How is this wrong?

How is this wrong?

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youre a faggot, thats how

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You completely ignored your paub for . 6

How the fuck did I not only answer these types of questions in high school, but had one of the highest math grades in my entire class? I look at this now and see miscellaneous characters.

A U B would be 1.2 retard

If they're completely enclosed such that A = B, then #3 would be true. So, it's not wrong.
For some reason your teacher didn't mark you wrong for the next one too, so it was inconsistent. Talk to your teacher, he messed up grading. Explain your reasoning.

What kind of math is this?

voodoo math

>samefag, cont.
Your example shows they can't be mutually exclusive since P(A)+P(B) = 1.2??? So, you'd need the P(A U B) = P(A) + P(B) - P(A N B) and your example would have P(A) = 0.6, P(B) = 0.6, P(A N B) = 0.6, resulting in P(A U B) = 0.6.
That's how to explain it.

That's what I thought, but my T.A is a street shitter and said that it was wrong and that I didn't show work.

Probabilities can't go above 1
Except the probability that you're literally retarded

Talk to the professor directly, man. Since you said he's indian, the TA probably hasn't taken the class in 4 years and knows nothing about what the class is. Just skip the middle man.

Bruh, when I was in college I did calculus by hand for orbits in Astro.

Now I'm on /b and I'm a retard and can't do shit.

Discrete Math or Probability

Probabilities should add up to one

u must be 18 2 post here user. real men are feminist

Probabilities add up to 1 only for P(A U B) + P(A U B)', those are mutually exclusive. P(A) and P(B) are not mutually exclusive.

It's prob

I just noticed the question says Les than 1

But being a stats enthusiast drunk as i am right now I think you're right. Whoever wrote the question is a retard

What kind of math is this? Can someone explain? How do you come up with numbers from random letters?

this is post-high school stats in usa.

Set theory. The big U is for Union.

To find union you just multiply A and B
.6 * .6 isnt .6 retard, how are you taking a probability unit in math and on chan

It's statistical probabilities.

P(A) = % Probability of Event A
P(B) = % Probability of Event B
P(AUB)= Probability of A *OR* B
P(AUB)' = Whatever value will give you a total of 1, which represents 100%. It has to be 100% because you can't have a "120% chance" at something; It's impossible. It is also important to know that P(A) and P(B) are not mutually exclusive, which means they are independent of each other. A formula like P(ANB) (N is A AND B) is mutually exclusive because

Example: Flip a coin ten times and record what face the coin falls on. Let's say Heads (A) hits 4 times and Tails (B) hits 6 times. Your probabilities would be:

Heads = 0.4
Tails = 0.6
Heads OR tails = 0.24
Heads AND tails = 0.76
Heads OR tails prime =

P((AUB)') = 1 - P(AUB)
= 1 - 0.76
= 0.24

>mutually exclusive because

My bad. I went to check my reasoning and forgot to fucking type the fact in. Derp.

I'll just post a picture people like those. In the picture, King of Hearts is NOT mutually exclusive to either Hearts or Kings. But the King of Spades, for example, is independent from Hearts. If you were measuring probabilities for Spades in the same situation, you'd get the same math answer. Calculations don't always reflect the reality. Chance.

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As stated, it's not wrong. But the person who is grading the question is probably working under an unstated assumption that A and B are completely independent variables. However, you answer could be correct if A and B were 100% dependent.

If P(A) and P(B) are both .6,
P(AUB) is .84.

The only way you *DON'T* get the AUB event is if *NEITHER* of them happen.

P(A)' = .4
P(B)' = .4

Both of these have to happen.

P(A)'×P(B)'

Or .42 or .16... meaning the AUB event has a chance of .84.

That assumes that A and B are independent variables which was not stated in the question.

This is a bad example.

If you flip a coin, your chances of getting heads or tails is 1.

You, like the person grading the question, and possibly the person who wrote the question are making the assumption that A and B are independent. It's perfectly possible for there to be completely dependent variables with the probabilities given by the OP, or even two partially dependent variables with the values:
P(A) = 0.6
P(B) = 0.6
P(A U B) = 0.2
P(!(AUB)) = 0.8

If the measured probability of "A or B" differed from P(AUB), that would be a good argument for dependence, but that means your statistical model is flawed, it doesn't change the calculated value of P(AUB).

I think you'll find that if you step outside a basic stats class that the notation P(A U B) is taken to mean the actual probability given the complete probability space regardless of dependence/independence and simply calculating P(A U B) given only P(A) and P(B) is improper unless you know that they are independent variables. "The probability of 'A or B'" and "P(A U B)" mean the same thing.

This is like saying x=1 and y=1, but x+y=3 is true because in the modeled system when X and Y get together they have a baby.

I mean, yeah, sure, okay, nice nonlinear thinking, but you just look like retard who can't math good.

discord gg/bARc5p

>This is like saying x=1 and y=1, but x+y=3 is true because in the modeled system when X and Y get together they have a baby.
It's probability, not simple algebra. Different rules.

Gross conceptual error.
> simply calculating P(A U B) given only P(A) and P(B) is improper unless you know that they are independent variables
When you step outside of a basic statistics class, how would you prove that the A variable and B variable aren't independent in the system being studied?
By showing P(AUB) does not correctly predict the measured probability.

discord gg/bARc5p

You didn't catch enough cum in your butt op that's fucking how you limey cunt nugget

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Addition and Union have different rules, sure, but if you aren't following the rules of the operator, you're not performing the specified operation.

No it isn't. There are all kinds of things that follow this kind of behavior. But usually you talk about these kinds of dependent variables as being "correlated".

For an example of two variables that match the OP's given distributions, consider
P(A) = the probability a d10 rolled value is less than 7
P(B) = the probability the same d10's rolled value is not greater than 6.
You might say this example is silly because the two events are just two ways to say the same thing, but that's entirely the point. Events with 100% dependence tend to just be different ways to measure the same state, but recognizing that the two are actually the same is sometimes not obvious.


But I did make a mistake on my provided alternate example. The values I gave aren't possible. P(AUB) must be at least .6 (in the case they were 100% correlated and as much as 1.0 if they were maximally anti-correlated.

strongnet.org/cms/lib6/OH01000884/Centricity/Domain/308/Venn Diagrams.pdf
The "U" operator is a set operation. See the 8th slide for the real way to calculate unioned probabilities whether independent or dependent.

It just so happens that when two variables are independent P(A) + P(B) - P(A&&B) = P(A)*P(B).

Try out the exercises on this page.
siyavula.com/read/maths/grade-11/probability/10-probability-03

scores matter more than actual comprehension so of course nothing sticks

lmao I bet you're the kind of retard to say that flipping a coin 3 times and having it all be heads is the least likely outcome