Statistics and Probability - Conditional probability – Independent events

In a group of 100 students, 60 study Mathematics ($M$) and 40 study Physics ($P$). If 24 students study both subjects, find the probability that a student studies Physics given that they study Mathematics. Determine if the events $M$ and $P$ are independent.

First, we find the individual probabilities: $P(M) = \frac{60}{100} = 0.6$ $P(P) = \frac{40}{100} = 0.4$ $P(M \cap P) = \frac{24}{100} = 0.24$

To find $P(P|M)$: $P(P|M) = \frac{P(P \cap M)}{P(M)} = \frac{0.24}{0.6} = 0.4$

To check for independence, we check if $P(M \cap P) = P(M) \times P(P)$: $P(M) \times P(P) = 0.6 \times 0.4 = 0.24$ Since $P(M \cap P) = 0.24$ and $P(M) \times P(P) = 0.24$, the events are independent.

We use the definition of conditional probability to find $P(P|M)$. Since $P(P|M) = P(P)$, it confirms independence. The calculation of the intersection matching the product of individual probabilities also proves independence.

Events $A$ and $B$ are such that $P(A) = 0.7$ and $P(B) = 0.2$. If $A$ and $B$ are independent, find $P(A \cup B)$.

Since $A$ and $B$ are independent: $P(A \cap B) = P(A) \times P(B) = 0.7 \times 0.2 = 0.14$

Now use the addition rule: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ Substituting the values: $P(A \cup B) = 0.7 + 0.2 - 0.14$ Summing the first two terms: $\begin{array}{r} 0.70 \\ + 0.20 \\ \hline 0.90 \end{array}$ Subtracting the intersection: $\begin{array}{r} 0.90 \\ - 0.14 \\ \hline 0.76 \end{array}$ So, $P(A \cup B) = 0.76$.

Independence allows us to calculate the intersection $P(A \cap B)$ by multiplying the given probabilities. We then apply the General Addition Rule to find the union.

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

Let $R_1$ be the event that the first marble is red, and $B_2$ be the event that the second marble is blue. Initially, there are $5 + 3 = 8$ marbles. If the first marble drawn is red, there are now $5 - 1 = 4$ red marbles and 3 blue marbles remaining in the bag. Total marbles remaining = $4 + 3 = 7$.

The probability of drawing a blue marble now is: $P(B_2 | R_1) = \frac{3}{7}$

This is a classic conditional probability problem where the sample space changes. Since the first marble was red and not replaced, the total count decreases by 1, but the count of blue marbles remains the same.