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Sagot :
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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