Multi-Event Probability - ISEE Upper Level: Quantitative Reasoning
Card 1 of 25
A fair die is rolled twice. What is the probability of at least one $6$?
A fair die is rolled twice. What is the probability of at least one $6$?
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$\frac{11}{36}$. Complement of no sixes: $1 - \left(\frac{5}{6}\right)^2 = 1 - \frac{25}{36}$.
$\frac{11}{36}$. Complement of no sixes: $1 - \left(\frac{5}{6}\right)^2 = 1 - \frac{25}{36}$.
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State the multiplication rule for independent events $A$ and $B$.
State the multiplication rule for independent events $A$ and $B$.
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$P(A \cap B)=P(A)P(B)$. For independent events, the probability of their intersection equals the product of their individual probabilities.
$P(A \cap B)=P(A)P(B)$. For independent events, the probability of their intersection equals the product of their individual probabilities.
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State the general multiplication rule for events $A$ and $B$ using conditional probability.
State the general multiplication rule for events $A$ and $B$ using conditional probability.
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$P(A \cap B)=P(A)P(B\mid A)$. The general rule expresses the joint probability as the product of one event's probability and the conditional probability of the other given the first.
$P(A \cap B)=P(A)P(B\mid A)$. The general rule expresses the joint probability as the product of one event's probability and the conditional probability of the other given the first.
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State the addition rule for any events $A$ and $B$.
State the addition rule for any events $A$ and $B$.
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$P(A\cup B)=P(A)+P(B)-P(A\cap B)$. The addition rule accounts for overlap by subtracting the intersection probability from the sum of individual probabilities.
$P(A\cup B)=P(A)+P(B)-P(A\cap B)$. The addition rule accounts for overlap by subtracting the intersection probability from the sum of individual probabilities.
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State the addition rule for mutually exclusive events $A$ and $B$.
State the addition rule for mutually exclusive events $A$ and $B$.
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$P(A\cup B)=P(A)+P(B)$. Mutually exclusive events have no overlap, so their union probability is simply the sum of their individual probabilities.
$P(A\cup B)=P(A)+P(B)$. Mutually exclusive events have no overlap, so their union probability is simply the sum of their individual probabilities.
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State the complement rule for an event $A$.
State the complement rule for an event $A$.
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$P(A^c)=1-P(A)$. The complement rule states that the probability of an event not occurring is one minus the probability of it occurring.
$P(A^c)=1-P(A)$. The complement rule states that the probability of an event not occurring is one minus the probability of it occurring.
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State the conditional probability formula for $P(A\mid B)$.
State the conditional probability formula for $P(A\mid B)$.
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$P(A\mid B)=\frac{P(A\cap B)}{P(B)}$. Conditional probability is the joint probability divided by the probability of the conditioning event.
$P(A\mid B)=\frac{P(A\cap B)}{P(B)}$. Conditional probability is the joint probability divided by the probability of the conditioning event.
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What condition must hold for events $A$ and $B$ to be independent?
What condition must hold for events $A$ and $B$ to be independent?
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$P(A\cap B)=P(A)P(B)$. Independence requires that the joint probability equals the product of the marginal probabilities.
$P(A\cap B)=P(A)P(B)$. Independence requires that the joint probability equals the product of the marginal probabilities.
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What condition must hold for events $A$ and $B$ to be mutually exclusive?
What condition must hold for events $A$ and $B$ to be mutually exclusive?
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$P(A\cap B)=0$. Mutually exclusive events cannot occur together, so their intersection probability is zero.
$P(A\cap B)=0$. Mutually exclusive events cannot occur together, so their intersection probability is zero.
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State De Morgan's law for the complement of a union: $(A\cup B)^c$.
State De Morgan's law for the complement of a union: $(A\cup B)^c$.
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$(A\cup B)^c=A^c\cap B^c$. De Morgan's law equates the complement of a union to the intersection of the complements.
$(A\cup B)^c=A^c\cap B^c$. De Morgan's law equates the complement of a union to the intersection of the complements.
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State De Morgan's law for the complement of an intersection: $(A\cap B)^c$.
State De Morgan's law for the complement of an intersection: $(A\cap B)^c$.
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$(A\cap B)^c=A^c\cup B^c$. De Morgan's law equates the complement of an intersection to the union of the complements.
$(A\cap B)^c=A^c\cup B^c$. De Morgan's law equates the complement of an intersection to the union of the complements.
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If $P(A)=0.3$ and $P(B)=0.5$ and $A,B$ are independent, what is $P(A\cap B)$?
If $P(A)=0.3$ and $P(B)=0.5$ and $A,B$ are independent, what is $P(A\cap B)$?
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$0.15$. Since events are independent, multiply their probabilities: $0.3 \times 0.5$.
$0.15$. Since events are independent, multiply their probabilities: $0.3 \times 0.5$.
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If $P(A)=0.6$ and $P(B\mid A)=0.2$, what is $P(A\cap B)$?
If $P(A)=0.6$ and $P(B\mid A)=0.2$, what is $P(A\cap B)$?
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$0.12$. Apply the multiplication rule: $P(A) \times P(B \mid A) = 0.6 \times 0.2$.
$0.12$. Apply the multiplication rule: $P(A) \times P(B \mid A) = 0.6 \times 0.2$.
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If $P(A)=0.4$, $P(B)=0.5$, and $P(A\cap B)=0.1$, what is $P(A\cup B)$?
If $P(A)=0.4$, $P(B)=0.5$, and $P(A\cap B)=0.1$, what is $P(A\cup B)$?
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$0.8$. Use the addition rule: $0.4 + 0.5 - 0.1$.
$0.8$. Use the addition rule: $0.4 + 0.5 - 0.1$.
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If $P(A)=0.35$, what is $P(A^c)$?
If $P(A)=0.35$, what is $P(A^c)$?
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$0.65$. Apply the complement rule: $1 - 0.35$.
$0.65$. Apply the complement rule: $1 - 0.35$.
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If $P(A\cap B)=0.12$ and $P(B)=0.3$, what is $P(A\mid B)$?
If $P(A\cap B)=0.12$ and $P(B)=0.3$, what is $P(A\mid B)$?
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$0.4$. Use conditional probability: $\frac{0.12}{0.3}$.
$0.4$. Use conditional probability: $\frac{0.12}{0.3}$.
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A fair coin is flipped twice. What is the probability of getting $2$ heads?
A fair coin is flipped twice. What is the probability of getting $2$ heads?
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$\frac{1}{4}$. Each flip is independent with $P(H)=\frac{1}{2}$, so multiply: $\left(\frac{1}{2}\right)^2$.
$\frac{1}{4}$. Each flip is independent with $P(H)=\frac{1}{2}$, so multiply: $\left(\frac{1}{2}\right)^2$.
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A fair die is rolled twice. What is the probability both rolls are even?
A fair die is rolled twice. What is the probability both rolls are even?
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$\frac{1}{4}$. Each roll has $P(\text{even})=\frac{3}{6}=\frac{1}{2}$, and independent, so $\left(\frac{1}{2}\right)^2$.
$\frac{1}{4}$. Each roll has $P(\text{even})=\frac{3}{6}=\frac{1}{2}$, and independent, so $\left(\frac{1}{2}\right)^2$.
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A fair coin is flipped $3$ times. What is the probability of at least one head?
A fair coin is flipped $3$ times. What is the probability of at least one head?
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$\frac{7}{8}$. Complement of all tails: $1 - \left(\frac{1}{2}\right)^3 = 1 - \frac{1}{8}$.
$\frac{7}{8}$. Complement of all tails: $1 - \left(\frac{1}{2}\right)^3 = 1 - \frac{1}{8}$.
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A card is drawn from a $52$-card deck. What is $P(\text{heart or king})$?
A card is drawn from a $52$-card deck. What is $P(\text{heart or king})$?
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$\frac{4}{13}$. Addition rule: $P(\text{heart}) + P(\text{king}) - P(\text{heart and king}) = \frac{13}{52} + \frac{4}{52} - \frac{1}{52} = \frac{16}{52}$.
$\frac{4}{13}$. Addition rule: $P(\text{heart}) + P(\text{king}) - P(\text{heart and king}) = \frac{13}{52} + \frac{4}{52} - \frac{1}{52} = \frac{16}{52}$.
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Two cards are drawn without replacement. What is $P(\text{both aces})$?
Two cards are drawn without replacement. What is $P(\text{both aces})$?
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$\frac{1}{221}$. Multiplication rule without replacement: $\frac{4}{52} \times \frac{3}{51} = \frac{12}{2652} = \frac{1}{221}$.
$\frac{1}{221}$. Multiplication rule without replacement: $\frac{4}{52} \times \frac{3}{51} = \frac{12}{2652} = \frac{1}{221}$.
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Two cards are drawn without replacement. What is $P(\text{first ace, second king})$?
Two cards are drawn without replacement. What is $P(\text{first ace, second king})$?
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$\frac{4}{663}$. Sequential probabilities: $P(\text{first ace}) = \frac{4}{52}$, $P(\text{second king} \mid \text{first ace}) = \frac{4}{51}$, product simplifies to $\frac{4}{663}$.
$\frac{4}{663}$. Sequential probabilities: $P(\text{first ace}) = \frac{4}{52}$, $P(\text{second king} \mid \text{first ace}) = \frac{4}{51}$, product simplifies to $\frac{4}{663}$.
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A bag has $3$ red and $2$ blue marbles. Two are drawn without replacement. What is $P(\text{both red})$?
A bag has $3$ red and $2$ blue marbles. Two are drawn without replacement. What is $P(\text{both red})$?
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$\frac{3}{10}$. Multiplication rule: $P(\text{first red}) = \frac{3}{5}$, $P(\text{second red} \mid \text{first red}) = \frac{2}{4}$, product is $\frac{3}{10}$.
$\frac{3}{10}$. Multiplication rule: $P(\text{first red}) = \frac{3}{5}$, $P(\text{second red} \mid \text{first red}) = \frac{2}{4}$, product is $\frac{3}{10}$.
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A bag has $3$ red and $2$ blue marbles. Two are drawn without replacement. What is $P(\text{one red and one blue})$?
A bag has $3$ red and $2$ blue marbles. Two are drawn without replacement. What is $P(\text{one red and one blue})$?
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$\frac{3}{5}$. Sum of red-then-blue and blue-then-red: $\frac{3}{5} \times \frac{2}{4} + \frac{2}{5} \times \frac{3}{4} = \frac{3}{10} + \frac{3}{10} = \frac{3}{5}$.
$\frac{3}{5}$. Sum of red-then-blue and blue-then-red: $\frac{3}{5} \times \frac{2}{4} + \frac{2}{5} \times \frac{3}{4} = \frac{3}{10} + \frac{3}{10} = \frac{3}{5}$.
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If $P(A)=0.25$ and $P(B\mid A)=0.6$, what is $P(B\cap A)$?
If $P(A)=0.25$ and $P(B\mid A)=0.6$, what is $P(B\cap A)$?
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$0.15$. Apply the multiplication rule: $0.25 \times 0.6$.
$0.15$. Apply the multiplication rule: $0.25 \times 0.6$.
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