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Sagot :
Using conditional probability, it is found that there is a 0.9556 = 95.56% probability that Juan chose to get home from work by bus, given that he arrived home after 7 p.m.
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: Arrived home after 7 p.m.
- Event B: Got home by bus.
The percentages associated with arriving home after 7 p.m. are:
- 14% of 17%(by car).
- 62% of 83%(by bus).
Hence:
[tex]P(A) = 0.14(0.17) + 0.62(0.83) = 0.5384[/tex]
The probability of both arriving home after 7 p.m. and using bus is:
[tex]P(A \cap B) = 0.62(0.83)[/tex]
Hence, the conditional probability is:
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.62(0.83)}{0.5384} = 0.9556[/tex]
0.9556 = 95.56% probability that Juan chose to get home from work by bus, given that he arrived home after 7 p.m.
You can learn more about conditional probability at https://brainly.com/question/14398287
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