Statistics and Probability - Conditional probability, including addition and multiplication rules-extended

In a group of 30 students, 18 play football ($F$), 15 play basketball ($B$), and 8 play both. If a student is chosen at random, find the probability that: i) They play football or basketball. ii) They play basketball, given that they play football.

i) $P(F \cup B) = P(F) + P(B) - P(F \cap B) = \frac{18}{30} + \frac{15}{30} - \frac{8}{30} = \frac{25}{30} = \frac{5}{6}$.

ii) $P(B|F) = \frac{P(B \cap F)}{P(F)} = \frac{8/30}{18/30} = \frac{8}{18} = \frac{4}{9}$.

To find 'or' ($A \cup B$), we use the addition rule to avoid double-counting the overlap. For the conditional probability, we restrict our sample space to only those who play football ($18$ students) and then see how many of those also play basketball ($8$ students).

A bag contains 5 red marbles and 3 blue marbles. Two marbles are drawn one after another without replacement. Find the probability that both marbles are red.

Let $R_1$ be the event the first marble is red, and $R_2$ be the event the second marble is red. $P(R_1) = \frac{5}{8}$ Since the marble is not replaced, if the first was red, there are now 4 red and 7 total marbles left. $P(R_2|R_1) = \frac{4}{7}$ $P(R_1 \cap R_2) = P(R_1) \times P(R_2|R_1) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}$.

This is an application of the Multiplication Rule for dependent events. The probability of the second event changes based on the outcome of the first event.

If $P(A) = 0.6$, $P(B) = 0.5$, and $A$ and $B$ are independent events, find $P(A \cup B)$.

Since $A$ and $B$ are independent: $P(A \cap B) = P(A) \times P(B) = 0.6 \times 0.5 = 0.30$. Now, use the addition rule: $P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.6 + 0.5 - 0.3 = 0.8$.

For independent events, we first find the intersection (the 'and' part) by multiplying the individual probabilities, then apply the general addition rule.