There are two stations from A to B and two stations from B to C. Suppose that each station is closed with probability p and that the state of each station is independent of the others. What condition on p ensures that the probability that I cannot travel from A to C on train is no more than 1/2?

Can some please guide me on doing this probability problem.

2. The probability that any station is open is (1-p).

The probability that all stations are open is $(1-p)^4$.

You want the likelihood that you can travel to be

$(1-p)^4 \geq \frac{1}{2}$

3. [quote=Henderson;209102]The probability that any station is open is (1-p).

The probability that all stations are open is $(1-p)^4$.

You want the likelihood that you can travel to be

$(1-p)^4 \geq \frac{1}{2}$[/quote}

So what will the probability that i can not travel be? I don't understand?

4. Hello, maths2008!

There are two stations from $A$ to $B$ and two stations from $B$ to $C$.
. .
Are there three stations or four?
Suppose that each station is closed with probability $p$
and that the state of each station is independent of the others.
What condition on $p$ ensures that the probability that
I cannot travel from $A$ to $C$ on train is no more than 1/2?
I will assume there are three stations.
Code:
      * - - - - - * - - - - - - *
A           B             C

We cannot make the trip if one or more stations are closed.
The only way to make the trip is when all three stations are open.

Let $q \:=\:1-p$ be the probability that a station is open.

The probability that all three stations are open is: $q^3$
. . and we want this probability to be greater than one-half.

So we have: . $q^3 \:> \:\frac{1}{2} \quad\Rightarrow\quad q \:>\:\frac{1}{\sqrt[3]{2}}$

Hence: . $p \;\geq\;1-q \;=\;1-\frac{1}{\sqrt[3]{2}} \;=\;0.206299474$

Therefore: . $p \:\geq \:0.2063$

5. Originally Posted by Soroban
Hello, maths2008!

I will assume there are three stations.
Code:
      * - - - - - * - - - - - - *
A           B             C
We cannot make the trip if one or more stations are closed.
The only way to make the trip is when all three stations are open.

Let $q \:=\:1-p$ be the probability that a station is open.

The probability that all three stations are open is: $q^3$
. . and we want this probability to be greater than one-half.

So we have: . $q^3 \:> \:\frac{1}{2} \quad\Rightarrow\quad q \:>\:\frac{1}{\sqrt[3]{2}}$

Hence: . $p \;\geq\;1-q \;=\;1-\frac{1}{\sqrt[3]{2}} \;=\;0.206299474$

Therefore: . $p \:\geq \:0.2063$
Thanks for the help but actually there are 4 stations

Shouldn't i caluclate the probabilty that i can not travel to A to C as the question suggests that What condition on P ensures that the probability that
I cannot travel from a to c is no more than 1/2?

6. Hello again, maths2008!

So there are four stations.
Code:
      * - - - - - * - * - - - - - - *
1           2   3             4

We cannot make the trip if one or more stations are closed.
The only way to make the trip is when all four stations are open.

Let $q \:=\:1-p$ be the probability that a station is open.

The probability that all four stations are open is: $q^4$
. . and we want this probability to be greater than one-half.

So we have: . $q^4 \:> \:\frac{1}{2} \quad\Rightarrow\quad q \:>\:\frac{1}{\sqrt[4]{2}}$

Hence: . $p \;\leq\;1-q \;=\;1-\frac{1}{\sqrt[4]{2}} \;=\;0.159103585$

Therefore: . $p \:\leq \:0.1591$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

You can solve the problem "head on" if you dare.

But there are fifteen cases you must consider . . .

. . $C \:=\:\text{Closed, }\;O \:=\:\text{O{p}en}$

. . $\begin{array}{|c|c|c|c| } 1&2&3&4 \\ \hline \hline C&C&C&C \\ C&C&C&O \\ C&C&O&C \\ C&C&O&O \\ C&O&C&C \\ C&O&C&O \\ C&O&O&C \\ C&O&O&O \\ \end{array}$. . . $\begin{array}{|c|c|c|c|} 1&2&3&4 \\ \hline O&C&C&C \\ O&C&C&O \\ O&C&O&C \\ O&C&O&O \\ O&O&C&C \\ O&O&C&O \\ O&O&O&C \end{array}$

7. Originally Posted by Soroban
Hello again, maths2008!

So there are four stations.
Code:
      * - - - - - * - * - - - - - - *
1           2   3             4
We cannot make the trip if one or more stations are closed.
The only way to make the trip is when all four stations are open.

Let $q \:=\:1-p$ be the probability that a station is open.

The probability that all four stations are open is: $q^4$
. . and we want this probability to be greater than one-half.

So we have: . $q^4 \:> \:\frac{1}{2} \quad\Rightarrow\quad q \:>\:\frac{1}{\sqrt[4]{2}}$

Hence: . $p \;\leq\;1-q \;=\;1-\frac{1}{\sqrt[4]{2}} \;=\;0.159103585$

Therefore: . $p \:\leq \:0.1591$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

You can solve the problem "head on" if you dare.

But there are fifteen cases you must consider . . .

. . $C \:=\:\text{Closed, }\;O \:=\:\text{O{p}en}$

. . $\begin{array}{|c|c|c|c| } 1&2&3&4 \\ \hline \hline C&C&C&C \\ C&C&C&O \\ C&C&O&C \\ C&C&O&O \\ C&O&C&C \\ C&O&C&O \\ C&O&O&C \\ C&O&O&O \\ \end{array}$. . . $\begin{array}{|c|c|c|c|} 1&2&3&4 \\ \hline O&C&C&C \\ O&C&C&O \\ O&C&O&C \\ O&C&O&O \\ O&O&C&C \\ O&O&C&O \\ O&O&O&C \end{array}$

Hi Soroban

Thanks for the help.