Probability - Partition of a sample space

A bag contains $4$ red and $4$ black balls, another bag contains $2$ red and $6$ black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag.

Let $E_1$ be the event of choosing the first bag and $E_2$ be the event of choosing the second bag. These form a partition since $E_1 \cap E_2 = \emptyset$ and $P(E_1) + P(E_2) = 1$. Let $A$ be the event of drawing a red ball. $P(E_1) = \frac{1}{2}, P(E_2) = \frac{1}{2}$ $P(A|E_1) = \frac{4}{8} = \frac{1}{2}, P(A|E_2) = \frac{2}{8} = \frac{1}{4}$ Using the Theorem of Total Probability for the denominator in Bayes' Theorem: $P(A) = P(E_1)P(A|E_1) + P(E_2)P(A|E_2) = \left(\frac{1}{2} \times \frac{1}{2}\right) + \left(\frac{1}{2} \times \frac{1}{4}\right) = \frac{1}{4} + \frac{1}{8} = \frac{3}{8}$ By Bayes' Theorem: $P(E_1|A) = \frac{P(E_1)P(A|E_1)}{P(A)} = \frac{1/4}{3/8} = \frac{1}{4} \times \frac{8}{3} = \frac{2}{3}$

We first define the partition $(E_1, E_2)$ based on the selection of bags. Then we use the Theorem of Total Probability to find the total probability of drawing a red ball $P(A)$, which serves as the base for finding the posterior probability $P(E_1|A)$.

In a factory, machine $M_1$ produces $60\%$ of the items and machine $M_2$ produces $40\%$. $2\%$ of items from $M_1$ are defective and $1\%$ of items from $M_2$ are defective. What is the total probability that an item selected at random is defective?

Let $E_1$ be the event that the item is produced by $M_1$ and $E_2$ by $M_2$. $E_1$ and $E_2$ partition the production. $P(E_1) = 0.60, P(E_2) = 0.40$ Let $D$ be the event that the item is defective. $P(D|E_1) = 0.02, P(D|E_2) = 0.01$ Total probability $P(D)$ is: $P(D) = P(E_1)P(D|E_1) + P(E_2)P(D|E_2)$ $P(D) = (0.60 \times 0.02) + (0.40 \times 0.01)$ $P(D) = 0.012 + 0.004 = 0.016$

The total probability is found by summing the weighted probabilities of defects from each machine in the partition.