Question

Amy commutes to work by two different routes A and B. If she comes home by route A, then she will be home no later than 6 PM, with probability .8, but if she comes home by route B, then she will be home no later than 6PM, with probability .7. In the past, the proportion of times that Amy chose route A is .4.

(a) What proportion of times is Amy home no later than 6PM?

(b) If Amy is home after 6PM today, what is the probability that she took route B?

Answer 1

I'm a bit unsure on how to approach A.

Answer 2

P(route A) = 0.4
P(route B) = 1 - 0.4 = 0.6
The probability of choosing route A and being home no later than 6PM = 0.4 * 0.8
The probability of choosing route B and being home no later than 6PM = 0.6 * 0.7
The probability of being home no later than 6PM = (0.4 * 0.8) + (0.6 * 0.7) = ?

Answer 3

The answer would be .74 but I'm confused as to where you are getting .6 from.

Answer 4

oh wait nevermind I see now.

Answer 5

Good :)

Answer 6

Okay, so I think I've finally figured out the second part. The process is smililaar to that of the first step accept this time I'm including the rule of conditional probability.

If Amy comes home after 6 PM by choosing route A she will have had a probability of .2, where as if she had taken route B, a .3 chance. So the probability of her chooing route A after 6 PM is (.2)(.4)=.8 and the probability of her choosing B after 6 is (.3)(.6)=.18. So the probability of her choosing route B when she has arrived after 6PM is:
P(B|B)=P(B)/P(A intersection B) = .18/.26 = .6923

Answer 7

@JerJasonThe probability of Amy arriving home after 6PM is 1 - 0.74 = 0.26
The probability that Amy arrived home after 6PM and took route B = 0.26 * 0.6 = ?

Answer 8

It would be .156