Question

A survey of a group of college students was done to find out how students get to school for the school year. 15% of those surveyed were from out of state. Of those that were in-state, 56% used a car as their primary form of transport to school, 13% used a train and 18% used a bus. Of those that were from out of state, 29% used an airplane, 31% used a car, and 12% used the train.

1. What is the probability that a respondent uses the train?

2. What is the probability a randomly chosen respondent is from out of state and uses the bus as his primary transport?

3. What is the probability that a respondent is from in state, takes an airplane or both?

4. If a respondent is chosen and that person uses a car, what is the probability the respondent is from out of state?  

5. Are primary form of transportation to school and in state/out of state statistically independent?

Answer 1

Let O is the event that student is from out of state and I shows the event that student is from in-state.

So we have

P(O) = 0.15, P(I) = 1 - P(O) = 1-0.15 = 0.85

Let C shows the event that student use car, A shows the event that student use airplane, B shows the event that student use bus and T shows the event that student use train. So

P(C |O) = 0.31, P(T|O) = 0.12, P(A|O) = 0.29, P(B|O) = 1 - 0.31 - 0.12 - 0.29 = 0.28

and

P(C |I) = 0.56, P(T|I) = 0.13, P(B|I) = 0.18, P(A|I) = 1 - 0.56- 0.13 - 0.18 = 0.13

1:

By the law of total probability, the probability that  a respondent uses the train is

P(T) = P(T|I) P(I) + P(T|O)P(O) = 0.13 * 0.85 + 0.12 * 0.15 = 0.1285

2:

The probability a randomly chosen respondent is from out of state and uses the bus as his primary transport is

P(O and B) = P(B|O) P(O) = 0.28 * 0.15 = 0.042

3:

In is not given that what second transport other than airplane need to use.

4:

P(O|C) = [ P(C|O) P(O) ] / [ P(C|O) P(O) +P(C|I) P(I) ] = [0.31 * 0.15] / [ 0.31*0.15 + 0.56 * 0.85] = 0.088995

5:

Since

P(O|C) is not eqaul to P(O)

so primary form of transportation to school and in state/out of state are not statistically independent.