# What is my mistake? Dependent Probability does not add up to 100%

I have a question.

There is some event 'A' and an event 'B'. Event 'B' only takes place when event 'A' is true. (The outcome of event 'B' is otherwise not dependent on the outcome of event 'A').

event A:

event A: false = 58.2%
event A: true  = 41.8%

if (event A = true) then event B:

event B: true  = 23.8%
event B: false = 41.8% × (1 - 23.8%) = 31.9%


BUT: My intuition tells me that the chances for 'event A: false' + 'event B: true' + 'event B: false' should be 100% because they encompass the total of the 3 possible outcomes. Yet, with the above, the sum equals 113.81%

where am I wrong?

Update:

To clear up what I meant above, it should (probably) be written like:

P(A)   = 41.8%
P(B|A) = 23.8%


I tried to calculate the probability of B not occurring given that A has already occurred:

P( {not B}|A ) 41.8% × (1 - 23.8%) = 31.9%


Since the possible outcomes are:
A:false or (A:true and (B:false or B:true))

I figured that the sum of the probabilities for A:false + B:false + B:true should be 100%.
BUT: 58.2% + 23.8% + 31.9% = 113.9% instead.

Hence my question: where am I wrong?

(I can't get the dash formatted to appear above a letter, sorry)

• $B$ completely depends on $A$. $\neg A \wedge B$ is impossible ($P(\bar A \cap B) = 0$) and if $B$ is known to be true, $A$ must have occured: $P(A|B) = 1$ Sep 1, 2014 at 12:21
• Sorry, I need a more elaborate explanation (please explain in common English if at all possible)?
– B.T.
Sep 1, 2014 at 12:23
• Rereading your statement, I think that you should make clear wich probabilites exactly you are given. If possible, use the notations $P(A)$ for probability of $A$ occuring and $P(B|A)$ for probability of $B$ occurring given that $A$ has already occured. Additionally $\bar B$ means "not $B$", i.e. $P(\bar B) = 1- P(B)$. Note that $A$ and $B$ are statements, either false or true. If your $B$ can assume more states (for example $false, true, ignored$), you need to craft statements like $P(B = true)$, $P(b = ignored)$ etc. Sep 1, 2014 at 12:41
• @Alexr, I updated the question.
– B.T.
Sep 1, 2014 at 14:48

Think of your sample space (i.e. the set of possible outcomes) as:

{{A and B}, {A and Not B}, {Not A and B}, {Not A and Not B}}. We can assign probabilities to each event in the sample space.

We obviously assign a probability of $0$ to the event {Not A and B}.

The probability of {A and B} is $(.418)(.238) = .099$

The probability of {A and Not B} is $(.418)(1-.238) = .319$

The probability of {Not A and Not B} is $.582$

The sum of these probabilities equals $1$.