Conditional probability for dependent events

Question

There are two bags, the first bag contains 2 white balls and 5 black balls and the second contains 3 white balls and 4 black balls. What is the probability that the first bag was chosen given a white ball was chosen? Now suppose the ball was put back into its original bag, and another ball is picked from the same
bag randomly. What is the probability that the second ball is also white?

For the first question I got 2/7 by doing probability of the first bag and a red ball being chosen divided by the probability of the first bag being chosen. ((1/2)*(2/7))/(1/2). (Don't know if this is right)

I'm not sure how to go about doing the second part.

Answer 1

Before answering your question, I am assuming that where you meant white where you've written red.

When dealing with probabilities it helps to think of actions as events, and assigning an event a probability. In your example,

Let $P(W)=5/14$ be the probability of drawing a white ball, as 5 of 14 balls are while.

$P(W')=9/14$, i.e. the probability of drawing a black ball.

Let $B_1$, and $B_2$ be the event that the ball was taken from bag 1, or bag 2, respectively.

$P(W\cap B_1)= 2/14$, and $P(W\cap B_2)=3/14$.

$P(W'\cap B_1)=5/14$, and $P(W'\cap B_2)=4/14$.

All of this information is given to you in the context of the exercise. The conditional probability, "what is the probability event $B$ happens, given that the event $A$ has happened" is written as $$P(B|A)=\dfrac{P(B\cap A)}{P(A)}$$ Use this, along with the information above to answer question 1.

Question two resets by putting the white ball back into it's bag (i.e. bag 1), and asks what the probability of getting a white ball from bag one. In this case, just ignore bag two.