Question

Each time a shopper purchases a tube of toothpaste, he chooses either brand A or brand B. Suppose that for each purchase after the first, the probability is 1/3 that he will choose the same brand that he chose on his preceding purchase and the probability is 2/3 that he will switch brands. If he is equally likely to choose either brand A or brand B on his first purchase, what is the probability that both his first and second purchases will be brand A and both his third and fourth purchases will be brand B

Answer 1

Answer:

$\dfrac{1}{27}$

Step-by-step explanation:

Let the required probability be denoted by P(1A 2A 3B 4B).

This means a shopper chooses brand A first. Then by choosing brand A as the second purchase, the same brand is used. The third purchase is brand B; hence he switches brand. The fourth purchase is also brand B, maintaining the same brand as the third.

On the first purchase, the probabilities of A and B are both equal. Hence, each probability = 1/2

$P(\text{1A 2A 3B 4B}) = P(\text{A}) \times P(\text{same brand}) \times P(\text{different brand}) \times P(\text{same brand})$

$P(\text{1A 2A 3B 4B}) = \dfrac{1}{2}\times\dfrac{1}{3}\times\dfrac{2}{3}\times\dfrac{1}{3} = \dfrac{1}{27}$