Single and Compound Probability - ISEE Upper Level: Mathematics Achievement

$\frac{7}{12}$. Adds probabilities since events are mutually exclusive, with no overlap.

$\frac{3}{13}$. Twelve face cards ($3$ per suit across $4$ suits) out of $52$ total cards.

$\frac{1}{13}$. Four aces (one per suit) out of $52$ cards in the deck.

$\frac{1}{2}$. There are $26$ red cards (hearts and diamonds) out of $52$ total cards.

$\frac{1}{3}$. Two favorable outcomes ($2$ or $5$) out of six, and events are mutually exclusive.

$\frac{5}{6}$. Uses the complement rule: $1$ minus the probability of rolling a $6$ ($\frac{1}{6}$).

$\frac{1}{2}$. Two equally likely outcomes (heads or tails) on a fair coin, with one favorable.

$\frac{1}{6}$. One favorable outcome ($6$) out of six equally likely faces on the die.

$\frac{1}{4}$. There are $13$ hearts out of $52$ cards, yielding the ratio of favorable to total outcomes.

$\frac{1}{2}$. Three even numbers ($2,4,6$) out of six possible outcomes on the die.

$P(A)=\frac{\text{favorable}}{\text{total}}$. Defines probability as the ratio of favorable outcomes to total possible outcomes in a sample space.

$P(A^c)=1-P(A)$. Calculates the probability of the complement by subtracting the event's probability from 1, as they are mutually exclusive and exhaustive.

$P(A\cup B)=P(A)+P(B)$. Adds probabilities directly for mutually exclusive events since their intersection is empty.

$P(A\cup B)=P(A)+P(B)-P(A\cap B)$. Accounts for overlap by subtracting the intersection probability from the sum of individual probabilities.

$P(A\mid B)=\frac{P(A\cap B)}{P(B)}$. Divides the joint probability by the probability of the conditioning event to find the likelihood given that event.

$\frac{3}{10}$. Multiplies $P(A)$ by conditional $P(B\mid A)$ for joint probability of dependent events.

$\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}$. Independent draws with replacement, each with $\frac{1}{2}$ probability of red.

$\frac{4}{52}\cdot\frac{3}{51}=\frac{1}{221}$. Multiplies probabilities of drawing first ace then second without replacement.

$\frac{13}{52}\cdot\frac{12}{51}=\frac{1}{17}$. Multiplies conditional probabilities for first and second heart without replacement.

$\frac{1}{6}$. Six double outcomes out of $36$ total combinations for two dice.

$\frac{1}{6}$. Six ways to get sum $7$ out of $36$ possible outcomes for two dice.

$\frac{1}{2}$. Two favorable outcomes (HT, TH) out of four possible results for two coins.

$\frac{1}{4}$. Independent coin flips each with $\frac{1}{2}$ probability of heads, multiplied for both.

$P(A\cap B)=P(A)P(B\mid A)$. Expresses joint probability using the initial event and the conditional probability for dependent events.