communication system

**jut** · September 24th 2009, 12:33 AM

Need assistance with this hw prob:

A communication network has 5 links (see attach.)

The prob. that each link is working is 0.9. What is the prob of being able to transmit a message from point A to point B.

My attempt was to say, that the prob. of transmitting (TX) a message is the prob. of the top series link + prob. of middle + prob. of bottom - prob. of all 3. That is,

$Pr(TX) = 0.9*0.9 + 0.9 + 0.9*0.9 - (0.9)^5$

But that is > 1. Doh.

**tanujkush** · September 24th 2009, 01:58 AM

Originally Posted by jut

Need assistance with this hw prob:

A communication network has 5 links (see attach.)

The prob. that each link is working is 0.9. What is the prob of being able to transmit a message from point A to point B.

My attempt was to say, that the prob. of transmitting (TX) a message is the prob. of the top series link + prob. of middle + prob. of bottom - prob. of all 3. That is,

$Pr(TX) = 0.9*0.9 + 0.9 + 0.9*0.9 - (0.9)^5$

But that is > 1. Doh.

From a lay point of view, it would be the probability that one of the three routes is chosen (the top, middle or bottom) times the probability of the signal going through. The probability of the signal going through would be just p for the first link encountered but for the second link it would be a conditional probability. i.e. probability that link 2 works given link 1 has worked. Hope this helps!

**mr fantastic** · September 24th 2009, 02:35 AM

Originally Posted by jut

Need assistance with this hw prob:

A communication network has 5 links (see attach.)

The prob. that each link is working is 0.9. What is the prob of being able to transmit a message from point A to point B.

My attempt was to say, that the prob. of transmitting (TX) a message is the prob. of the top series link + prob. of middle + prob. of bottom - prob. of all 3. That is,

$Pr(TX) = 0.9*0.9 + 0.9 + 0.9*0.9 - (0.9)^5$

But that is > 1. Doh.

Let A be the event successful transmission through the top series link, B be the event successful transmission through the middle series link and C be the event successful transmission through the bottom series link.

$\Pr(A \cup B \cup C) = \Pr(A) + \Pr(B) + \Pr(C)$ $- \Pr(A \cap B) - \Pr(A \cap C) - \Pr(B \cap C) + \Pr(A \cap B \cap C)$ .

However, the approach I'd use is

Pr(successful transmission of message) = 1 - Pr(no transmission of message) = $1 - \Pr(A' \cap B' \cap C')$

and the last probability should be simple to calculate since (I assume) each series link is independent.

**jut** · September 24th 2009, 10:56 AM

Thank you sir.

I understand most of this formula:

$\Pr(A \cup B \cup C) = \Pr(A) + \Pr(B) + \Pr(C)$ $- \Pr(A \cap B) - \Pr(A \cap C) - \Pr(B \cap C) + \Pr(A \cap B \cap C)<br />$

Since the Pr(A), Pr(B), Pr(C) are NOT mutually exclusive, you subtract their overlap (ie, the intersect). But why did you add the intersection of all three paths?

**mr fantastic** · September 24th 2009, 02:11 PM

Originally Posted by jut

Thank you sir.

I understand most of this formula:

$\Pr(A \cup B \cup C) = \Pr(A) + \Pr(B) + \Pr(C)$ $- \Pr(A \cap B) - \Pr(A \cap C) - \Pr(B \cap C) + \Pr(A \cap B \cap C)<br />$

Since the Pr(A), Pr(B), Pr(C) are NOT mutually exclusive, you subtract their overlap (ie, the intersect). But why did you add the intersection of all three paths?

Draw a Venn diagram and it will be obvious (it gets subtracted twice and so has to be added).

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