Question

A commuter crosses one of three bridges, A, B, or C, to go home from work. The commuter crosses bridge A with probability 1/3, crosses bridge B with probability 1/6, and crosses bridge C with probability 1/2. The commuter arrives home by 6pm with probability 75%, 60%, and 80% by crossing bridge A, B, or C, respectively. If the commuter arrives home by 6pm, find the probability that bridge B was used.

Answer 1

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

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ 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

$P(A) = 0.75*\frac{1}{3} + 0.6*\frac{1}{6} + 0.8*\frac{1}{2} = 0.75$

Probability of arriving home by 6 pm using Bridge B:

60% of 1/6. So

$P(A \cap B) = 0.6*\frac{1}{6} = 0.1$

Find the probability that bridge B was used.

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.1}{0.75} = 0.1333$

0.1333 = 13.33% probability that bridge B was used.