Statistics Flashcards: Conditional Probability And Independence

What this deck covers

This deck focuses on Conditional Probability And Independence, 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: State an equivalent independence condition using only $P(A\cap B)$, $P(A)$, and $P(B)$.

Answer: $A$ and $B$ are independent if $P(A\cap B)=P(A)P(B)$. Product rule holds when events don't influence each other.

Flashcard 2: Decide whether $A$ and $B$ are independent: $P(A)=0.5$, $P(B)=0.4$, $P(A\cap B)=0.2$.

Answer: Independent, since $P(A\cap B)=P(A)P(B)=0.2$. Checks if $P(A\cap B)=P(A)P(B)$: $0.2=0.5\times 0.4$ ✓.

Flashcard 3: State the multiplication rule that expresses $P(A\cap B)$ using $P(A\mid B)$ and $P(B)$.

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

Flashcard 4: Find $P(B\mid A)$ if $A$ and $B$ are independent and $P(B)=0.15$.

Answer: $P(B\mid A)=0.15$. Independence means $P(B\mid A)=P(B)$.

Flashcard 5: Identify the error: A student writes $P(A\mid B)=\frac{P(A\cup B)}{P(B)}$. What is the correction?

Answer: Use $P(A\mid B)=\frac{P(A\cap B)}{P(B)}$. Union $\cup$ should be intersection $\cap$.

Flashcard 6: What does the notation $P(A\mid B)$ mean in words?

Answer: Probability that $A$ occurs given that $B$ has occurred. The vertical bar means "given" or "assuming."

Flashcard 7: State the independence rule that connects $P(A\cap B)$, $P(A)$, and $P(B)$.

Answer: If independent, $P(A\cap B)=P(A)P(B)$. For independent events, joint probability equals product of individual probabilities.

Flashcard 8: Find $P(B)$ given $P(A\cap B)=0.18$ and $P(A\mid B)=0.30$.

Answer: $P(B)=0.6$. Rearranges $P(A\cap B)=P(A\mid B)P(B)$ to solve for $P(B)=\frac{0.18}{0.30}$.

Flashcard 9: Decide whether $A$ and $B$ are independent if $P(A)=0.40$ and $P(A\mid B)=0.25$.

Answer: Not independent. Since $0.25\neq 0.40$, conditioning changes probability.

Flashcard 10: Find $P(A\cap B)$ given $P(A\mid B)=0.25$ and $P(B)=0.60$.

Answer: $P(A\cap B)=0.15$. Uses multiplication rule: $P(A\cap B)=P(A\mid B)P(B)=0.25\times 0.60$.

Flashcard 11: Find $P(A\cap B)$ if $P(A\mid B)=0.25$ and $P(B)=0.60$.

Answer: $0.15$. Multiply: $P(A\cap B)=0.25\times 0.60=0.15$.

Flashcard 12: What is the definition of independence using conditional probability $P(B\mid A)$?

Answer: If $P(A)>0$, independence means $P(B\mid A)=P(B)$. Independence means knowing $A$ doesn't change the probability of $B$.

Flashcard 13: What is the definition of independence using conditional probability $P(A\mid B)$?

Answer: If $P(B)>0$, independence means $P(A\mid B)=P(A)$. Independence means knowing $B$ doesn't change the probability of $A$.

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

Answer: $P(A\cap B)=P(B\mid A)P(A)$. Alternative form using reversed conditioning.

Flashcard 15: Which option is equivalent to independence: $P(A\mid B)=P(A)$ or $P(A\mid B)=P(B)$?

Answer: $P(A\mid B)=P(A)$. Independence means conditioning doesn't change probability, not that they're equal.

Flashcard 16: Identify the correct relationship when $A$ and $B$ are independent and $P(B)>0$.

Answer: $P(A\cap B)=P(A\mid B)P(B)=P(A)P(B)$. Substitutes $P(A)$ for $P(A\mid B)$ when events are independent.

Flashcard 17: State the formula for conditional probability $P(A\mid B)$ in terms of $P(A\cap B)$ and $P(B)$.

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 18: What condition must be true for $P(A\mid B)$ to be defined?

Answer: $P(B)>0$. Can't condition on an impossible event.

Flashcard 19: Find $P(A\mid B)$ if $A$ and $B$ are independent and $P(A)=0.35$.

Answer: $P(A\mid B)=0.35$. For independent events, $P(A\mid B)=P(A)$ regardless of $B$.

Flashcard 20: State the independence condition using conditional probability of $A$ given $B$.

Answer: $A$ and $B$ are independent if $P(A\mid B)=P(A)$. Knowing $B$ doesn't change $A$'s probability.

Flashcard 21: Find $P(A\mid B)$ if $A$ and $B$ are independent and $P(A)=0.72$.

Answer: $P(A\mid B)=0.72$. Independence means $P(A\mid B)=P(A)$.

Flashcard 22: Find $P(A\mid B)$ given $P(A\cap B)=0.12$ and $P(B)=0.30$.

Answer: $P(A\mid B)=0.4$. Uses $P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{0.12}{0.30}$.

Flashcard 23: Find $P(A\mid B)$ if $P(A\cap B)=0.12$ and $P(B)=0.30$.

Answer: $0.4$. Apply $P(A\mid B)=\frac{0.12}{0.30}=0.4$.

Flashcard 24: Find $P(B)$ if $P(A\cap B)=0.09$ and $P(A\mid B)=0.30$.

Answer: $0.3$. Rearrange: $P(B)=\frac{0.09}{0.30}=0.3$.

Flashcard 25: Find $P(B\mid A)$ if $A$ and $B$ are independent and $P(B)=0.80$.

Answer: $P(B\mid A)=0.80$. For independent events, $P(B\mid A)=P(B)$ regardless of $A$.

Flashcard 26: Identify the value of $P(A\cap B)$ from $P(A)=0.3$, $P(B)=0.5$, assuming independence.

Answer: $P(A\cap B)=0.15$. Uses independence formula: $P(A\cap B)=P(A)P(B)=0.3\times 0.5$.

Flashcard 27: Compute $P(A\mid B)$ from the table: $A\cap B=15$, $A^c\cap B=35$.

Answer: $P(A\mid B)=\frac{15}{50}=0.3$. Total in $B$ is $15+35=50$, so $P(A\mid B)=\frac{15}{50}$.

Flashcard 28: Identify the correct expression for $P(A\cap B)$ in terms of $P(A)$ and $P(B\mid A)$.

Answer: $P(A\cap B)=P(A)P(B\mid A)$. Applies the multiplication rule correctly.

Flashcard 29: Find $P(A\cap B)$ if $A$ and $B$ are independent, $P(A)=0.30$, and $P(B)=0.40$.

Answer: $0.12$. For independent events, multiply probabilities.

Flashcard 30: Which probability is updated by new information: $P(A)$ or $P(A\mid B)$?

Answer: $P(A\mid B)$. Conditioning incorporates new information.

Flashcard 31: Decide whether $A$ and $B$ are independent if $P(A)=0.50$ and $P(A\mid B)=0.50$.

Answer: Independent. Since $P(A\mid B)=P(A)$, they're independent.

Flashcard 32: State the independence condition using conditional probability of $B$ given $A$.

Answer: $A$ and $B$ are independent if $P(B\mid A)=P(B)$. Knowing $A$ doesn't change $B$'s probability.

Flashcard 33: Decide whether $A$ and $B$ are independent: $P(A)=0.6$, $P(B)=0.5$, $P(A\cap B)=0.25$.

Answer: Not independent, since $P(A)P(B)=0.3\ne 0.25$. Checks if $P(A\cap B)=P(A)P(B)$: $0.25\ne 0.6\times 0.5=0.3$ ✗.

Flashcard 34: Find $P(A)$ assuming independence, given $P(A\cap B)=0.14$ and $P(B)=0.70$.

Answer: $P(A)=0.2$. Uses independence: $P(A)=\frac{P(A\cap B)}{P(B)}=\frac{0.14}{0.70}$.

Flashcard 35: Decide whether $A$ and $B$ are independent if $P(A\cap B)=0.10$, $P(A)=0.20$, and $P(B)=0.50$.

Answer: Independent. Check: $0.10=0.20\times 0.50$ confirms independence.