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Deconstructing the Question

Topic: Sequential Probability

Bransen visits the shop 3 times. There are 3 chairs available each time. We need the probability that he is assigned to a different chair in each visit.

1. Break down the visits
  • 1st Visit: He can sit in any of the 3 chairs.
    Probability = \(3/3 = 1\)
  • 2nd Visit: To be different from the first, he must sit in one of the 2 remaining chairs.
    Probability = \(2/3\)
  • 3rd Visit: To be different from the previous two, he must sit in the last remaining chair.
    Probability = \(1/3\)

2. Total Probability
Since these are independent sequential events, we multiply them:
\(P = 1 * (2/3) * (1/3) = \) 2/9

Alternative Method (Counting Outcomes):
* Total possible assignments: \(3 * 3 * 3 = 27\)
* Favorable outcomes (permutations of 3 chairs): \(3! = 3 * 2 * 1 = 6\)
* \(P = 6 / 27 = 2 / 9\)

The correct answer is D.
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