# Probability question

• B
How do I express the following scenario mathematically?

I have access to CNN news online from two different internet sites, each of which have about an 80% reliability of actually providing the feed when I log onto the site. If I only had access to one of the sites, I'd know that I had an 80% chance that I'd get a live feed when I logged in. How does this figure change when I have two sites, both with that 80% reliability.

The temptation is to multiply the two probabilities together, i.e., .8 * .8 = .64 But that can't be right because the probability has to be greater than .8

Can someone set this scenario up for me mathematically?

BiGyElLoWhAt
Gold Member
If you multiply them, you're asking for the probability of getting a news feed from one site and then the other.
I think I'm missing some detail, though. So you have to choose a site, and then log in? Or what? Do both sites pop up when you log onto whatever you're logging into?

BiGyElLoWhAt
Gold Member
If that's the case, you're looking for the probability of A and/or B = A or B + A and B. At least I would think so. You only care that you get at least one, but it's also a possibility that you get two, so you need the probability of getting 1 + the probability of getting 2.

BiGyElLoWhAt
Gold Member
Ok, so this is what I'm thinking, I get a value of 96% when I do this.
##A## is getting the feed from A ##A'## is not getting the feed from A.

##P = AB' + A'B + AB##

stevendaryl
Staff Emeritus
Let $A$ be the outcome "The first site works". Let $B$ be the outcome "The second site works". Then there are 4 possible joint outcomes: $A \wedge B, A \wedge \neg B, \neg A \wedge B, \neg A \wedge \neg B$ (where $\neg$ means "not", and $\wedge$ means "and"). So the probability that at least one of the sites works is:

$P(A \vee B) = P(A \wedge B) + P(A \wedge \neg B) + P(\neg A \wedge B)$

where $\vee$ means "or".

So do you know how to compute those three probabilities on the right side of the equals?

So do you know how to compute those three probabilities on the right side of the equals?

No I don't. But for some reason I thought intuitively the answer was what Bigyellowhat stated, 96%

Ok, so this is what I'm thinking, I get a value of 96% when I do this.

I don't know where I came up with that figure, and when I tried to think about how the probability would be characterized mathematically, I drew a blank. This is especially disturbing since I just took the GRE last Summer and had these type of probability questions drilled into my head. I've already forgotten how to do them BiGyElLoWhAt
Gold Member
What's the probability of A but not B?

As for where I got that number, it's the same as what Steven Darryl is doing.

• DiracPool
stevendaryl
Staff Emeritus
No I don't. But for some reason I thought intuitively the answer was what Bigyellowhat stated, 96%

You have to use some laws of probability:
$P(\neg X) = 1 - P(X)$
$P(X \wedge Y) = P(X) \cdot P(Y)$ (if the probabilities are independent)

• DiracPool
Add them up to get whatever you get.

That looks like it adds up to .96! If there's one thing I still know how to do, it's add decimals .64 + .16 + .16 =.96 Am I right? I think the probabilities are independent. If you have cable or satellite, you pretty much have a 100% reliability that you will get a CNN feed. But the two sites I get my CNN news feed from have essentially an 80% reliability independent of one another. I was just wondering how much closer to 100% my reliability was by having the two options.

pbuk
Gold Member
These answers seem somewhat over-complicated. Assuming the probabilities of NOT getting a feed from one site are independently 1 - 0.8 = 0.2, the probability of not getting a feed from both sites is 0.2 x 0.2 = 0.04. So the probability of this not happening (i.e. getting a feed from at least one site) is 1 - 0.04 = 0.96.

[edit - added this]And if you have three sites available, the probability of not getting a feed from any is 1 - 0.23 = 99.2%.

• DiracPool and BiGyElLoWhAt
PeroK
Homework Helper
Gold Member
2020 Award
That looks like it adds up to .96! If there's one thing I still know how to do, it's add decimals .64 + .16 + .16 =.96 Am I right? I think the probabilities are independent. If you have cable or satellite, you pretty much have a 100% reliability that you will get a CNN feed. But the two sites I get my CNN news feed from have essentially an 80% reliability independent of one another. I was just wondering how much closer to 100% my reliability was by having the two options.
Yes, it's 96% if the two are independent. The simplest calculation is how often you have neither. That's 0.2 x 0.2 = 0.04.

0.8 x 0.8 = 0.64 is the probability you have both.

And 0.8 x 0.2 = 0.16 is the probability you have the first but not the second. Similarly it's the same probability you have the second but not the first. Hence:

64% both feeds available
32% only one feed available
4% neither feed available

• DiracPool
BiGyElLoWhAt
Gold Member
These answers seem somewhat over-complicated. Assuming the probabilities of NOT getting a feed from one site are independently 1 - 0.8 = 0.2, the probability of not getting a feed from both sites is 0.2 x 0.2 = 0.04. So the probability of this not happening (i.e. getting a feed from at least one site) is 1 - 0.04 = 0.96.
I suppose I overlooked that, no? lol

64% both feeds available
32% only one feed available
4% neither feed available

I like how you broke that down PeroK, thanks.

This is studied in reliability analysis. Basically, you have some subdevices (here: the two sites) which are operational which probability ##p##. What is asked here is what is the probability that the entire device is operational.

If the devices are connected in series, then the probability is ##p^n## where ##n## is the number of devices.
If the devices are connected in parallel, then the probability is ##1 - (1- p)^n##.
You can generalize this to ##k## out of ##n## systems where the device works if ##k## out of ##n## subdevices work. This requires the binomial distribution. Of course, you can go even more complicated than this.