Statistics Flashcards: Finding Conditional Probability In Models

What this deck covers

This deck focuses on Finding Conditional Probability In Models, giving you a quick way to review the definitions, rules, and examples that matter most for Statistics.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

Flashcard 1: Find $P(A\mid B)$ if $|A\cap B|=18$ and $|B|=60$ in an equally likely model.

Answer: $0.30$. Divide favorable count by total in $B$: $\frac{18}{60}=0.30$.

Flashcard 2: Find and correct the error: $P(A\mid B)=\frac{P(A)}{P(B)}$.

Answer: Correct: $P(A\mid B)=\frac{P(A\cap B)}{P(B)}$. Must use intersection, not just $P(A)$.

Flashcard 3: What is $P(A\mid B)$ if $B\subseteq A$ and $P(B)>0$?

Answer: $1$. All of $B$ is in $A$, so $P(A\cap B)=P(B)$.

Flashcard 4: In a two-way table, which cell count is the numerator for $P(A\mid B)$?

Answer: The intersection count $\#(A\cap B)$. Counts outcomes satisfying both conditions.

Flashcard 5: State the formula for $P(A\mid B)$ using probabilities (not outcome counts).

Answer: $P(A\mid B)=\frac{P(A\cap B)}{P(B)}$, with $P(B)>0$. Divides joint probability by the condition's probability.

Flashcard 6: If $A\subseteq B$ and $P(B)>0$, what is $P(A\mid B)$ in terms of $P(A)$ and $P(B)$?

Answer: $\frac{P(A)}{P(B)}$. When $A\subseteq B$, $P(A\cap B)=P(A)$.

Flashcard 7: What is the relationship between $P(A\cap B)$ and $P(A\mid B)$?

Answer: $P(A\cap B)=P(A\mid B)\,P(B)$. Multiplication rule for joint probabilities.

Flashcard 8: A bag has $10$ red and $6$ blue marbles; $4$ red are large and $3$ blue are large. Find $P(\text{red}\mid \text{large})$.

Answer: $\frac{4}{7}$. Among $7$ large marbles, $4$ are red.

Flashcard 9: A class has $20$ students; $8$ are in band, and $5$ are in band and play a sport. Find $P(\text{sport}\mid \text{band})$.

Answer: $\frac{5}{8}$. Among $8$ band students, $5$ play sports.

Flashcard 10: Find $P(A\cap B)$ if $P(A\mid B)=0.20$ and $P(B)=0.50$.

Answer: $0.10$. Multiply: $0.20 \times 0.50 = 0.10$.

Flashcard 11: Identify the condition required for $P(A\mid B)$ to be defined.

Answer: $P(B)>0$. Cannot divide by zero probability.

Flashcard 12: What does $P(A\mid B)$ mean in words in terms of outcomes in $B$?

Answer: Probability of $A$ among outcomes restricted to $B$. Restricts sample space to only outcomes in $B$.

Flashcard 13: State the fraction interpretation of $P(A\mid B)$ using counts $|A\cap B|$ and $|B|$.

Answer: $P(A\mid B)=\frac{|A\cap B|}{|B|}$ (equally likely outcomes). Counts favorable outcomes within the restricted set $B$.

Flashcard 14: State the multiplication rule that rewrites $P(A\cap B)$ using $P(A\mid B)$.

Answer: $P(A\cap B)=P(A\mid B)P(B)$. Rearranges conditional probability formula.

Flashcard 15: Which expression equals $P(A\mid B)$: $\frac{P(A\cap B)}{P(B)}$ or $\frac{P(A\cup B)}{P(B)}$?

Answer: $\frac{P(A\cap B)}{P(B)}$. Uses intersection, not union, in numerator.

Flashcard 16: If events $A$ and $B$ are disjoint and $P(B)>0$, what is $P(A\mid B)$?

Answer: $0$. Disjoint events have empty intersection.

Flashcard 17: Identify the correct equality for independent events using conditional probability.

Answer: If independent, then $P(A\mid B)=P(A)$. Independence means conditioning doesn't change probability.

Flashcard 18: Find $P(B)$ if $P(A\cap B)=0.18$ and $P(A\mid B)=0.60$.

Answer: $0.30$. Rearrange: $P(B)=\frac{0.18}{0.60}=0.30$.

Flashcard 19: State the formula for conditional probability $P(A\mid B)$ using intersection and $P(B)$.

Answer: $P(A\mid B)=\frac{P(A\cap B)}{P(B)}$ for $P(B)>0$. Divides joint probability by the condition's probability.

Flashcard 20: Choose the correct interpretation: If $P(A\mid B)=0.7$, what does $0.7$ represent?

Answer: Among outcomes in $B$, the fraction that also satisfy $A$ is $0.7$. Proportion of $B$ outcomes that also satisfy $A$.

Flashcard 21: What does $P(A\mid B)$ mean in words in a probability model?

Answer: Probability that $A$ occurs given that $B$ has occurred. Updates probability based on knowing $B$ happened.

Flashcard 22: What is the definition of conditional probability $P(A\mid B)$ in terms of outcomes?

Answer: $P(A\mid B)=\frac{\#(A\cap B)}{\#(B)}$ for equally likely outcomes. Counts favorable outcomes in $B$ that also satisfy $A$.

Flashcard 23: A jar has $10$ marbles: $6$ red, $4$ blue; $3$ red are striped, $1$ blue is striped. Find $P(\text{striped}\mid \text{red})$.

Answer: $\frac{3}{6}=\frac{1}{2}$. Among $6$ red marbles, $3$ are striped.

Flashcard 24: Compute $P(A\mid B)$ from a table where $|A\cap B|=25$ and $|B|=100$.

Answer: $0.25$. Fraction of $B$'s outcomes in $A$: $\frac{25}{100}=0.25$.

Flashcard 25: A table shows $\#(A\cap B)=9$ and $\#(B)=15$. What is $P(A\mid B)$?

Answer: $\frac{9}{15}=\frac{3}{5}$. Direct calculation using table counts.

Flashcard 26: Compute $P(A\mid B)$ if $A$ and $B$ are independent and $P(A)=0.35$.

Answer: $0.35$. Independence implies $P(A\mid B)=P(A)$.

Flashcard 27: If $B\subseteq A$ and $P(B)>0$, what is $P(A\mid B)$?

Answer: $1$. If $B\subseteq A$, all $B$ outcomes are in $A$.

Flashcard 28: State the multiplication rule that rewrites $P(A\cap B)$ using $P(B\mid A)$.

Answer: $P(A\cap B)=P(B\mid A)P(A)$. Same rule with roles of $A$ and $B$ reversed.

Flashcard 29: In a class, $12$ students are left-handed; $5$ are left-handed and wear glasses. Find $P(\text{glasses}\mid \text{left})$.

Answer: $\frac{5}{12}$. Among $12$ left-handed, $5$ wear glasses.

Flashcard 30: What condition must hold for $P(A\mid B)$ to be defined?

Answer: $P(B)>0$. Cannot divide by zero probability.

Flashcard 31: In a two-way table, which total count is the denominator for $P(A\mid B)$?

Answer: The $B$ row (or $B$ column) total $\#(B)$. Total count of the given condition.

Flashcard 32: Find $P(A\mid B)$ if $\#(A\cap B)=12$ and $\#(B)=30$ (equally likely outcomes).

Answer: $\frac{12}{30}=\frac{2}{5}$. Direct application of outcome-based formula.

Flashcard 33: Identify the correct interpretation: If $P(A\mid B)=0.70$, what does $0.70$ describe?

Answer: $70\%$ of outcomes in $B$ are also in $A$. Fraction of $B$'s outcomes that belong to $A$.

Flashcard 34: What is $P(B\mid A)$ in terms of $P(A\cap B)$ and $P(A)$?

Answer: $P(B\mid A)=\frac{P(A\cap B)}{P(A)}$, with $P(A)>0$. Swaps roles of $A$ and $B$ in conditional formula.

Flashcard 35: What is $P(A\mid B)$ if $A$ and $B$ are disjoint and $P(B)>0$?

Answer: $0$. Disjoint means $A\cap B=\emptyset$, so $P(A\cap B)=0$.

Flashcard 36: Find $P(A\cap B)$ if $P(A\mid B)=0.25$ and $P(B)=0.40$.

Answer: $0.25\times 0.40=0.10$. Uses multiplication rule $P(A\cap B)=P(A\mid B)P(B)$.

Flashcard 37: Choose the correct denominator for $P(A\mid B)$ in a two-way table: $|A\cap B|$, $|A|$, or $|B|$?

Answer: Denominator is $|B|$. Restricts to outcomes in condition $B$.

Flashcard 38: Identify the sample space used when computing $P(A\mid B)$ from equally likely outcomes.

Answer: The restricted sample space consisting only of outcomes in $B$. Conditioning restricts to outcomes where $B$ occurs.