conditional probability of several events I'm having a hard time understanding what this question wants:
A person initially purchases either type A or type B. She will choose either type A or type B with an equal probability on her first purchase. Every purchase afterwards, she will either buy the same type (with a probability of 1/3) or will buy the other type (with a probability of 2/3). What probability will her first and second item be type B and her third and fourth item be type A?

I was thinking of drawing a tree and multiplying the probabilities, but I was wondering if I am working correctly.
P(1st purchase is type B)=1/2
P(B2)=P(B1)*P(B2|B1)+P(A1)*P(B2|A1)=1/2
Following the same logic, I get that 
P(A3)=1/2 and P(A4) is also 1/2. But for some reason this seems rather odd that the probabilities are all the same.
I was wondering then, if it was wanting P(B1∩B2) and P(A3∩A4) instead? Even if it wasn't, I would find such a probability by multiplying across, correct? Meaning that P(B1∩B2)=(1/2)*(1/3)?
 A: We want the probability of "BBAA."
The probability the first item bought is of Type B is $\frac{1}{2}$. Given that the first item is of Type B, the probability the second item is of Type B is $\frac{1}{3}$.
Thus the probability the first two items are of type B is $\frac{1}{2}\cdot \frac{1}{3}$.
Given that the second item is of Type B, the probability the third item is of Type A is $\frac{2}{3}$. Thus the probability the first two items are of Type B and the third is of Type A is $\frac{1}{2}\cdot\frac{1}{3}\cdot\frac{2}{3}$.
Continue. 
Remark: A tree diagram is useful here, though not really necessary since we are chasing down a single path. 
A: A bit more formally you need to calculate (index means the order of purchase)
$$P\{B_1,B_2,A_3,A_4\}$$
$$=P\{A_4|B_1,B_2,A_3\}\cdot P\{B_1,B_2,A_3\}=\dots$$
$$=P\{A_4|B_1,B_2,A_3\}\cdot P\{A_3|B_1,B_2\}\cdot P\{B_2|,B_1\}\cdot P\{B_1\}$$
Noticing that only the last purchase matters the last formula becomes
$$=P\{A_4|A_3\}P\{A_3|B_2\}P\{B_2|B_1\}P\{B_1\}=\frac{1}{3}\cdot\frac{2}{3}\cdot\frac{1}{3}\cdot\frac{1}{2}$$
A: The probabilities you calculated are all $1/2$ simply from the symmetry of the problem. You don't need the unconditional probabilities of the second purchase being of type $B$ etc.; all you need to do is to multiply up the probabilities for the individual steps occurring as prescribed: The probability for the first item to be $B$ is $1/2$. If this has occurred, the probability for the second item to also be $B$ is $1/3$. If this has occurred, the probability for the third item to be $A$ is $2/3$. And if this has occurred, the probability for the fourth item to also be $A$ is $1/3$. So the desired probability is
$$1/2\cdot1/3\cdot2/3\cdot1/3=1/27\;.$$
A: There is only one favorable sequence BBAA with Pr = $\frac{1}{2}\times\frac{1}{3}\times\frac{2}{3}\times\frac{1}{3} = \frac{1}{27}$ 
