# B Probability question

#### DiracPool

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?

#### DiracPool

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.

#### 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

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%.

#### PeroK

Homework Helper
Gold Member
2018 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

#### 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

#### DiracPool

64% both feeds available
32% only one feed available
4% neither feed available
I like how you broke that down PeroK, thanks.

#### micromass

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.

"Probability question"

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