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MC1103: A travel agent books passages on three different tours, with half of her customers choosing tour T1, one-third choosing tour T2, and the remaining one-sixth choosing tour T3. The agent has noted that 3/4 of those who take T1 return to book passage again, 2/3 of those who take T2 return, and 1/2 of those who take T3 return. If a customer does return, what is the probability that the person first went on tour T2

Answer :

Using conditional probability, it is found that there is a 0.3266 = 32.66% probability that the person first went on tour T2.

Conditional Probability

[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]

In which

  • P(B|A) is the probability of event B happening, given that A happened.
  • [tex]P(A \cap B)[/tex] is the probability of both A and B happening.
  • P(A) is the probability of A happening.

In this problem:

  • Event A: Customer returns.
  • Event B: Customer first went on tour T2.

The percentages associated with a return are:

  • 3/4 = 0.75 of 0.5(choosing T1).
  • 2/3 = 0.6667 of 1/3 = 0.3333(choosing T2).
  • 1/2 = 0.5 of 1/6 = 0.1667(choosing T3).

Hence:

[tex]P(A) = 0.75(0.5) + 0.667(0.3333) + 0.5(0.1667) = 0.6806611[/tex]

The probability of both returning and first going on T2 is:

[tex]P(A \cap B) = 0.667(0.3333)[/tex]

Hence, the conditional probability is:

[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.667(0.3333)}{0.6806611} = 0.3266[/tex]

0.3266 = 32.66% probability that the person first went on tour T2.

A similar problem is given at https://brainly.com/question/14398287

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