What is the Probability of Waiting Additional Time at a Bus Stop?

[Linder88]

Homework Statement

If you already have waited five minutes at the bus stop, what is the probability of having to wait additionally five minutes or more?

Homework Equations

Declare a random time T, together with the CDF shown below, which specify the time (in minutes) that a frozen traveller has to wait at the bus stop before the bus arrives
\begin{equation}
F_T(t)=
\begin{cases}
\bigg (\frac{t}{2}\bigg ]^2,0\leq t\leq 1\\
\frac{t}{4},1\leq t\leq 2\\
\frac{1}{2},2\leq t \leq 10\\
\frac{t}{20},10\leq t\leq 20\\
1,t\geq 20
\end{cases}
\end{equation}

The Attempt at a Solution

We are looking for $P[T\geq 5|T>5]$
\begin{equation}
P[T\geq 5|T>5]\frac{P[T\geq 5,T\geq 5]}{p[T=5]}=\frac{P[T\geq 5]P[T>5]}{P[T=5]}=\frac{1}{2}
\end{equation}

 

[andrewkirk]

The formula for conditional probability is
$$P(B|A)=\frac{P(B\cap A)}{P(A)}$$
$A$ is the event on which you are conditioning. In this case, it is that the bus has not arrived in the first five minutes.
$B$ is the event that the bus has not arrived in the first ten minutes.

In your calculation you have not made any reference to ten minutes. Using the above, set up the formulas again and see what happens.

Hint: what is the relationship between $P(B\cap A)$ and $P(B)$? Can that relationship help simplify the calc?

 

[Ray Vickson]

Homework Statement

If you already have waited five minutes at the bus stop, what is the probability of having to wait additionally five minutes or more?

Homework Equations

Declare a random time T, together with the CDF shown below, which specify the time (in minutes) that a frozen traveller has to wait at the bus stop before the bus arrives
\begin{equation}
F_T(t)=
\begin{cases}
\bigg (\frac{t}{2}\bigg ]^2,0\leq t\leq 1\\
\frac{t}{4},1\leq t\leq 2\\
\frac{1}{2},2\leq t \leq 10\\
\frac{t}{20},10\leq t\leq 20\\
1,t\geq 20
\end{cases}
\end{equation}

The Attempt at a Solution

We are looking for $P[T\geq 5|T>5]$
\begin{equation}
P[T\geq 5|T>5]\frac{P[T\geq 5,T\geq 5]}{p[T=5]}=\frac{P[T\geq 5]P[T>5]}{P[T=5]}=\frac{1}{2}
\end{equation}

You are NOT looking for $P(T \geq 5 | T > 5)$; this equals 1 automatically. You should be looking for $P(T \geq 10 | T > 5)$.

 

[Linder88]

Okey, thanks for the replies. Using your advices leads me to
\begin{equation}
P(T\geq 10|T>5)=\frac{P(T\geq 10\cap T>5)}{P(T>5)}=\frac{F_T(T\geq 10)F_T(T>5)}{F_T(T>5)}\\
=F_T(T\geq 10)=\frac{10}{20}=\frac{1}{2}
\end{equation}
I know this isn't rigth but I can't handle the fact that they are non-independent.

 

[Ray Vickson]

Okey, thanks for the replies. Using your advices leads me to
\begin{equation}
P(T\geq 10|T>5)=\frac{P(T\geq 10\cap T>5)}{P(T>5)}=\frac{F_T(T\geq 10)F_T(T>5)}{F_T(T>5)}\\
=F_T(T\geq 10)=\frac{10}{20}=\frac{1}{2}
\end{equation}
I know this isn't rigth but I can't handle the fact that they are non-independent.

Independence has nothing to do with anything in this problem: you have a singe random variable, $T$, and are told exactly what is its distribution. Why would you suppose independence has any relevance?

Why do you write $P(T > 5 \; \cap \; T \geq 10) = P(T > 5) P(T \geq 1)$? Never mind probabilities for the moment; just look at the event $\{ T > 5 \: \& \: T \geq 10\}$. That is saying something about $T$. Can you write this more simply? (If you need to do so, draw a number line for the values of $T$ and sketch the event on that line.)

 

[Linder88]

Are you suggesting somethin like in the equation under?
\begin{equation}
P(A\cap B)=P(B)P(A|B)
\end{equation}

 

[Ray Vickson]

Are you suggesting somethin like in the equation under?
\begin{equation}
P(A\cap B)=P(B)P(A|B)
\end{equation}

No, I am not suggesting that. Go back and re-read my suggestion: "Never mind probabilities for the moment; just look at the event $\{T>5\: \&\:T≥10 \}$. That is saying something about $T$. Can you write this more simply? (If you need to do so, draw a number line for the values of $T$ and sketch the event on that line.)"

Did you not see where I said to forget about probabilities for the moment? Until you understand the nature of the event $\{ T > 5\} \cap \{ T \geq 10 \}$ there is no hope of calculating its probability correctly. So, let me repeat: understand the event first, then worry later about its probability!

That is my very last word on this topic.

 

[Linder88]

I don't understand why you have to be so mysterious about the answer, this is not a homework question in that sense. It is a question from my last exam which I'm trying to figure out what I did wrong. To be totally honest I don't understand what you mean, English is not my first language, nor my second, but unfortunately the one I have to use the most.

 

[andrewkirk]

\begin{equation}
\frac{P(T\geq 10\cap T>5)}{P(T>5)}=\frac{F_T(T\geq 10)F_T(T>5)}{F_T(T>5)}\end{equation}

This is incorrect, because
$P(T\geq 10\cap T>5)$ is not equal to $P(T\geq 10)P(T>5)$
In fact $P(T\geq 10\cap T>5)=P(T\geq 10)$.

By the way $F_T(T<5)$ is incorrect notation that does not mean anything. The correct notation is $F_T(5)$ (which is equal to $P(T<5)$). That may seem pedantic but I find that keeping one's notation correct helps a lot in avoiding confusion in a difficult topic like probability.

Note also that $P(T>5)$ is equal to $1-P(T\leq 5)= 1-F_T(5)$, not $F_T(5)$.

 

[Ray Vickson]

This is incorrect, because
$P(T\geq 10\cap T>5)$ is not equal to $P(T\geq 10)P(T>5)$
In fact $P(T\geq 10\cap T>5)=P(T\geq 10)$.

By the way $F_T(T<5)$ is incorrect notation that does not mean anything. The correct notation is $F_T(5)$ (which is equal to $P(T<5)$). That may seem pedantic but I find that keeping one's notation correct helps a lot in avoiding confusion in a difficult topic like probability.

Note also that $P(T>5)$ is equal to $1-P(T\leq 5)= 1-F_T(5)$, not $F_T(5)$.

I was attempting to get the OP to think about the nature of the event $\{T > 5\} \cap \{T \geq 10\}$; in particular, I wanted him/her to conclude that
\{T &gt; 5\} \cap \{T \geq 10\} = \{ T \geq 10 \}
However, he/she did not seem willing to cooperate with me.

 

[Linder88]

This is incorrect, because
$P(T\geq 10\cap T>5)$ is not equal to $P(T\geq 10)P(T>5)$
In fact $P(T\geq 10\cap T>5)=P(T\geq 10)$.

By the way $F_T(T<5)$ is incorrect notation that does not mean anything. The correct notation is $F_T(5)$ (which is equal to $P(T<5)$). That may seem pedantic but I find that keeping one's notation correct helps a lot in avoiding confusion in a difficult topic like probability.

Note also that $P(T>5)$ is equal to $1-P(T\leq 5)= 1-F_T(5)$, not $F_T(5)$.

So
\begin{equation}
P(T\geq 10| T>5)=\frac{P(T\geq 10)}{1-P(T\leq 5)}
\end{equation}
The probability that the traveller will have to wait at least 10 minutes
\begin{equation}
P(T\geq 10)=\int_{10}^{\infty}f_T(t)dt=\int_{10}^{20}\frac{1}{20}dt=\frac{1}{2}
\end{equation}
the probability that the traveller will have to wait more than 5 minutes
\begin{equation}
P(T>5)=1-P(T\leq 5)=1-F_T(5)=1-\frac{1}{2}=\frac{1}{2}
\end{equation}
Equations (14) and (15) in equation (13)
\begin{equation}
P(T\geq 10|T>5)=\frac{\frac{1}{2}}{\frac{1}{2}}=1
\end{equation}
Thank you for the help, this concludes this topic.

 

[Samy_A]

Never mind, my mistake.