Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

1 Answer 1

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$\ldots$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.