Applying the Multiplication Rule for Probability - Statistics
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If $A$ and $B$ are independent, what does the Multiplication Rule simplify to?
If $A$ and $B$ are independent, what does the Multiplication Rule simplify to?
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$P(A \cap B)=P(A)P(B)$. Independence means conditional equals marginal probability.
$P(A \cap B)=P(A)P(B)$. Independence means conditional equals marginal probability.
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Identify the correct interpretation of $P(A)P(B\mid A)$ in context.
Identify the correct interpretation of $P(A)P(B\mid A)$ in context.
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Probability $A$ occurs, then $B$ occurs given $A$ occurred. Multiplication rule gives sequential probability interpretation.
Probability $A$ occurs, then $B$ occurs given $A$ occurred. Multiplication rule gives sequential probability interpretation.
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If $P(A \cap B)=0$, what must be true about $A$ and $B$?
If $P(A \cap B)=0$, what must be true about $A$ and $B$?
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$A$ and $B$ are mutually exclusive (cannot occur together). Zero intersection means events have no common outcomes.
$A$ and $B$ are mutually exclusive (cannot occur together). Zero intersection means events have no common outcomes.
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In a uniform model with $#(S)=30$, $#(A)=12$, $#(B)=10$, $#(A \cap B)=4$, find $P(A)P(B\mid A)$.
In a uniform model with $#(S)=30$, $#(A)=12$, $#(B)=10$, $#(A \cap B)=4$, find $P(A)P(B\mid A)$.
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$\frac{2}{5}\cdot\frac{1}{3}=\frac{2}{15}$. $P(A)=\frac{12}{30}=\frac{2}{5}$, $P(B|A)=\frac{4}{12}=\frac{1}{3}$.
$\frac{2}{5}\cdot\frac{1}{3}=\frac{2}{15}$. $P(A)=\frac{12}{30}=\frac{2}{5}$, $P(B|A)=\frac{4}{12}=\frac{1}{3}$.
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Find and correct the error: "$P(A \cap B)=P(A)+P(B\mid A)$".
Find and correct the error: "$P(A \cap B)=P(A)+P(B\mid A)$".
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Correct: $P(A \cap B)=P(A)P(B\mid A)$. Should multiply, not add, for joint probability.
Correct: $P(A \cap B)=P(A)P(B\mid A)$. Should multiply, not add, for joint probability.
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In a uniform model with $#(A)=20$ and $#(A \cap B)=5$, find $P(B\mid A)$.
In a uniform model with $#(A)=20$ and $#(A \cap B)=5$, find $P(B\mid A)$.
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$\frac{1}{4}$. Count $B$ outcomes within $A$: $\frac{5}{20} = \frac{1}{4}$.
$\frac{1}{4}$. Count $B$ outcomes within $A$: $\frac{5}{20} = \frac{1}{4}$.
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In a uniform model with $#(S)=50$, $#(A)=20$, $#(A \cap B)=5$, find $P(A \cap B)$.
In a uniform model with $#(S)=50$, $#(A)=20$, $#(A \cap B)=5$, find $P(A \cap B)$.
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$\frac{1}{10}$. Count favorable outcomes over total: $\frac{5}{50} = \frac{1}{10}$.
$\frac{1}{10}$. Count favorable outcomes over total: $\frac{5}{50} = \frac{1}{10}$.
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What is $P(A \cap B)$ if $P(B)=0.20$ and $P(A\mid B)=0.50$?
What is $P(A \cap B)$ if $P(B)=0.20$ and $P(A\mid B)=0.50$?
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$0.10$. Use $P(A \cap B) = P(B)P(A|B) = 0.20 \times 0.50$.
$0.10$. Use $P(A \cap B) = P(B)P(A|B) = 0.20 \times 0.50$.
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What is $P(B\mid A)$ if $P(A \cap B)=0.12$ and $P(A)=0.30$?
What is $P(B\mid A)$ if $P(A \cap B)=0.12$ and $P(A)=0.30$?
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$0.40$. Divide joint probability by marginal: $0.12 \div 0.30 = 0.40$.
$0.40$. Divide joint probability by marginal: $0.12 \div 0.30 = 0.40$.
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What is $P(A \cap B)$ if $P(A)=\frac{1}{4}$ and $P(B\mid A)=\frac{1}{2}$?
What is $P(A \cap B)$ if $P(A)=\frac{1}{4}$ and $P(B\mid A)=\frac{1}{2}$?
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$\frac{1}{8}$. Apply multiplication rule: $\frac{1}{4} \times \frac{1}{2} = \frac{1}{8}$.
$\frac{1}{8}$. Apply multiplication rule: $\frac{1}{4} \times \frac{1}{2} = \frac{1}{8}$.
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In a uniform probability model, how do you compute $P(B\mid A)$ from counts?
In a uniform probability model, how do you compute $P(B\mid A)$ from counts?
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$P(B\mid A)=\frac{#(A \cap B)}{#(A)}$. Restricts sample space to outcomes where $A$ occurred.
$P(B\mid A)=\frac{#(A \cap B)}{#(A)}$. Restricts sample space to outcomes where $A$ occurred.
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In a uniform probability model, how do you compute $P(E)$ from counts?
In a uniform probability model, how do you compute $P(E)$ from counts?
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$P(E)=\frac{#(E)}{#(S)}$. Uniform model means all outcomes are equally likely.
$P(E)=\frac{#(E)}{#(S)}$. Uniform model means all outcomes are equally likely.
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State the general Multiplication Rule for the probability of $A \cap B$.
State the general Multiplication Rule for the probability of $A \cap B$.
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$P(A \cap B)=P(A)P(B\mid A)=P(B)P(A\mid B)$. Expresses joint probability using conditional probability.
$P(A \cap B)=P(A)P(B\mid A)=P(B)P(A\mid B)$. Expresses joint probability using conditional probability.
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What does the notation $P(B\mid A)$ mean in words?
What does the notation $P(B\mid A)$ mean in words?
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Probability that $B$ occurs given that $A$ occurred. The vertical bar means "given" or "conditional on."
Probability that $B$ occurs given that $A$ occurred. The vertical bar means "given" or "conditional on."
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Which formula gives $P(B\mid A)$ in terms of $P(A \cap B)$ and $P(A)$?
Which formula gives $P(B\mid A)$ in terms of $P(A \cap B)$ and $P(A)$?
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$P(B\mid A)=\frac{P(A \cap B)}{P(A)}$. Rearranges the multiplication rule to isolate conditional probability.
$P(B\mid A)=\frac{P(A \cap B)}{P(A)}$. Rearranges the multiplication rule to isolate conditional probability.
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Which expression equals the probability that both events occur: $P(A \text{ and } B)$?
Which expression equals the probability that both events occur: $P(A \text{ and } B)$?
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$P(A \cap B)$. "And" in probability means intersection.
$P(A \cap B)$. "And" in probability means intersection.
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A fair die is rolled. Let $A={\text{even}}$ and $B={\ge 5}$. Find $P(A \cap B)$.
A fair die is rolled. Let $A={\text{even}}$ and $B={\ge 5}$. Find $P(A \cap B)$.
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$P(A \cap B)=\frac{1}{6}$. Only outcome 6 is both even and $\ge 5$.
$P(A \cap B)=\frac{1}{6}$. Only outcome 6 is both even and $\ge 5$.
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A fair die is rolled. Let $A={\text{even}}$ and $B={\ge 5}$. Find $P(B\mid A)$.
A fair die is rolled. Let $A={\text{even}}$ and $B={\ge 5}$. Find $P(B\mid A)$.
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$P(B\mid A)=\frac{1}{3}$. Given even (2,4,6), only 6 satisfies $\ge 5$.
$P(B\mid A)=\frac{1}{3}$. Given even (2,4,6), only 6 satisfies $\ge 5$.
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In a uniform probability model with $N$ equally likely outcomes, what is $P(E)$ for an event with $|E|$ outcomes?
In a uniform probability model with $N$ equally likely outcomes, what is $P(E)$ for an event with $|E|$ outcomes?
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$P(E)=\frac{|E|}{N}$. Favorable outcomes divided by total outcomes in uniform model.
$P(E)=\frac{|E|}{N}$. Favorable outcomes divided by total outcomes in uniform model.
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Identify the required condition for $P(B\mid A)=\frac{P(A \cap B)}{P(A)}$ to be defined.
Identify the required condition for $P(B\mid A)=\frac{P(A \cap B)}{P(A)}$ to be defined.
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$P(A)>0$. Division by zero is undefined when $P(A)=0$.
$P(A)>0$. Division by zero is undefined when $P(A)=0$.
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State the definition of conditional probability $P(B\mid A)$ in terms of $P(A \cap B)$ and $P(A)$.
State the definition of conditional probability $P(B\mid A)$ in terms of $P(A \cap B)$ and $P(A)$.
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$P(B\mid A)=\frac{P(A \cap B)}{P(A)}$. Ratio of joint probability to marginal probability of $A$.
$P(B\mid A)=\frac{P(A \cap B)}{P(A)}$. Ratio of joint probability to marginal probability of $A$.
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What equivalent Multiplication Rule expression gives $P(A \cap B)$ starting with $P(B)$?
What equivalent Multiplication Rule expression gives $P(A \cap B)$ starting with $P(B)$?
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$P(A \cap B)=P(B)P(A\mid B)$. Alternative form using $P(B)$ first, then $P(A|B)$.
$P(A \cap B)=P(B)P(A\mid B)$. Alternative form using $P(B)$ first, then $P(A|B)$.
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State the general Multiplication Rule for the intersection $P(A \cap B)$ using conditional probability.
State the general Multiplication Rule for the intersection $P(A \cap B)$ using conditional probability.
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$P(A \cap B)=P(A)P(B\mid A)$. Multiply $P(A)$ by the probability of $B$ given $A$ occurred.
$P(A \cap B)=P(A)P(B\mid A)$. Multiply $P(A)$ by the probability of $B$ given $A$ occurred.
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Two fair coins are flipped. Let $A={\text{first is H}}$ and $B={\text{exactly one H}}$. Find $P(A \cap B)$.
Two fair coins are flipped. Let $A={\text{first is H}}$ and $B={\text{exactly one H}}$. Find $P(A \cap B)$.
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$P(A \cap B)=\frac{1}{4}$. Only HT satisfies both conditions out of 4 outcomes.
$P(A \cap B)=\frac{1}{4}$. Only HT satisfies both conditions out of 4 outcomes.
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Two fair coins are flipped. Let $A={\text{first is H}}$ and $B={\text{exactly one H}}$. Find $P(B\mid A)$.
Two fair coins are flipped. Let $A={\text{first is H}}$ and $B={\text{exactly one H}}$. Find $P(B\mid A)$.
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$P(B\mid A)=\frac{1}{2}$. Given first H, only HT has exactly one H out of HH, HT.
$P(B\mid A)=\frac{1}{2}$. Given first H, only HT has exactly one H out of HH, HT.
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A deck has 52 cards. Draw 1 card. Let $A={\text{heart}}$ and $B={\text{face card}}$. Find $P(A \cap B)$.
A deck has 52 cards. Draw 1 card. Let $A={\text{heart}}$ and $B={\text{face card}}$. Find $P(A \cap B)$.
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$P(A \cap B)=\frac{3}{52}$. Three face cards (J,Q,K) in hearts out of 52 cards.
$P(A \cap B)=\frac{3}{52}$. Three face cards (J,Q,K) in hearts out of 52 cards.
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A deck has 52 cards. Draw 1 card. Let $A={\text{heart}}$ and $B={\text{face card}}$. Find $P(B\mid A)$.
A deck has 52 cards. Draw 1 card. Let $A={\text{heart}}$ and $B={\text{face card}}$. Find $P(B\mid A)$.
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$P(B\mid A)=\frac{3}{13}$. Three face cards out of 13 hearts in the suit.
$P(B\mid A)=\frac{3}{13}$. Three face cards out of 13 hearts in the suit.
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In a uniform model, what is $P(B\mid A)$ in terms of counts $|A\cap B|$ and $|A|$?
In a uniform model, what is $P(B\mid A)$ in terms of counts $|A\cap B|$ and $|A|$?
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$P(B\mid A)=\frac{|A\cap B|}{|A|}$. Count outcomes in both $A$ and $B$ divided by outcomes in $A$.
$P(B\mid A)=\frac{|A\cap B|}{|A|}$. Count outcomes in both $A$ and $B$ divided by outcomes in $A$.
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What condition makes the simplified rule $P(A \cap B)=P(A)P(B)$ valid?
What condition makes the simplified rule $P(A \cap B)=P(A)P(B)$ valid?
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$A$ and $B$ are independent. Independence allows factoring joint probability into product.
$A$ and $B$ are independent. Independence allows factoring joint probability into product.
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Identify the independence criterion stated using conditional probability.
Identify the independence criterion stated using conditional probability.
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$P(B\mid A)=P(B)$ (with $P(A)>0$). Independence means conditioning doesn't change probability.
$P(B\mid A)=P(B)$ (with $P(A)>0$). Independence means conditioning doesn't change probability.
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