Can you please provide all the work for parts A,B and C please? In a big city wi
ID: 3272549 • Letter: C
Question
Can you please provide all the work for parts A,B and C please?
In a big city with good public transit, when commuting to work, the probability of being delayed by more than 30 minutes is estimated to be the following based on your mode of commute: if taking the subway is .05: if taking the bus is .25 if driving into work this is .6. (a) Assume Brad is equally likely to use any of the above three options. Find the probability that Brad is not delayed by more than 30 minutes. (b) Suppose Brad is delayed by more than 30 minutes. What is the chance he took the subway? (c) Suppose Brad is delayed by more than 30 minutes. What is the chance he took the bus?Explanation / Answer
Hi, I've tried explaining with theory and number calculations, please refer below answer and let me know in case you don't understand any part of it:
a.
P( he not delayed) = P( he takes subway and is not delayed) + P(he takes bus and is not delayed) + P( he drives and is not delayed)
Now, since he choosen each of 3 options with an equal probability ( as given in the question), each
vehicle is taken with a probability of 1/3
Hence, P( he isn't delayed) = 1/3 (1-.05) + 1/3(1-.25) + 1/3( 1-.6) = 1/3 * ( 3-.9) = .70
So, there is .7 probability of 70% chance that he is not delayed
b. Given he was late, chaace of him taking the subway is ?
= (1/3 * 0.05) / (1/3 (.05) + 1/3(.25) + 1/3(.6) = 1/3 * (.9))
= .05 / (.05+.25+.6)
= .056
c.Given he was late, chaace of him taking the bus is ?
= (1/3 * 0.25) / (1/3 (.05) + 1/3(.25) + 1/3(.6) = 1/3 * (.9))
= .25 / (.05+.25+.6)
= .28