Conditional Probability and the Rules of Probability: Using Permutations and Combinations in Probability (CCSS.S-CP.9)
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Common Core High School Statistics And Probability › Conditional Probability and the Rules of Probability: Using Permutations and Combinations in Probability (CCSS.S-CP.9)
In a school survey of 200 students, 80 participate in sports, 50 are in band, and 30 are in both activities.
What is the probability that a student is in band given they are in sports?
$3/10$
$3/8$
$1/4$
$3/5$
Explanation
Use $P(\text{Band}|\text{Sports})=\frac{P(\text{Band}\cap\text{Sports})}{P(\text{Sports})}$. So $\frac{30/200}{80/200}=\frac{30}{80}=\frac{3}{8}$.
A disease affects 2% of a population. A test correctly identifies the disease 95% of the time when a person has it (sensitivity), and correctly returns negative 90% of the time when a person does not have it (specificity). A person tests positive.
What is the probability the person actually has the disease?
\$0.95$
\$0.02$
\$0.50$
\$0.16$
Explanation
By Bayes: $P(D|+)=\frac{P(+|D)P(D)}{P(+|D)P(D)+P(+|\neg D)P(\neg D)}=\frac{0.95\cdot0.02}{0.95\cdot0.02+0.10\cdot0.98}\approx0.16$.
Two cards are drawn at random without replacement from a standard 52-card deck.
What is the probability that both cards are aces?
$1/221$
$1/169$
$1/663$
$1/26$
Explanation
$P=\frac{\binom{4}{2}}{\binom{52}{2}}=\frac{6}{1326}=\frac{1}{221}$.
In a group, some students are in sports and some are in band.
Which calculation gives $P(\text{Band}|\text{Sports})$?
$\frac{P(\text{Band})}{P(\text{Sports})}$
$\frac{P(\text{Band}\cup \text{Sports})}{P(\text{Band})}$
$\frac{P(\text{Band}\cap \text{Sports})}{P(\text{Sports})}$
$\frac{P(\text{Sports})}{P(\text{Band}\cap \text{Sports})}$
Explanation
By definition, $P(A|B)=\frac{P(A\cap B)}{P(B)}$, so $P(\text{Band}|\text{Sports})=\frac{P(\text{Band}\cap\text{Sports})}{P(\text{Sports})}$.
A 5-card hand is dealt from a standard 52-card deck.
Which expression correctly computes the probability of getting exactly two queens?
$\frac{\binom{4}{2}}{\binom{52}{5}}$
$\frac{\binom{4}{2}\binom{48}{3}}{\binom{52}{5}}$
$\frac{\binom{4}{3}\binom{48}{2}}{\binom{52}{5}}$
$\frac{\binom{4}{2}\binom{48}{2}}{\binom{52}{4}}$
Explanation
Choose 2 of the 4 queens and 3 of the 48 non-queens: $\binom{4}{2}\binom{48}{3}$, over all hands $\binom{52}{5}$.
In a school of 300 students, 120 play sports, 45 are in band, and 25 are in both. What is the probability that a student is in band given they are in sports?
1/12
3/8
5/24
5/9
Explanation
Use $P(\text{Band}\mid\text{Sports})=\frac{P(\text{Band}\cap\text{Sports})}{P(\text{Sports})}$. Here $\frac{25}{120}=\frac{5}{24}$.
A disease affects 4% of a population. A test has 90% sensitivity and 95% specificity. If a person tests positive, what is the probability they actually have the disease?
0.43
0.9
0.04
0.69
Explanation
By Bayes: $P(D\mid +)=\dfrac{0.04\cdot 0.90}{0.04\cdot 0.90+0.96\cdot 0.05}=\dfrac{0.036}{0.084}\approx 0.4286\approx 0.43$.
A club has 8 seniors and 6 juniors. A 4-student team is chosen at random. Given that the team has at least one senior, what is the probability it has exactly two seniors?
420/1001
28/1001
420/986
210/493
Explanation
Total teams: $\binom{14}{4}=1001$. With no seniors: $\binom{6}{4}=15$. So denominator $=1001-15=986$. Exactly two seniors: $\binom{8}{2}\binom{6}{2}=420$. Thus $\dfrac{420}{986}=\dfrac{210}{493}$.
A jar has 5 red and 7 blue marbles. Two marbles are drawn without replacement. Given that at least one is red, what is the probability both are red?
5/33
2/9
15/22
7/12
Explanation
$P(RR)=\frac{5}{12}\cdot\frac{4}{11}=\frac{5}{33}$. Also $P(\text{at least one red})=1-\frac{7}{12}\cdot\frac{6}{11}=\frac{15}{22}$. So $P(RR\mid \text{at least one red})=\dfrac{\frac{5}{33}}{\frac{15}{22}}=\dfrac{2}{9}$.
A bag contains 4 green and 6 yellow balls. Two balls are drawn without replacement. Let $A$ be the event "both balls are green" and $B$ be "at least one ball is green." Which calculation gives $P(A \mid B)$?
$\frac{\binom{4}{2}}{\binom{10}{2}}$
$\frac{\binom{4}{2}}{\binom{10}{2}-\binom{6}{2}}$
$\frac{\binom{4}{2},\binom{6}{0}}{\binom{10}{2}}$
$\frac{\binom{4}{1}}{\binom{10}{1}}$
Explanation
$P(A)=\dfrac{\binom{4}{2}}{\binom{10}{2}}$. $P(B)=1-\dfrac{\binom{6}{2}}{\binom{10}{2}}=\dfrac{\binom{10}{2}-\binom{6}{2}}{\binom{10}{2}}$. Hence $P(A\mid B)=\dfrac{P(A)}{P(B)}=\dfrac{\binom{4}{2}}{\binom{10}{2}-\binom{6}{2}}$.