Statistics and Probability - Conditional probability

In a class of 30 students, 18 study Biology ($B$), 15 study Chemistry ($C$), and 8 study both. A student is selected at random. Find the probability that the student studies Biology, given that they study Chemistry.

We are looking for $P(B|C)$. From the information given: $n(C) = 15$ $n(B \cap C) = 8$ Using the formula for conditional probability based on frequencies: $P(B|C) = \frac{n(B \cap C)}{n(C)}$ $P(B|C) = \frac{8}{15}$ $P(B|C) \approx 0.533$

To find $P(B|C)$, we only consider the group of students who study Chemistry (the denominator). Within that specific group, we count how many also study Biology (the numerator).

Given that $P(A) = 0.6$, $P(B) = 0.5$, and $P(A \cup B) = 0.8$, determine if events $A$ and $B$ are independent.

First, find $P(A \cap B)$ using the addition rule: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ $0.8 = 0.6 + 0.5 - P(A \cap B)$ $P(A \cap B) = 1.1 - 0.8 = 0.3$

Now, check the condition for independence $P(A \cap B) = P(A) \times P(B)$: $P(A) \times P(B) = 0.6 \times 0.5 = 0.3$

Since $P(A \cap B) = 0.3$ and $P(A) \times P(B) = 0.3$, the values are equal.

Because the probability of the intersection is equal to the product of the individual probabilities, events $A$ and $B$ are independent.

A bag contains 5 red balls and 3 blue balls. Two balls are drawn one after the other without replacement. Find the probability that the second ball is blue, given that the first ball was red.

Let $R_1$ be the event that the first ball is red and $B_2$ be the event that the second ball is blue. Initially, there are $5 + 3 = 8$ balls. If the first ball is red, it is not replaced. Now the bag contains: $5 - 1 = 4$ red balls $3$ blue balls Total balls remaining $= 7$ $P(B_2|R_1) = \frac{3}{7}$

Without replacement means the total count and the count of the specific color change for the second draw. The probability is calculated based on the state of the bag after the first event has occurred.