Calculating Probability - ISEE Middle Level: Mathematics Achievement

$\frac{2}{5}$. The set has ten numbers, with four primes (2, 3, 5, 7) equally likely to be chosen.

$P(A\cup B)=P(A)+P(B)-P(A\cap B)$. The general addition rule subtracts the intersection to correct for overlapping outcomes in the union.

$P(A\cap B)=P(A)\cdot P(B)$. For independent events, the joint probability multiplies individual probabilities as one does not affect the other.

$\frac{1}{6}$. A fair die has six equally likely outcomes, with only one favorable for rolling a 5.

$\frac{1}{2}$. Even numbers (2, 4, 6) are three of the six equally likely outcomes on a fair die.

$\frac{1}{3}$. Numbers greater than 4 (5, 6) are two of the six equally likely outcomes on a fair die.

$\frac{1}{2}$. A fair coin has two equally likely outcomes, with heads as one of them.

$\frac{1}{2}$. Two coin flips yield four equally likely outcomes, with exactly one head in two of them (HT, TH).

$\frac{1}{4}$. Two coin flips yield four equally likely outcomes, with two heads (HH) in one of them.

$\frac{3}{4}$. Two coin flips yield four equally likely outcomes, with at least one head in three of them (HH, HT, TH).

$\frac{1}{2}$. A standard deck has 52 cards, with 26 red cards equally likely to be drawn.

$\frac{1}{13}$. A standard deck has 52 cards, with 4 aces equally likely to be drawn.

$\frac{1}{4}$. A standard deck has 52 cards, with 13 hearts equally likely to be drawn.

$\frac{3}{13}$. A standard deck has 52 cards, with 12 face cards (J, Q, K in 4 suits) equally likely to be drawn.

$\frac{3}{5}$. The complement rule gives $1 - \frac{2}{5} = \frac{3}{5}$ for the probability of not $A$.

$\frac{7}{12}$. For mutually exclusive events, add probabilities: $\frac{1}{4} + \frac{1}{3} = \frac{3}{12} + \frac{4}{12} = \frac{7}{12}$.

$\frac{2}{3}$. Use the general addition rule: $\frac{1}{2} + \frac{1}{3} - \frac{1}{6} = \frac{3}{6} + \frac{2}{6} - \frac{1}{6} = \frac{4}{6}$.

$\frac{6}{35}$. For independent events, multiply: $\frac{3}{5} \times \frac{2}{7} = \frac{6}{35}$.

$\frac{1}{4}$. The word MATH has four distinct letters, with one vowel (A) equally likely to be chosen.

$\frac{1}{3}$. The set has nine numbers, with three multiples of 3 (3, 6, 9) equally likely to be chosen.

$P(\text{event})=\frac{\text{favorable outcomes}}{\text{total outcomes}}$. This formula divides the number of favorable outcomes by the total number of equally likely outcomes in the sample space.

$0$. An impossible event has no favorable outcomes, resulting in zero probability.

$P(A^c)=1-P(A)$. The complement rule subtracts the probability of event $A$ from 1 to find the probability of not $A$.

$P(A\cup B)=P(A)+P(B)$. For mutually exclusive events, the union probability is the sum since there is no overlap.