Conditional Winning

[KenNoMath]

There are two situations with two events happening on each of them,
but "Event 2" is depended on "Event 1"
and
"Event 4" is depended on "Event 3"

1st Question:
What is the probability of Area being Secure?

Area Is Secured?
{
Event 1:
P(A)=25%
25% chance that it is secure

Event 2:
This event can only occur if 1st event is a fails.
P(B)=25%
25% chance that it is secure
}

Result of P(A) and P(B) combination where P(B) only has a chance to occur if P(A) already failed
------------------------------------------------------------------------------------------------------------
2nd Question:
What is the probability of Building being Secure?

Building Is Secured?
{
Event 3:
P(C)=25%
25% chance that it is secure

Event 4:
This event can only occur if 3rd event is a success.
P(D)=75%
75% chance that it is secure
}

Result of P(C) and P(D) combination where P(D) only has a chance to occur if P(C) already succeeded
--------------------------------------------------------------------------
This is probably pretty simple probability but i forgot so much...
I know there are concepts like "Dependend/Independend", "Inclusive/Exclusive", "Conditional Probability"
And i know some basic formulas like:
P(A∪B) = P(A) + P(B) - P(A∩B)
P(A | B) = P(A∩B) / P(B)
But Idon't know how to make use of them.
I have been looking up videos and checking wiki's etc. for hours now. I'm at this point mad at myself for not understanding simple probability...
Can you help me find probabilities of Securing the Area and the Building ?
Thanks

 

[Dr.Peterson]

Part of the problem, I think, is a matter of unclear notation and terminology, which makes it hard to think about. I'm not sure whether I fully understand the events.

Let's call the events A, B, C, and D rather than 1, 2, 3, 4 for consistency. It isn't quite clear what each one means, though.

P(A) = 0.25
P(B|not A) = 0.25 and P(B|A) = 0, so P(B) = P(B and not A) = P(B|not A)*P(not A) = 0.25(1-0.25) = 0.1875

(But as I understand it, A is an event that must not be true in order for B to occur, and B means area is secure, right? You write as if A and B both mean "area is secure", which doesn't make sense. Might you mean that either A or B must occur to be secure, like maybe A = outer gate locked and B = inner gate locked, so that B only matters if A fails, but only A being true would be okay?)

P(C) = 0.25
P(D|C) = 0.75 and P(D|not C) = 0, so P(D) = P(D and C) = P(D|C)*P(C) = 0.75*0.25 = 0.1875

(This time, you seem to be saying that C and D are two events that must happen in succession, and together mean the building is secure. Right?)

By the way, I think you probably mean "dependent", not "depended" or "dependend".

 

[Romsek]

I can make sense out of (1). For example let the event A be the security system power supply stays operational.
This system provides 25% chance of detecting an intrusion.

Now let's say there is a backup generator that only turns itself on when the primary fails,
with 25% of success of doing so. (really need to get a better system designer)

Now the system has power and the security system provides it's 25% of detection.

Question 2 makes no sense imo.

 

[KenNoMath]

I like the image you created @Romsek and @Dr. Peterson , maybe I can explain it better now:

-----------------------------------------------------
For 1st question:

There is an "Area" and i need to get inside it
There are two doors Front door and Back door
There is 25% chance that Front door is unlocked and i can pass through no problem. Then I don't need to try the back door.
If Front door fails, I "have to" try the Back door.
Back door also has 25% chance of being unlocked.

So I imagine "probability of getting into the area" will be something like this ? :
[probability of front door opening] + [probability of back door opening given that front door was tried and it was locked]

This should have probability of >25% because there are 2 choices which have 25% chance and you can go in with one of them being right, if first one is wrong you will have a 2nd chance.

For 2nd question:

There is a "Building" and i need to get inside it but the security is really tougher now.
I need to go through 2 doors to get inside the "inner" building. Front door and Inner door.
There is 25% chance that Front door is unlocked and I can pass through no problem.
But I still have to pass though another door.
If the Front door fails. I am done.
If the Front door works now there is the inner door which has 75% chance to be unlocked.

So i imagine "probability of getting into the building" will be something like this ? :
[probability of front door unlocked]+[probability of inner door opening given that front door was unlocked and I passed through it]

--------------------------------------------------
As @Dr. Peterson said, this is a really simplified version of a game code I am trying to understand. So this is not a "problem" from a textbook. That is probably one of the reasons I am having so much trouble understanding the situation.

"
(But as I understand it, A is an event that must not be true in order for B to occur, and B means area is secure, right? You write as if A and B both mean "area is secure", which doesn't make sense. Might you mean that either A or B must occur to be secure, like maybe A = outer gate locked and B = inner gate locked, so that B only matters if A fails, but only A being true would be okay?)

"
"/Might you mean that either A or B must occur to be secure/" Yes! but you have to check the "Front Door(A)" first then "Back Door(B)", more then inner and outer doors, then your front and back entrances. Inner outer is the second question.

P(B|not A)*P(not A) = 0.25(1-0.25) = 0.1875
If I try to translate this for myself I get:
[Probability of B happening given that A didn't happen] * [Probability that A is not happening]
I understand the first Probability, we have the Front door locked and we are trying Back door.
I don't understand the difference between "*" and "+" here, how would I choose , what do they mean ?
Probability of Front door not opening, but I think we need to use the probability of Front door opening right ?
Do we multiply "not happening and happening" at the same time to get the "total probability of getting inside the area"

P(D) = P(D and C) = P(D|C)*P(C) = 0.75*0.25 = 0.1875
Oh I see you can use probability formulas like this, good to know.
P(D|C)*P(C)
[probability of D given that C happened]*[Probability of C]
or
[Probability of Inner door being open given that front door was unlocked]
and
[probability of front door is unlocked]
This seems correct! But Im not sure. I got this one but i used a different (probably a wrong method)
I just did
P(D)*P(C)=0.1875
since I though I needed both of them to be right but i didn't account for the order of operations
like first it needs to check if front door was unlocked or not
then the inner door check should happen
I made it like I was checking both doors at the same time.
Your explanation makes a lot more sense, I think I somehow got lucky with this one.


Thank you for your help and sorry about my English, It's not my main language.

 

[KenNoMath]

Im still looking for help about this. If anyone knows please provide some info.
This part make sense and it fits into my head:

P(C) = 0.25
P(D|C) = 0.75 and P(D|not C) = 0, so P(D) = P(D and C) = P(D|C)*P(C) = 0.75*0.25 = 0.1875

but not the first part the "Area" one

 

[Dr.Peterson]

For 1st question:

There is an "Area" and i need to get inside it
There are two doors Front door and Back door
There is 25% chance that Front door is unlocked and i can pass through no problem. Then I don't need to try the back door.
If Front door fails, I "have to" try the Back door.
Back door also has 25% chance of being unlocked.

So I imagine "probability of getting into the area" will be something like this ? :
[probability of front door opening] + [probability of back door opening given that front door was tried and it was locked]

This should have probability of >25% because there are 2 choices which have 25% chance and you can go in with one of them being right, if first one is wrong you will have a 2nd chance.

Really what you want is P(A unlocked or B unlocked); you know that P(A unlocked) = 0.25 and P(B unlocked | A locked) = 0.25. This is

P(A unlocked or B unlocked) = P(A unlocked) + P(A locked and B unlocked) = P(A unlocked) + P(B unlocked | A locked)*P(A locked) = 0.25 + 0.25(1 - 0.25) = 0.4375​

Your description sounds like you really know that P(B unlocked) = 0.25 and A and B are independent. In this form, you could do the following:

P(A unlocked or B unlocked) = P(not(A locked and B locked)) = 1 - P(A locked and B locked) = 1 - P(A locked)*P(B locked) = 1 - 0.75*0.75 = 0.4375​

For 2nd question:

There is a "Building" and i need to get inside it but the security is really tougher now.
I need to go through 2 doors to get inside the "inner" building. Front door and Inner door.
There is 25% chance that Front door is unlocked and I can pass through no problem.
But I still have to pass though another door.
If the Front door fails. I am done.
If the Front door works now there is the inner door which has 75% chance to be unlocked.

So i imagine "probability of getting into the building" will be something like this ? :
[probability of front door unlocked]+[probability of inner door opening given that front door was unlocked and I passed through it]

What you want here is P(C unlocked and D unlocked); you know that P(C unlocked) = 0.25 and P(D unlocked) = 0.75; presumably you know that C and D are independent? This is

P(C unlocked and D unlocked) = P(C unlocked) * P(D unlocked) = 0.25*0.75 = 0.1875​

If you don't know independence, but only that P(D unlocked | C unlocked) = 0.75, then it's

P(C unlocked and D unlocked) = P(C unlocked) * P(D unlocked | C unlocked) = 0.25*0.75 = 0.1875​