Question

"In a tri-state conference, 60% attendees are from California, 25% from Oregon, and 15% from Washington. As it turns out 6 % of the attendees from California, 17% of the attendees from Oregon, and 12% of the attendees from Washington came to the conference by train. If an attendee is selected at random and found to have arrived by train, what is the probability that the person is from Washington?"

Answer 1

Answer:

18.65% probability that the person is from Washington

Step-by-step explanation:

Bayes Theorem:

Two events, A and B.

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

In which P(B|A) is the probability of B happening when A has happened and P(A|B) is the probability of A happening when B has happened.

In this question:

Event A: Arrived by train.

Event B: From Washington.

15% are from Washington

This means that $P(B) = 0.15$

12% of the attendees from Washington came to the conference by train.

This means that $P(A|B) = 0.12$

Probability of arriving by train:

6% of 60%(from California), 17% of 25%(from Oregon) or 12% of 15%(from Washington). So

$P(A) = 0.06*0.6 + 0.17*0.25 + 0.12*0.15 = 0.0965$

If an attendee is selected at random and found to have arrived by train, what is the probability that the person is from Washington?"

$P(B|A) = \frac{0.15*0.12}{0.0965} = 0.1865$

18.65% probability that the person is from Washington