Introduction to Probability: Practice

There are $13$ hearts and $12$ face cards, and three cards (the jack, queen, and king of hearts) belong to both groups. Adding $13$ and $12$ would count those three twice, so the addition rule subtracts them once.

$P(\text{heart} \cup \text{face}) = \frac{13}{52} + \frac{12}{52} - \frac{3}{52} = \frac{22}{52} = \frac{11}{26}$

The value $\frac{25}{52}$ is exactly the answer you get by forgetting the correction term, and it claims $25$ favourable cards when only $22$ exist.

There are $26$ red cards and $4$ kings, and two cards are both red and kings (the king of hearts and the king of diamonds).

$P(\text{red} \cup \text{king}) = \frac{26}{52} + \frac{4}{52} - \frac{2}{52} = \frac{28}{52} = \frac{7}{13}$

The value $\frac{15}{26} = \frac{30}{52}$ is what you get by skipping the subtraction, which double counts the two red kings.

Use complementary counting, since "at least one" is the opposite of "none". For neither die to show a $4$, each die must land on one of the five other faces.

$P(\text{no } 4) = \frac{5 \times 5}{36} = \frac{25}{36}$

Subtract from $1$.

$P(\text{at least one } 4) = 1 - \frac{25}{36} = \frac{11}{36}$

The answer $\frac{1}{3} = \frac{12}{36}$ comes from adding $\frac{1}{6} + \frac{1}{6}$, which double counts the roll $(4,4)$.

The sample space has $2^3 = 8$ equally likely outcomes. The complement of "at least one head" is "no heads at all", which happens only for TTT, a single outcome.

$P(\text{no heads}) = \frac{1}{8}$

Subtract from $1$.

$P(\text{at least one head}) = 1 - \frac{1}{8} = \frac{7}{8}$

Test independence by comparing the intersection with the product of the two probabilities.

$P(A)\,P(B) = 0.6 \times 0.5 = 0.3 = P(A \cap B)$

The two sides agree, so the events are independent. They are not mutually exclusive, because $P(A \cap B) = 0.3 \ne 0$. The conditional probability is $P(A \mid B) = \frac{0.3}{0.5} = 0.6$, matching $P(A)$ as independence requires, and the union is $0.6 + 0.5 - 0.3 = 0.8$, not $1.1$.

A committee is a set, so order does not matter and the counts are combinations. The sample space is every three-person committee from the $10$ people.

$\lvert S \rvert = \binom{10}{3} = 120, \qquad \lvert A \rvert = \binom{6}{3} = 20$

Divide.

$P(\text{all men}) = \frac{20}{120} = \frac{1}{6}$

The value $\frac{27}{125} = \left(\frac{3}{5}\right)^3$ would be right only if each pick were made from the full group of $10$ with replacement, which would allow the same person to be chosen twice.

"At least one green" is the complement of "no green at all", and no green means both marbles are red, which was computed as $\frac{5}{14}$.

$P(\text{at least one green}) = 1 - P(\text{both red}) = 1 - \frac{5}{14} = \frac{9}{14}$

Counting the cases directly (exactly one green, or two green) would give the same answer with three times the work.

The sample space is the ordered triples of faces, so $\lvert S \rvert = 6^3 = 216$. To avoid a $6$, each roll must land on one of the five other faces.

$\lvert A \rvert = 5 \times 5 \times 5 = 125$

Divide.

$P(\text{no } 6) = \frac{125}{216} \approx 0.579$

This is the complement of the previous question, where $P(\text{no } 6) = \frac{125}{216}$.

$P(\text{at least one } 6) = 1 - \frac{125}{216} = \frac{91}{216} \approx 0.421$

Adding $\frac{1}{6}$ three times would give $\frac{1}{2}$, which is wrong because it double counts the rolls that contain more than one $6$.

Mutually exclusive means $A \cap B = \emptyset$, so the intersection has probability $0$. Since $P(B) = 0.4 > 0$, the conditional probability is defined.

$P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{0}{0.4} = 0$

Learning that $B$ happened rules $A$ out completely, which is why the probability collapses from $0.3$ all the way to $0$.

Conditioning on $B$ shrinks the sample space to the multiples of $3$ on the spinner.

$B = \{3, 6\}, \qquad A \cap B = \{6\}$

One of those two outcomes is even.

$P(A \mid B) = \frac{\lvert A \cap B \rvert}{\lvert B \rvert} = \frac{1}{2}$

Since $P(A) = \frac{4}{8} = \frac{1}{2}$ as well, the conditional probability equals the unconditional one, so $A$ and $B$ happen to be independent here.

Knowing the card is a heart shrinks the sample space from $52$ cards to the $13$ hearts. Inside that world, the face cards are the jack, queen, and king of hearts.

$P(\text{face} \mid \text{heart}) = \frac{\lvert \text{face} \cap \text{heart} \rvert}{\lvert \text{heart} \rvert} = \frac{3}{13}$

The value $\frac{3}{52}$ is $P(\text{face and heart})$, the numerator before the sample space was shrunk.