A fair coin is tossed, and a single 6-sided die is rolled. Find the probability of obtaining a head on the coin and rolling a 3 on the die.

Solution:

$P(\text{head}) = \frac{1}{2}$

$P(3) = \frac{1}{6}$

$P(\text{head and 3}) = P(\text{head}) \times P(3) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}$

A card is chosen at random from a deck of 52 cards. It is then replaced, and a second card is chosen. What is the probability of choosing a jack and then an eight?

Solution:

$P(\text{jack}) = \frac{4}{52}$

$P(\text{eight}) = \frac{4}{52}$

$P(\text{jack and 8}) = P(\text{jack}) \times P(8) = \frac{4}{52} \times \frac{4}{52} = \frac{16}{2704} = \frac{1}{169}$

Dependent events are those in which the occurrence of one event affects the occurrence of the other. Two events A and B are dependent if:

$P(A \cap B) = P(A|B) \times P(B)$

This is called the multiplicative rule for dependent events. $P(A|B)$ is the conditional probability, read as "probability of A given that B has occurred."

For example, if cards are drawn from a pack of 52 playing cards without replacement, the event of getting a diamond in the first draw will impact the probability of getting a diamond in the second draw. The probability of getting a diamond in the first draw will be $\frac{13}{52}$, while the probability of getting a diamond in the second draw will be $\frac{12}{51}$.

If A and B are dependent events such that $P(B) = \frac{2}{3}$ and $P(A \cap B) = \frac{1}{4}$, find $P(A|B)$.

Solution:

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

$P(A|B) = \frac{1/4}{2/3}$

$P(A|B) = \frac{1}{4} \times \frac{3}{2}$

$P(A|B) = \frac{3}{8}$

Select two cards from a standard deck of 52 cards without replacement. Find the probability of selecting two kings.

Solution:

Let A be the event that the first card is a king, and B be the event that the second card is a king.

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

$P(B|A) = \frac{3}{51} = \frac{1}{17}$

$P(A \cap B) = P(A) \times P(B|A) = \frac{1}{13} \times \frac{1}{17} = \frac{1}{221}$

A committee consists of four women and three men. The committee will randomly select two people to attend a conference in Nairobi. The first person selected will not be considered in the second selection. Find the probability that both are women.

Solution:

Let A be the event that the first person is a woman, and B be the event that the second person is a woman.

$P(A) = \frac{4}{7}$

$P(B|A) = \frac{3}{6} = \frac{1}{2}$

$P(A \cap B) = P(A) \times P(B|A) = \frac{4}{7} \times \frac{1}{2} = \frac{2}{7}$