So there are 3 bags: bag A, bag B, and bag C.
- Bag A contains 2 red marbles and 3 white marbles.
- Bag B contains 3 red marbles and 5 white marbles.
- Bag C contains 4 red marbles and 6 white marbles.
Suppose a bag was randomly selected and marble was taken out.
What is the probability of taking a red marble from bag B?
So I've tried the conditional probability
https://i.sstatic.net/YSEky.png, but kinda confused.
If we make it like $P(A)$ is the probability of bag B chosen out of 3 bags and $P(A ∩ B)$ is the probability of getting red marbles out of marbles in bag B, making $P(A | B)$ the result.
But if we make it like $P(A)$ is the probability of red marbles get chosen out of all marbles, and $P(A | B)$ is the probability of only red marbles from bag B get chosen out of all marbles and making $P(A ∩ B)$ the result, it's going to be different. Which one is the right calculation (or maybe both are wrong…)?
Best Answer
This is a funny situation. Considering my comment, while the OP has clearly shown work, I am still not sure which question is being asked.
So I will answer each question separately.
Let $E_1$ be the event that bag B is chosen.
Let $E_2$ be the event that a red marble is chosen.
Q1: Compute $p(E_2 | E_1)$.
Q2: Compute $p(E_1 ~\text{and} ~E_2 ~\text{both occurring})$.
Q3: Compute $p(E_1 | E_2)$.
Answer-1:
Given that bag B is chosen, the answer is immediately seen to be $\frac{3}{8}$.
Answer-2:
$p(E_1 ~\text{and} ~E_2 ~\text{both occurring})$ is (using Answer-1)
$$p(E_1) \times p(E_2|E_1) = \frac{1}{3} \times \frac{3}{8} = \frac{1}{8}.$$
Answer-3:
$$p(E_2) = \left[\frac{1}{3} \times \frac{2}{5}\right] + \left[\frac{1}{3} \times \frac{3}{8}\right] + \left[\frac{1}{3} \times \frac{4}{10}\right] = \frac{47}{120}. $$
Using Answer-2,
$$p(E_1 | E_2) = \frac{p(E_1 ~\text{and} ~E_2 ~\text{both occurring})}{p(E_2)}$$
$$= \frac{(1/8)}{(47/120)} = \frac{15}{47}.$$