Probability Help!

[Daspurs1]

I have a probability problem that I'm not sure how to solve. If anyone can help i would appreciate it. For A I got 1/2 but not sure if that's right and for part B I got 25% not sure about that either.

 

[DaveSimmons]

I'd ask the professor whether an "originating" node also sends a message through all of its outgoing links, or if that's just "intermediate" nodes.

Then I'd do my own homework.

 

[Leros]

I remember doing problems like this.

 

[Hacp]

Isn't part B 0%?

 

[darkxshade]

welcome to the forums

 

[Inspire]

For the first part, 50%. For the second, 100% - The message has to originate at A or B, and if it didn't get to E, there's no way it originated at A.

Interesting thought exercise - if the message originated at B, but never made it to E, and the message is perpetually propogated (meaning the message can hit the same node more than once - which isn't clear from the question), then it essentially is caught in the C<->D loop, and the probability of such a scenario quickly regresses to zero. Given the nature of the exam, I expect your professor thinks himself rather clever for this irony.

 

[TuxDave]

if it didn't get to E, there's no way it originated at A.

Uh, each link has a 50% chance of failing. So it could've been sent from A, flipped badly on a coin and not get to E. The end.

 

[Inspire]

Yeah, I didn't read it that way. I read that it had a 50&#37; chance of being passed on, not of being recieved. If you read it that way, you take for granted that the point of origination successfully passes on the message at first.

Otherwise, if it's as you say, I'd have to work through it.

 

[esun]

I'm pretty sure both of your answers are wrong, although I just did a back of the envelope calculation and haven't really checked it carefully.

I'll give you some hints.

(a)

You want P(orig A | arrive E). Use Bayes' theorem and total probability to write this as

P(arrive E | orig A) * P(orig A) / [P(arrive E | orig A) P(orig A) + P(arrive E | orig B) P(orig B)]

The only tricky expression you need is P(arrive E | orig B). Everything else is obvious. You'll have to work that one out on your own, though (I personally found it easier to calculate P(not arrive E | orig B) so you may want to try that).

(b) Trivial once you have the result from (a).

 

[Daspurs1]

Thanks for the replies guys, really appreciate it.

ESUN - i initally thought about Bayes' Thm but since the problem says that the probability independence exists, i thought that P(orig A | arrive E) = P(orig A). Let me know if I'm not looking at this right.

 

[esun]

Thanks for the replies guys, really appreciate it.

ESUN - i initally thought about Bayes' Thm but since the problem says that the probability independence exists, i thought that P(orig A | arrive E) = P(orig A). Let me know if I'm not looking at this right.

The problem does not say that "orig A" and "arrive E" are independent events, and they most certainly are not. Read it more carefully. I do think the method I suggested is the most straightforward way of solving it.

Here's another hint:

Just consider the C, D, and E nodes. Assume the message starts at C (or D, doesn't matter since it is symmetric with respect to those nodes). What is the probability the message never arrives at E? This involves a convergent infinite series.

Once you get that, the overall solution should be very simple to get.

 

[JMapleton]

Drop out and become a Chippendale's stripper like me.

 

[szechuanpork]

the answer is 'c'