Introduction to Probability: Additional Practice (Set 1)

There are $4$ kings and $13$ hearts, and one card, the king of hearts, is both. The addition rule removes that double count.

$P(\text{king} \cup \text{heart}) = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13}$

The value $\frac{17}{52}$ counts the king of hearts twice, once as a king and once as a heart.

Mutual exclusivity kills the correction term, since $P(A \cap B) = 0$.

$P(A \cup B) = P(A) + P(B)$

Substitute and solve.

$0.7 = 0.25 + P(B) \quad \Longrightarrow \quad P(B) = 0.45$

The value $0.175$ is the product $0.25 \times 0.7$, which answers nothing here, and $0.95$ comes from adding instead of subtracting.

Multiply both sides of the conditional definition by $P(B)$ to get the multiplication rule.

$P(A \cap B) = P(B) \cdot P(A \mid B)$

Substitute the given values.

$P(A \cap B) = 0.5 \times 0.4 = 0.2$

Dividing instead of multiplying gives $0.8$, which is the reversed operation and would make the intersection larger than $P(B)$, an impossibility since $A \cap B \subseteq B$.

A product is even as soon as at least one factor is even, so the complement is the tidy event: the product is odd exactly when both dice are odd.

$P(\text{both odd}) = \frac{3 \times 3}{36} = \frac{9}{36} = \frac{1}{4}$

Subtract from $1$.

$P(\text{product even}) = 1 - \frac{1}{4} = \frac{3}{4}$

Guessing $\frac{1}{2}$ assumes even and odd products are equally likely, but three of the four parity combinations produce an even product.

Count the complement, where all $23$ birthdays are different. Assign days one person at a time, each new person avoiding the days already used.

$P(\text{all different}) = \frac{365 \times 364 \times \cdots \times 343}{365^{23}} \approx 0.4927$

Subtract from $1$.

$P(\text{at least two share}) \approx 1 - 0.4927 = 0.5073$

The value near $0.06$ answers a different question, the chance that someone shares your birthday, which really is small. The room has $\binom{23}{2} = 253$ pairs of people, not $22$, and that is why the answer is so much larger than it feels.

"Different colours" is the complement of "the same colour", which was $\frac{19}{66}$.

$P(\text{different}) = 1 - \frac{19}{66} = \frac{47}{66} \approx 0.712$

Counting head on would mean three separate cases (red with blue, red with green, blue with green), giving $4 \cdot 5 + 4 \cdot 3 + 5 \cdot 3 = 47$ pairs, the same answer with three times the work.

The sample space is every four-person committee drawn from the nine people.

$\lvert S \rvert = \binom{9}{4} = 126$

Build a favourable committee by choosing $2$ of the $4$ women and then $2$ of the $5$ men, and multiply the two counts.

$\lvert A \rvert = \binom{4}{2}\binom{5}{2} = 6 \times 10 = 60$

Divide.

$P(\text{exactly } 2 \text{ women}) = \frac{60}{126} = \frac{10}{21} \approx 0.476$

Conditioning shrinks the sample space to the six rolls whose sum is at least $10$.

$B = \{(4,6), (5,5), (6,4), (5,6), (6,5), (6,6)\}, \qquad \lvert B \rvert = 6$

Exactly one of those, $(6,6)$, sums to $12$.

$P(\text{sum} = 12 \mid \text{sum} \ge 10) = \frac{1}{6}$

The unconditional probability of a sum of $12$ is $\frac{1}{36}$, so the information multiplied it by six.

Independence is the numerical test $P(A \cap B) = P(A)P(B)$, so check the pairs rather than trusting intuition. For a heart and a queen, the intersection is the single queen of hearts.

$P(A \cap B) = \frac{1}{52} \qquad \text{and} \qquad P(A)\,P(B) = \frac{1}{4} \cdot \frac{1}{13} = \frac{1}{52}$

They agree, so suit and rank are independent: every rank appears once in each suit, so learning the rank leaves the suits in the same proportions.

The others all fail. A heart is always red, so $P(\text{heart} \cap \text{red}) = \frac{1}{4}$ while the product is only $\frac{1}{8}$. A heart and a spade are mutually exclusive, so the intersection has probability $0$ while the product is $\frac{1}{16}$. And every king is a face card, so $P(\text{face} \cap \text{king}) = \frac{1}{13}$ while the product is $\frac{3}{13} \cdot \frac{1}{13} = \frac{3}{169}$.

Conditioning removes the six doubles, so the sample space shrinks from $36$ rolls to $30$.

$\lvert B \rvert = 36 - 6 = 30$

The $15$ rolls with the second entry larger all survive, because none of them was a tie.

$P(\text{second} > \text{first} \mid \text{different}) = \frac{15}{30} = \frac{1}{2}$

Now the symmetry argument works exactly. Once ties are ruled out, every roll has either the first or the second entry larger, and swapping the two dice pairs the two cases off perfectly.

The sample space is every pair of socks, so $\lvert S \rvert = \binom{10}{2} = 45$. A matching pair is two blacks or two whites, and those cases are mutually exclusive, so add their counts.

$\binom{6}{2} + \binom{4}{2} = 15 + 6 = 21$

Divide.

$P(\text{match}) = \frac{21}{45} = \frac{7}{15} \approx 0.467$

So a matching pair is slightly less likely than a mismatched one, whose probability is $\frac{8}{15}$. The value $\frac{13}{25}$ assumes the first sock is put back, which changes the answer.

Disjoint events have an empty intersection, and since $P(B) > 0$ the conditional probability is defined.

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

Learning that $B$ occurred does not leave $A$ unchanged; it rules $A$ out entirely, dropping its probability from a positive number to zero. That is why the events are not independent: the product $P(A)P(B)$ is positive while $P(A \cap B)$ is zero, so the defining equation of independence fails.

$P(A \cap B) = 0 \ne P(A)\,P(B)$

Mutually exclusive events with positive probability are as dependent as two events can be.