Introduction to Probability: Practice

The sample space is $S = \{1,2,3,4,5,6\}$, six equally likely outcomes, so $\lvert S \rvert = 6$. The event $A$ holds three of them.

$P(A) = \frac{\lvert A \rvert}{\lvert S \rvert} = \frac{3}{6} = \frac{1}{2}$

Always reduce, so the answer is $\frac{1}{2}$.

Every outcome is either in $A$ or in $A^{c}$ and never in both, so the two probabilities fill up the whole and add to $1$. That is the complement rule.

$P(A^{c}) = 1 - P(A) = 1 - 0.3 = 0.7$

Subtract from $1$; never double the probability or flip it.

The deck has four suits of $13$ cards each, so there are $13$ spades among $52$ equally likely cards.

$P(\text{spade}) = \frac{13}{52} = \frac{1}{4}$

Each suit takes one quarter of the deck.

Start from the general addition rule, then use the fact that disjoint events have an empty overlap.

$P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.4 + 0.3 - 0 = 0.7$

The correction term is zero here, so the probabilities simply add. The value $0.12$ is the product $P(A)P(B)$, which answers a different question.

Adding $P(A)$ and $P(B)$ counts the overlap twice, so the addition rule subtracts it once.

$P(A \cup B) = 0.5 + 0.4 - 0.2 = 0.7$

The value $0.9$ is what you get by forgetting the correction, and it overstates the union by exactly the size of the overlap.

The union is every outcome in either event, so combine the two sets.

$A \cup B = \{2, 4, 6\} \cup \{5, 6\} = \{2, 4, 5, 6\}$

That is four outcomes out of six, giving $\frac{4}{6} = \frac{2}{3}$. The addition rule agrees, once the shared face $6$ is subtracted.

$P(A \cup B) = \frac{3}{6} + \frac{2}{6} - \frac{1}{6} = \frac{4}{6} = \frac{2}{3}$

Independence means that learning $B$ leaves the probability of $A$ unchanged, $P(A \mid B) = P(A)$. Substituting that into the multiplication rule $P(A \cap B) = P(B)P(A \mid B)$ gives the product form.

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

The equation $P(A \cap B) = 0$ says the events have an empty intersection here, which is mutual exclusivity, a different (and in fact opposing) condition.

The denominator is the whole bag, not just the other colour, so count every marble.

$\lvert S \rvert = 4 + 6 = 10$

Four of those ten are red.

$P(\text{red}) = \frac{4}{10} = \frac{2}{5}$

Conditioning shrinks the sample space. The odd faces are ruled out, so the world is now $B = \{2, 4, 6\}$, with $\lvert B \rvert = 3$, and only one of those faces is a $2$.

$P(\text{roll is } 2 \mid \text{even}) = \frac{\lvert \{2\} \rvert}{\lvert B \rvert} = \frac{1}{3}$

The unconditional probability was $\frac{1}{6}$, so the information doubled it.

The equally likely sample space is the set of ordered pairs, and the multiplication principle sizes it.

$\lvert S \rvert = 6 \times 6 = 36$

Only the pair $(6,6)$ gives a sum of $12$, so one outcome is favourable.

$P(\text{sum} = 12) = \frac{1}{36}$

The answer $\frac{1}{11}$ comes from treating the eleven possible sums as equally likely, which they are not.

A card cannot be both a heart and a spade, so the two events are mutually exclusive and the correction term in the addition rule is zero.

$P(\text{heart} \cup \text{spade}) = \frac{13}{52} + \frac{13}{52} - 0 = \frac{26}{52} = \frac{1}{2}$

Counting directly agrees, since $13 + 13 = 26$ of the $52$ cards qualify.

Clubs are black, so no card is both red and a club. The intersection is empty and the events are mutually exclusive.

$P(\text{red} \cap \text{club}) = 0 \quad \text{while} \quad P(\text{red})\,P(\text{club}) = \frac{1}{2} \cdot \frac{1}{4} = \frac{1}{8}$

Since $0 \ne \frac{1}{8}$, they are certainly not independent. They are not complementary either, because the complement of "red" is "black", which includes the spades as well as the clubs.