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In the following problem:

The probability that a shopper will choose the same brand of toothpaste that he chose on his preceding purchase is 1/3 and the probability that he will switch brands is 2/3. Suppose that on his first purchase the probability that he will choose brand A is 1/4 and the probability that he will choose brand B is 3/4. What is the probability that his second purchase will be brand B.

I have the following probabilities:

P(A) = 1/4 P(B) = 3/4

P(A | A) or P(B | B) = 1/3

P(A | B) or P(B | A) = 3/4

So the probability that he will choose brand B on his second purchase is:

P(A) * P(B|A) + P(B) * P(B|B) = 1/4 * 3/4 + 3/4 * 1/3

Is this correct?

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Converting my (now-deleted) comment into an answer: No, your solution is not quite right.

It is helpful to use $B_1$ to denote the event that the first choice is $B$, and $B_2$ to denote the event that the second choice is $B$, (and similarly for $A_1$ and $A_2$) so as to avoid writing things such as $P(B\mid B ) = 1/3$. The quantity $P(B\mid B )$ should have no value other than $1$ in any probability calculation.

So, you are given $P(A_1) = \frac{1}{4}, P(B_1) = 1 - P(A_1) = \frac{3}{4}$,
$ P(\text{same as last time}) = P(B_2 \mid B_1) = P(A_2 \mid A_1) = \frac{1}{3} $, $P(\text{different from last time}) = P(B_2 \mid A_1) = P(A_2 \mid B_1) = \frac{2}{3} = 1 - P(\text{same as last time})$. The law of total probability says $$ P(B_2) = P(B_2\mid B_1)P(B_1) + P(B_2\mid A_1)P(A_1) $$ Can you take it from here? A useful check on the analysis and the arithmetic is to compute $P(A_2)$ using the same method. If you don't get $P(A_2) = 1 - P(B_2)$, something is awry$\ldots$

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