Question

Each time that a shopper purchases a tube of toothpaste, she chooses either brand A or brand B. Suppose that the probability is 1/3 that she will choose the same brand chosen on her previous purchase, and the probability is 2/3 that she will switch brands.

a. If her first purchase is brand A, what is the probability that her fifth purchase will be brand B?

b. If her first purchase is brand B, what is the probability that her fifth purchase will be brand B?

Short answer

1. The probability that fifth purchase is brand B given that the first purchase is brand A is\(\frac{{40}}{{81}}\).
2. The probability that fifth purchase is brand B given that first purchase was brand B is\(\frac{{41}}{{81}}\).

Step-by-step solution

Explaining the law of probability:

Let\({E_{1,}}...{E_n}\)are n mutually exclusive and collect any event A, then\({\rm P}\left( A \right)\)is calculated as:

\({\rm P}\left( A \right) = \sum\limits_{i = 1}^n {{\rm P}\left( {A|{E_i}} \right) \times } {\rm P}\left( {{E_i}} \right)\)

Drawing a transition matrix

Consider a shopper choice to purchase either brand A or brand B in successive purchase forms a Markov chain with transition probability matrix as shown below.

Hence the transition matrix is as given:

	Brand A	Brand B
Brand A	\(\frac{1}{3}\)	\(\frac{2}{3}\)
Brand B	\(\frac{2}{3}\)	\(\frac{1}{3}\)

Calculating the probability that the fifth purchase will be brand B when the 1st purchase is brand B

Let\({X_i} = 1\)when brand A is chosen time t and\({X_i} = 0\)when brand B is chosen.

Hence we need to compute \({\rm P}\left( {{x_5} = 0|{x_1} = 1} \right).\)

Let us compute the 5^th step transition matrix of the student.

\({p^{\left( 4 \right)}} = p \times p \times p \times p\)

\(\begin{aligned}{}{p^{\left( 4 \right)}} &= \left[ {\begin{aligned}{}{\frac{1}{3}}&{\frac{2}{3}}\\{\frac{2}{3}}&{\frac{1}{3}}\end{aligned}} \right] \times \left[ {\begin{aligned}{}{\frac{1}{3}}&{\frac{2}{3}}\\{\frac{2}{3}}&{\frac{1}{3}}\end{aligned}} \right] \times \left[ {\begin{aligned}{}{\frac{1}{3}}&{\frac{2}{3}}\\{\frac{2}{3}}&{\frac{1}{3}}\end{aligned}} \right] \times \left[ {\begin{aligned}{}{\frac{1}{3}}&{\frac{2}{3}}\\{\frac{2}{3}}&{\frac{1}{3}}\end{aligned}} \right]\\ &= \left[ {\begin{aligned}{}{\frac{5}{9}}&{\frac{4}{9}}\\{\frac{4}{9}}&{\frac{5}{9}}\end{aligned}} \right] \times \left[ {\begin{aligned}{}{\frac{5}{9}}&{\frac{4}{9}}\\{\frac{4}{9}}&{\frac{5}{9}}\end{aligned}} \right]\\ &= \left[ {\begin{aligned}{}{\frac{{41}}{{81}}}&{\frac{{40}}{{81}}}\\{\frac{{40}}{{811}}}&{\frac{{41}}{{81}}}\end{aligned}} \right]\end{aligned}\)

Therefore the 4^th step transition matrix describing probabilities \({\rm P}\left( {{x_{t + 4}}|{x_t}} \right)\) as:

	Brand A	Brand B
Brand A	\(\frac{{41}}{{81}}\)	\(\frac{{40}}{{81}}\)
Brand B	\(\frac{{40}}{{81}}\)	\(\frac{{41}}{{81}}\)

Hence the required probability is

\({\rm P}\left( {{x_5} = 0|{x_1} = 1} \right) = \frac{{40}}{{81}}\)

Hence the probability that fifth purchase is brand B given that the first purchase is brand A is\(\frac{{40}}{{81}}\).

Calculating the probability that the fifth purchase will be brand B when the 1st purchase is brand B

The required probability can be obtained using fourth step transition probability in table one subpart a.

\({\rm P}\left( {{X_5} = 0|{X_1} = 0} \right) = \frac{{41}}{{81}}\)

Hence the probability that fifth purchase is brand B given that first purchase was brand B is\(\frac{{41}}{{81}}\).