I have the following two problems: 1. A bag contains 5 blue balls and 3 pink balls. Two balls are chosen at random and not replaced. What is the probability of choosing a blue ball after choosing a pink ball?
2.Kaj is tossing two coins. What is the probability that he will toss 2 tails given that the first toss was a tail?
For the first one, i thought the answer is $\dfrac{3}{8}*\dfrac{5}{7}$, because the probability of picking a pink ball is $\dfrac{3}{8}$ and the probability of picking a blue ball after that is $\dfrac{5}{7}$. However the book says the answer is $\dfrac{5}{7}$. For the second one, the book says the answer is $\dfrac{1}{3}$ but since there are 4 options, {h, t}, {h, h}, {t, h}, {t, t}, out of which only two begin with a tail, and out of which only one contains another tail, i figure the probability should be $\dfrac{1}{2}$.
I tried using the formula $P(A|B) = \dfrac{P(A \text{ and } B)}{P(B)} $, but i do not understand what things such as P(pink and blue) or P(heads and tails) should mean. I think they should be 1, but then the probabilities would come out greater than 1. What are the correct solutions to these problems?
After you have chosen the pink ball there are $7$ balls left and $5$ of those are blue. Thus the probability is $5/7$. Note that the book is asking the conditional probablity $P( \text{Blue ball} \mid \text{Pink ball on first time})$.
Your conclusion of the second one seems ok to me.
For the second one we have
$$ P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{1/4}{1/2}=\frac{1}{2}. $$
This is because $P(A \cap B) = P( \text{two tails}, \text{first is a tail}) = 1/4$ as when we have two tails the first is automatically a tail.