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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

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adelekeademolu

Answer:

[tex]\dfrac{1}{27}[/tex]

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

[tex]P(\text{1A 2A 3B 4B}) = P(\text{A}) \times P(\text{same brand}) \times P(\text{different brand}) \times P(\text{same brand})[/tex]

[tex]P(\text{1A 2A 3B 4B}) = \dfrac{1}{2}\times\dfrac{1}{3}\times\dfrac{2}{3}\times\dfrac{1}{3} = \dfrac{1}{27}[/tex]

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