# probability of waiting for a bus

can anyone please check if i've done it correctly?

The 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 did:

1)i calculated the expected mean and variance in order to understand the deviation from the mean. so using the formulas of expected value and expected mean i obtained that: expected mean $= 11.8$, expected variance $= 14.76$, then took the root of the expected variance to find the normal distribution $(\sim3.84)$ and then i had to find the $z$ score, so i computed it and obtained $0.2$

2)i defined the events a - being late, $b - x>15$, and then used bayes decomposed formula $p(a|b)=\frac{p(b|a)p(a)}{p(b)}$ where $p(a) = 0.18, p(b) = 0.2$, and calculated $p(b|a)$ by using the mean 20 so that $\frac{0.18*0.9}{0.2}=0.81$

3)i subtracted the expected mean from 1) by $8$, i.e $11.8-8=3.8$, but i don't know how to continue in order to use the $z$-score here.

would appreciate your help and assistance with this. thank you very much for your help!

• For what reason you used the normal distribution? – callculus May 31 '18 at 13:08
• i used the normal distribution in order to understand the deviation from the mean – BeginningMath May 31 '18 at 15:54

## 1 Answer

• For the first question, I don't see a justification to use the normal distribution.

Using the notation in your second part (for your second part, being late should be being busy), \begin{align}P(B)&=P(B|A)P(A)+P(B|A^c)P(A^c) \\ &=\exp\left(-\frac{15}{20}\right)\cdot 0.18+\exp\left(-\frac{15}{10} \right)\cdot 0.82\end{align}

• For the second part, Bayes rule is indeed the right tool. You might like to change the numbers for $P(B|A)$ and $P(B)$.

• Guide for the third part, let $X$ denotes the waiting time.

\begin{align} 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)} \end{align}