How do i calculate the following properties of waiting time for bus? (almost finished, very urgent to me)

Question

he waiting time in minutes for a bus in a certain station(from getting to the station until the bus arrival) is a random exponential variable with a mean of 10. But if the the traffic is busy, then its mean is 20. The probability that the traffic will be busy when waiting for the bus is 0.18.

1)what is the probability that a man that arrives in random time will wait longer than 15 minutes?

2)if a man waits longer than 15 minutes,what is the probability that the traffic is busy?

3)if a man comes in a random day and after 8 minutes still waiting, what is the probability he'll have to wait for not longer than 15 minutes?

what i already did:

1) expected variance: 14.76, expected mean= 11.8. using z score the result is 0.2.

2)defined the events a - being late, b−x>15.p(a)=0.2, #P(B)=P(B|A)P(A)+P(B|A^c)P(A^c) =\exp(-\frac{15}{20})\cdot 0.18+\exp(-\frac{15}{10} )\cdot 0.82# and then used bayes #p(a|b)=(p(b|a)p(a))/(p(b))# (couldn't finish it unfortunately).

3)here i got only the formula right, but i don't know how to calculate it correctly: what i got is: #P(X \le 15|X>8) = \frac{P(8 < X \le 15)}{P(X>8)}=\frac{P(8 < X \le 15 | A)P(A)+P(8 < X \le 15 |A^C)P(A^C)}{P(8< X | A)P(A)+P(8< X |A^C)P(A^C)}#

Grace Asher · Accepted Answer

#"1) 0.268"#
#"2) 0.31727"#
#"3) 0.466"# #"We use the Poisson process :"# #P["k events in time t"] = ((lambda*t)^k exp(-lambda*t))/(k!)# #"For the exponential distribution, we have k=0 :"# #P["0 events in time t"] = ((lambda*t)^0 exp(-lambda*t))/(0!)# #= exp(-lambda*t)# #=> P["at least one event in time t"] = 1 - exp(-lambda*t)# #"No need to use mean, variances, or z scores !!!"# #"Here we have"# #lambda = 1/10 " with probability 0.82, and"# #lambda = 1/20 " with probability 0.18."# #"1) "t = 15, k = 0# #=> P["0 events in 15 min."] = 0.82*e^-1.5+0.18*e^-0.75# #= 0.268# #"2) "P["traffic busy | 0 events in 15 min."]# #= (P["traffic busy AND 0 events in 15 min."])/(P["0 events in 15 min."])# #= (0.18*e^-0.75) / 0.268# #= 0.31727# #"3) We use the fact that the exponential distribution is"# #"memoryless, so the remaining waiting time is not influenced"# #"by the time that we already are waiting."# #= P["he has to wait no longer than 7 extra minutes"]# #= P["at least 1 event in 7 minutes"]# #= 1 - P["0 events in 7 minutes"]# #= 1 - 0.82*e^(-7/10) - 0.18*e^(-7/20)# #= 1 - 0.82*e^-0.7 - 0.18*e^-0.35# #= 0.466#