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

The probability that all stations are open is .

You want the likelihood that you can travel to be

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- Oct 25th 2008, 06:27 AM #1

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## [SOLVED] help please

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.

- Oct 25th 2008, 08:05 AM #2

- Oct 25th 2008, 08:28 AM #3

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

The probability that all stations are open is .

You want the likelihood that you can travel to be

[/quote}

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

- Oct 25th 2008, 11:01 AM #4

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Hello, maths2008!

There are two stations from to and two stations from to .

. . Are there three stations or four?

Suppose that each station is closed with probability

and that the state of each station is independent of the others.

What condition on ensures that the probability that

I cannot travel from to on train is no more than 1/2?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 be the probability that a station is open.

The probability that all three stations are open is:

. . and we want this probability to be__greater__than one-half.

So we have: .

Hence: .

Therefore: .

- Oct 25th 2008, 11:24 AM #5

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- Oct 25th 2008, 12:00 PM #6

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Hello again, maths2008!

So there arestations.*four*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 be the probability that a station is open.

The probability that all four stations are open is:

. . and we want this probability to be__greater__than one-half.

So we have: .

Hence: .

Therefore: .

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

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

But there are**fifteen**cases you must consider . . .

. .

. . . . .

- Oct 25th 2008, 12:17 PM #7

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