Answer:
0.1333 = 13.33% probability that bridge B was used.
Step-by-step explanation:
Conditional Probability
We use the conditional probability formula to solve this question. It is
[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 question:
Event A: Arrives home by 6 pm
Event B: Bridge B used.
Probability of arriving home by 6 pm:
75% of 1/3(Bridge A)
60% of 1/6(Bridge B)
80% of 1/2(Bridge C)
So
[tex]P(A) = 0.75*\frac{1}{3} + 0.6*\frac{1}{6} + 0.8*\frac{1}{2} = 0.75[/tex]
Probability of arriving home by 6 pm using Bridge B:
60% of 1/6. So
[tex]P(A \cap B) = 0.6*\frac{1}{6} = 0.1[/tex]
Find the probability that bridge B was used.
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.1}{0.75} = 0.1333[/tex]
0.1333 = 13.33% probability that bridge B was used.
