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1) By 'conditional' does it mean perhaps B has some impact on A sometimes?

2) Could someone explain this with a simple number maths example please?

3) Does the 'theorem' part have anything to do with the above? There is no need to teach me this and I found a good example before of A being like a coin flip, B being like a dice role, and how they have no impact on each other.

There is no need to post the proof as it asks on the last line, proofs are the last of my worries.

Helpful university lecture slide

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    $\begingroup$ $B$ doesn't necessarily have an impact on $A$. Example,Rolling a dice twice, probability of the first roll is 5 given that the sum if 6. $\endgroup$ – user 42493 Aug 2 '19 at 0:15
  • $\begingroup$ From my impression the OP knows this, but is asking whether conditional probability is related to their example. $\endgroup$ – Toby Mak Aug 2 '19 at 0:16
  • $\begingroup$ This video helped me a lot, it uses some simple fractions and plugs them into the formula youtube.com/watch?v=zubsk1HhD-0 $\endgroup$ – plagiarism Aug 2 '19 at 17:01
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1) By 'conditional' does it mean perhaps B has some impact on A sometimes?

Somewhat, yes.   Conditional probability is the measure for probability under a specified condition.   $\mathsf P(A\mid B)$ is the probability measure for event A on condition that event $B$ occurs.   That $B$ does sometimes have an impact on the occurrence of $A$ is the point of taking such a measure.

2) Could someone explain this with a simple number maths example please?

There are 6 red and 4 blue marbles in a bag before I draw two of them out, one by one.   What is the probability that the second of them is red ($A$) under condition that the first of them is blue ($B$)?   Well, given that the first is blue, the second marble can be one of nine marbles, six of which are red. $$\mathsf P(A\mid B)=\tfrac 69$$Of course, the unconditional probability that the second marble is red is simply: $$\mathsf P(A)=\tfrac 6{10}$$

Does the 'theorem' part have anything to do with the above?

The theorem tells you that these events are not independent, because $\mathsf P(A)\neq \mathsf P(A\mid B)$

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  • $\begingroup$ "Of course, the unconditional probability that the second marble is red is simply:" I did not understand why this was 6/10 sorry, nor what you mean by unconditional means? $\endgroup$ – plagiarism Aug 2 '19 at 14:21
  • $\begingroup$ Pr[A|B] is not equal to P[A] because when we remove a marble from B (blue) they are no longer independent. $\endgroup$ – plagiarism Aug 2 '19 at 14:25
  • $\begingroup$ I was hoping for an A intersect B number example. $\endgroup$ – plagiarism Aug 2 '19 at 14:27
  • $\begingroup$ I was hoping for an A intersect B number example. Like how would 6/10 intersect 6/9 work? $\endgroup$ – plagiarism Aug 2 '19 at 14:34
  • $\begingroup$ By definition: $\mathsf P(A\cap B)=\mathsf P(B)\mathsf P(A\mid B) = \tfrac 4{10}\times\tfrac 69=\tfrac{24}{90}=\tfrac 4{15}$ @plagiarism $\endgroup$ – Graham Kemp Aug 2 '19 at 21:27
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1) Conditional probability means the probability of event $A$ happening given $B$ is happening. 'Condition' is actually the word to focus on here.

2) The probability that a person on any given day has a cough may be only be $5 \%$. But they will be much more likely to have a cough, given that they have a cold. For example, $P(\text{Cough}) = 5 \%$ but $P(\text{Cough}| \text{Sick}) = 75 \%$.

3) Yes. The coin flip and the dice roll are mutually exclusive, so they have no impact on each other.

The information for parts 1) and 2) was taken from Wikipedia.

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  • $\begingroup$ I am looking for a number example where I cannot do Pr[A n B] / Pr[B] because it will not divide properly. $\endgroup$ – plagiarism Aug 2 '19 at 14:28

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