In a neighborhood, 35% of households own a dog (event $D$), 25% own a cat (event $C$), and 9% own both. Are dog ownership and cat ownership independent?
Yes, because $P(D\cap C)=P(D)\cdot P(C)=0.35\times 0.25=0.09$.
No, because $P(D\cap C)=0.09$ is not $P(D)\cdot P(C)=0.35\times 0.25$.
Yes, because $P(D)+P(C)=0.60$.
No, because $P(D)>P(C)$.
Explanation
Independence requires $P(D\cap C)=P(D)\cdot P(C)$. Here $0.35\times 0.25=0.0875\ne 0.09$, so they are not independent.
Events A and B occur with $P(A)=0.4$, $P(B)=0.5$, and $P(A\cap B)=0.20$.
Are A and B independent?
Yes, because $0.20=0.4\times 0.5$.
Yes, because $0.20=0.4+0.5$.
No, because $0.20\neq 0.4\times 0.5$.
No, because any time $P(B)=0.5$ the events are dependent.
Explanation
Independent if and only if $P(A\cap B)=P(A)P(B)$. Here $0.20=0.4\times 0.5=0.20$, so they are independent.
In a school, the chance a student drinks coffee is $P(C)=0.6$, the chance a student brings an umbrella is $P(U)=0.3$, and the chance both happen is $P(C\cap U)=0.18$.
Which calculation checks independence of $C$ and $U$?
Check whether $0.18 = 0.6 + 0.3$.
Check whether $0.18 = 0.6\times 0.3$.
Check whether $P(C\cup U)=0.6\times 0.3$.
Check whether $P(C)=P(U)$.
Explanation
Independence is tested by comparing $P(C\cap U)$ to $P(C)P(U)$. If equal, they are independent.
At a high school, $P(\text{plays soccer})=0.25$, $P(\text{is in honors})=0.40$, and $P(\text{plays soccer and is in honors})=0.10$.
Which statement best describes independence in this context?
The events are independent because $0.10=0.25+0.40$.
The events are not independent because $0.10=0.25-0.40$.
Playing soccer does not change the chance of being in honors because $0.10=0.25\times 0.40$.
The events are independent because $P(\text{plays soccer})=P(\text{is in honors})$.
Explanation
Independence holds when $P(\text{soccer}\cap \text{honors})=P(\text{soccer})P(\text{honors})$. Here $0.10=0.25\times 0.40=0.10$.
Suppose $P(A)=0.7$, $P(B)=0.6$, and $P(A\cap B)=0.50$.
Are $A$ and $B$ independent?
Yes, because $0.50=0.7+0.6$.
Yes, because $0.50=0.7\times 0.6$.
No, because $0.50=0.7-0.6$.
No, because $0.50\neq 0.7\times 0.6$.
Explanation
Test independence with $P(A\cap B)$ vs. $P(A)P(B)$. Since $0.50\neq 0.7\times 0.6=0.42$, they are not independent.
In a population, 20% are smokers, 5% have lung cancer, and 3% are smokers who have lung cancer: $P(S)=0.20$, $P(L)=0.05$, $P(S\cap L)=0.03$.
Which statement describes independence in this context?
Smoking and lung cancer would be independent only if $0.03=0.20\times 0.05$; since $0.03\neq 0.01$, they are not independent.
Smoking and lung cancer are independent because $0.03=0.20+0.05$.
They are independent because $P(S)=P(L)$.
They are independent because $0.03=0.20\times 0.50$.
Explanation
Independence requires $P(S\cap L)=P(S)P(L)$. Here $0.03\neq 0.20\times 0.05=0.01$, so the events are not independent.